Related
Suppose I have a df which has columns of 'ID', 'col_1', 'col_2'. And I define a function :
f = lambda x, y : my_function_expression.
Now I want to apply the f to df's two columns 'col_1', 'col_2' to element-wise calculate a new column 'col_3' , somewhat like :
df['col_3'] = df[['col_1','col_2']].apply(f)
# Pandas gives : TypeError: ('<lambda>() takes exactly 2 arguments (1 given)'
How to do ?
** Add detail sample as below ***
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
#df['col_3'] = df[['col_1','col_2']].apply(get_sublist,axis=1)
# expect above to output df as below
ID col_1 col_2 col_3
0 1 0 1 ['a', 'b']
1 2 2 4 ['c', 'd', 'e']
2 3 3 5 ['d', 'e', 'f']
There is a clean, one-line way of doing this in Pandas:
df['col_3'] = df.apply(lambda x: f(x.col_1, x.col_2), axis=1)
This allows f to be a user-defined function with multiple input values, and uses (safe) column names rather than (unsafe) numeric indices to access the columns.
Example with data (based on original question):
import pandas as pd
df = pd.DataFrame({'ID':['1', '2', '3'], 'col_1': [0, 2, 3], 'col_2':[1, 4, 5]})
mylist = ['a', 'b', 'c', 'd', 'e', 'f']
def get_sublist(sta,end):
return mylist[sta:end+1]
df['col_3'] = df.apply(lambda x: get_sublist(x.col_1, x.col_2), axis=1)
Output of print(df):
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
If your column names contain spaces or share a name with an existing dataframe attribute, you can index with square brackets:
df['col_3'] = df.apply(lambda x: f(x['col 1'], x['col 2']), axis=1)
Here's an example using apply on the dataframe, which I am calling with axis = 1.
Note the difference is that instead of trying to pass two values to the function f, rewrite the function to accept a pandas Series object, and then index the Series to get the values needed.
In [49]: df
Out[49]:
0 1
0 1.000000 0.000000
1 -0.494375 0.570994
2 1.000000 0.000000
3 1.876360 -0.229738
4 1.000000 0.000000
In [50]: def f(x):
....: return x[0] + x[1]
....:
In [51]: df.apply(f, axis=1) #passes a Series object, row-wise
Out[51]:
0 1.000000
1 0.076619
2 1.000000
3 1.646622
4 1.000000
Depending on your use case, it is sometimes helpful to create a pandas group object, and then use apply on the group.
A simple solution is:
df['col_3'] = df[['col_1','col_2']].apply(lambda x: f(*x), axis=1)
A interesting question! my answer as below:
import pandas as pd
def sublst(row):
return lst[row['J1']:row['J2']]
df = pd.DataFrame({'ID':['1','2','3'], 'J1': [0,2,3], 'J2':[1,4,5]})
print df
lst = ['a','b','c','d','e','f']
df['J3'] = df.apply(sublst,axis=1)
print df
Output:
ID J1 J2
0 1 0 1
1 2 2 4
2 3 3 5
ID J1 J2 J3
0 1 0 1 [a]
1 2 2 4 [c, d]
2 3 3 5 [d, e]
I changed the column name to ID,J1,J2,J3 to ensure ID < J1 < J2 < J3, so the column display in right sequence.
One more brief version:
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'J1': [0,2,3], 'J2':[1,4,5]})
print df
lst = ['a','b','c','d','e','f']
df['J3'] = df.apply(lambda row:lst[row['J1']:row['J2']],axis=1)
print df
The method you are looking for is Series.combine.
However, it seems some care has to be taken around datatypes.
In your example, you would (as I did when testing the answer) naively call
df['col_3'] = df.col_1.combine(df.col_2, func=get_sublist)
However, this throws the error:
ValueError: setting an array element with a sequence.
My best guess is that it seems to expect the result to be of the same type as the series calling the method (df.col_1 here). However, the following works:
df['col_3'] = df.col_1.astype(object).combine(df.col_2, func=get_sublist)
df
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
Returning a list from apply is a dangerous operation as the resulting object is not guaranteed to be either a Series or a DataFrame. And exceptions might be raised in certain cases. Let's walk through a simple example:
df = pd.DataFrame(data=np.random.randint(0, 5, (5,3)),
columns=['a', 'b', 'c'])
df
a b c
0 4 0 0
1 2 0 1
2 2 2 2
3 1 2 2
4 3 0 0
There are three possible outcomes with returning a list from apply
1) If the length of the returned list is not equal to the number of columns, then a Series of lists is returned.
df.apply(lambda x: list(range(2)), axis=1) # returns a Series
0 [0, 1]
1 [0, 1]
2 [0, 1]
3 [0, 1]
4 [0, 1]
dtype: object
2) When the length of the returned list is equal to the number of
columns then a DataFrame is returned and each column gets the
corresponding value in the list.
df.apply(lambda x: list(range(3)), axis=1) # returns a DataFrame
a b c
0 0 1 2
1 0 1 2
2 0 1 2
3 0 1 2
4 0 1 2
3) If the length of the returned list equals the number of columns for the first row but has at least one row where the list has a different number of elements than number of columns a ValueError is raised.
i = 0
def f(x):
global i
if i == 0:
i += 1
return list(range(3))
return list(range(4))
df.apply(f, axis=1)
ValueError: Shape of passed values is (5, 4), indices imply (5, 3)
Answering the problem without apply
Using apply with axis=1 is very slow. It is possible to get much better performance (especially on larger datasets) with basic iterative methods.
Create larger dataframe
df1 = df.sample(100000, replace=True).reset_index(drop=True)
Timings
# apply is slow with axis=1
%timeit df1.apply(lambda x: mylist[x['col_1']: x['col_2']+1], axis=1)
2.59 s ± 76.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# zip - similar to #Thomas
%timeit [mylist[v1:v2+1] for v1, v2 in zip(df1.col_1, df1.col_2)]
29.5 ms ± 534 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
#Thomas answer
%timeit list(map(get_sublist, df1['col_1'],df1['col_2']))
34 ms ± 459 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
I'm sure this isn't as fast as the solutions using Pandas or Numpy operations, but if you don't want to rewrite your function you can use map. Using the original example data -
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
df['col_3'] = list(map(get_sublist,df['col_1'],df['col_2']))
#In Python 2 don't convert above to list
We could pass as many arguments as we wanted into the function this way. The output is what we wanted
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
I'm going to put in a vote for np.vectorize. It allows you to just shoot over x number of columns and not deal with the dataframe in the function, so it's great for functions you don't control or doing something like sending 2 columns and a constant into a function (i.e. col_1, col_2, 'foo').
import numpy as np
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
#df['col_3'] = df[['col_1','col_2']].apply(get_sublist,axis=1)
# expect above to output df as below
df.loc[:,'col_3'] = np.vectorize(get_sublist, otypes=["O"]) (df['col_1'], df['col_2'])
df
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
Here is a faster solution:
def func_1(a,b):
return a + b
df["C"] = func_1(df["A"].to_numpy(),df["B"].to_numpy())
This is 380 times faster than df.apply(f, axis=1) from #Aman and 310 times faster than df['col_3'] = df.apply(lambda x: f(x.col_1, x.col_2), axis=1) from #ajrwhite.
I add some benchmarks too:
Results:
FUNCTIONS TIMINGS GAIN
apply lambda 0.7 x 1
apply 0.56 x 1.25
map 0.3 x 2.3
np.vectorize 0.01 x 70
f3 on Series 0.0026 x 270
f3 on np arrays 0.0018 x 380
f3 numba 0.0018 x 380
In short:
Using apply is slow. We can speed up things very simply, just by using a function that will operate directly on Pandas Series (or better on numpy arrays). And because we will operate on Pandas Series or numpy arrays, we will be able to vectorize the operations. The function will return a Pandas Series or numpy array that we will assign as a new column.
And here is the benchmark code:
import timeit
timeit_setup = """
import pandas as pd
import numpy as np
import numba
np.random.seed(0)
# Create a DataFrame of 10000 rows with 2 columns "A" and "B"
# containing integers between 0 and 100
df = pd.DataFrame(np.random.randint(0,10,size=(10000, 2)), columns=["A", "B"])
def f1(a,b):
# Here a and b are the values of column A and B for a specific row: integers
return a + b
def f2(x):
# Here, x is pandas Series, and corresponds to a specific row of the DataFrame
# 0 and 1 are the indexes of columns A and B
return x[0] + x[1]
def f3(a,b):
# Same as f1 but we will pass parameters that will allow vectorization
# Here, A and B will be Pandas Series or numpy arrays
# with df["C"] = f3(df["A"],df["B"]): Pandas Series
# with df["C"] = f3(df["A"].to_numpy(),df["B"].to_numpy()): numpy arrays
return a + b
#numba.njit('int64[:](int64[:], int64[:])')
def f3_numba_vectorize(a,b):
# Here a and b are 2 numpy arrays with dtype int64
# This function must return a numpy array whith dtype int64
return a + b
"""
test_functions = [
'df["C"] = df.apply(lambda row: f1(row["A"], row["B"]), axis=1)',
'df["C"] = df.apply(f2, axis=1)',
'df["C"] = list(map(f3,df["A"],df["B"]))',
'df["C"] = np.vectorize(f3) (df["A"].to_numpy(),df["B"].to_numpy())',
'df["C"] = f3(df["A"],df["B"])',
'df["C"] = f3(df["A"].to_numpy(),df["B"].to_numpy())',
'df["C"] = f3_numba_vectorize(df["A"].to_numpy(),df["B"].to_numpy())'
]
for test_function in test_functions:
print(min(timeit.repeat(setup=timeit_setup, stmt=test_function, repeat=7, number=10)))
Output:
0.7
0.56
0.3
0.01
0.0026
0.0018
0.0018
Final note: things could be optimzed with Cython and other numba tricks too.
The way you have written f it needs two inputs. If you look at the error message it says you are not providing two inputs to f, just one. The error message is correct.
The mismatch is because df[['col1','col2']] returns a single dataframe with two columns, not two separate columns.
You need to change your f so that it takes a single input, keep the above data frame as input, then break it up into x,y inside the function body. Then do whatever you need and return a single value.
You need this function signature because the syntax is .apply(f)
So f needs to take the single thing = dataframe and not two things which is what your current f expects.
Since you haven't provided the body of f I can't help in anymore detail - but this should provide the way out without fundamentally changing your code or using some other methods rather than apply
Another option is df.itertuples() (generally faster and recommended over df.iterrows() by docs and user testing):
import pandas as pd
df = pd.DataFrame([range(4) for _ in range(4)], columns=list("abcd"))
df
a b c d
0 0 1 2 3
1 0 1 2 3
2 0 1 2 3
3 0 1 2 3
df["e"] = [sum(row) for row in df[["b", "d"]].itertuples(index=False)]
df
a b c d e
0 0 1 2 3 4
1 0 1 2 3 4
2 0 1 2 3 4
3 0 1 2 3 4
Since itertuples returns an Iterable of namedtuples, you can access tuple elements both as attributes by column name (aka dot notation) and by index:
b, d = row
b = row.b
d = row[1]
My example to your questions:
def get_sublist(row, col1, col2):
return mylist[row[col1]:row[col2]+1]
df.apply(get_sublist, axis=1, col1='col_1', col2='col_2')
It can be done in two simple ways:
Let's say, we want sum of col1 and col2 in output column named col_sum
Method 1
f = lambda x : x.col1 + x.col2
df['col_sum'] = df.apply(f, axis=1)
Method 2
def f(x):
x['col_sum'] = x.col_1 + col_2
return x
df = df.apply(f, axis=1)
Method 2 should be used when some complex function has to applied to the dataframe. Method 2 can also be used when output in multiple columns is required.
I suppose you don't want to change get_sublist function, and just want to use DataFrame's apply method to do the job. To get the result you want, I've wrote two help functions: get_sublist_list and unlist. As the function name suggest, first get the list of sublist, second extract that sublist from that list. Finally, We need to call apply function to apply those two functions to the df[['col_1','col_2']] DataFrame subsequently.
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
def get_sublist_list(cols):
return [get_sublist(cols[0],cols[1])]
def unlist(list_of_lists):
return list_of_lists[0]
df['col_3'] = df[['col_1','col_2']].apply(get_sublist_list,axis=1).apply(unlist)
df
If you don't use [] to enclose the get_sublist function, then the get_sublist_list function will return a plain list, it'll raise ValueError: could not broadcast input array from shape (3) into shape (2), as #Ted Petrou had mentioned.
If you have a huge data-set, then you can use an easy but faster(execution time) way of doing this using swifter:
import pandas as pd
import swifter
def fnc(m,x,c):
return m*x+c
df = pd.DataFrame({"m": [1,2,3,4,5,6], "c": [1,1,1,1,1,1], "x":[5,3,6,2,6,1]})
df["y"] = df.swifter.apply(lambda x: fnc(x.m, x.x, x.c), axis=1)
Suppose I have a df which has columns of 'ID', 'col_1', 'col_2'. And I define a function :
f = lambda x, y : my_function_expression.
Now I want to apply the f to df's two columns 'col_1', 'col_2' to element-wise calculate a new column 'col_3' , somewhat like :
df['col_3'] = df[['col_1','col_2']].apply(f)
# Pandas gives : TypeError: ('<lambda>() takes exactly 2 arguments (1 given)'
How to do ?
** Add detail sample as below ***
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
#df['col_3'] = df[['col_1','col_2']].apply(get_sublist,axis=1)
# expect above to output df as below
ID col_1 col_2 col_3
0 1 0 1 ['a', 'b']
1 2 2 4 ['c', 'd', 'e']
2 3 3 5 ['d', 'e', 'f']
There is a clean, one-line way of doing this in Pandas:
df['col_3'] = df.apply(lambda x: f(x.col_1, x.col_2), axis=1)
This allows f to be a user-defined function with multiple input values, and uses (safe) column names rather than (unsafe) numeric indices to access the columns.
Example with data (based on original question):
import pandas as pd
df = pd.DataFrame({'ID':['1', '2', '3'], 'col_1': [0, 2, 3], 'col_2':[1, 4, 5]})
mylist = ['a', 'b', 'c', 'd', 'e', 'f']
def get_sublist(sta,end):
return mylist[sta:end+1]
df['col_3'] = df.apply(lambda x: get_sublist(x.col_1, x.col_2), axis=1)
Output of print(df):
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
If your column names contain spaces or share a name with an existing dataframe attribute, you can index with square brackets:
df['col_3'] = df.apply(lambda x: f(x['col 1'], x['col 2']), axis=1)
Here's an example using apply on the dataframe, which I am calling with axis = 1.
Note the difference is that instead of trying to pass two values to the function f, rewrite the function to accept a pandas Series object, and then index the Series to get the values needed.
In [49]: df
Out[49]:
0 1
0 1.000000 0.000000
1 -0.494375 0.570994
2 1.000000 0.000000
3 1.876360 -0.229738
4 1.000000 0.000000
In [50]: def f(x):
....: return x[0] + x[1]
....:
In [51]: df.apply(f, axis=1) #passes a Series object, row-wise
Out[51]:
0 1.000000
1 0.076619
2 1.000000
3 1.646622
4 1.000000
Depending on your use case, it is sometimes helpful to create a pandas group object, and then use apply on the group.
A simple solution is:
df['col_3'] = df[['col_1','col_2']].apply(lambda x: f(*x), axis=1)
A interesting question! my answer as below:
import pandas as pd
def sublst(row):
return lst[row['J1']:row['J2']]
df = pd.DataFrame({'ID':['1','2','3'], 'J1': [0,2,3], 'J2':[1,4,5]})
print df
lst = ['a','b','c','d','e','f']
df['J3'] = df.apply(sublst,axis=1)
print df
Output:
ID J1 J2
0 1 0 1
1 2 2 4
2 3 3 5
ID J1 J2 J3
0 1 0 1 [a]
1 2 2 4 [c, d]
2 3 3 5 [d, e]
I changed the column name to ID,J1,J2,J3 to ensure ID < J1 < J2 < J3, so the column display in right sequence.
One more brief version:
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'J1': [0,2,3], 'J2':[1,4,5]})
print df
lst = ['a','b','c','d','e','f']
df['J3'] = df.apply(lambda row:lst[row['J1']:row['J2']],axis=1)
print df
The method you are looking for is Series.combine.
However, it seems some care has to be taken around datatypes.
In your example, you would (as I did when testing the answer) naively call
df['col_3'] = df.col_1.combine(df.col_2, func=get_sublist)
However, this throws the error:
ValueError: setting an array element with a sequence.
My best guess is that it seems to expect the result to be of the same type as the series calling the method (df.col_1 here). However, the following works:
df['col_3'] = df.col_1.astype(object).combine(df.col_2, func=get_sublist)
df
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
Returning a list from apply is a dangerous operation as the resulting object is not guaranteed to be either a Series or a DataFrame. And exceptions might be raised in certain cases. Let's walk through a simple example:
df = pd.DataFrame(data=np.random.randint(0, 5, (5,3)),
columns=['a', 'b', 'c'])
df
a b c
0 4 0 0
1 2 0 1
2 2 2 2
3 1 2 2
4 3 0 0
There are three possible outcomes with returning a list from apply
1) If the length of the returned list is not equal to the number of columns, then a Series of lists is returned.
df.apply(lambda x: list(range(2)), axis=1) # returns a Series
0 [0, 1]
1 [0, 1]
2 [0, 1]
3 [0, 1]
4 [0, 1]
dtype: object
2) When the length of the returned list is equal to the number of
columns then a DataFrame is returned and each column gets the
corresponding value in the list.
df.apply(lambda x: list(range(3)), axis=1) # returns a DataFrame
a b c
0 0 1 2
1 0 1 2
2 0 1 2
3 0 1 2
4 0 1 2
3) If the length of the returned list equals the number of columns for the first row but has at least one row where the list has a different number of elements than number of columns a ValueError is raised.
i = 0
def f(x):
global i
if i == 0:
i += 1
return list(range(3))
return list(range(4))
df.apply(f, axis=1)
ValueError: Shape of passed values is (5, 4), indices imply (5, 3)
Answering the problem without apply
Using apply with axis=1 is very slow. It is possible to get much better performance (especially on larger datasets) with basic iterative methods.
Create larger dataframe
df1 = df.sample(100000, replace=True).reset_index(drop=True)
Timings
# apply is slow with axis=1
%timeit df1.apply(lambda x: mylist[x['col_1']: x['col_2']+1], axis=1)
2.59 s ± 76.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# zip - similar to #Thomas
%timeit [mylist[v1:v2+1] for v1, v2 in zip(df1.col_1, df1.col_2)]
29.5 ms ± 534 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
#Thomas answer
%timeit list(map(get_sublist, df1['col_1'],df1['col_2']))
34 ms ± 459 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
I'm sure this isn't as fast as the solutions using Pandas or Numpy operations, but if you don't want to rewrite your function you can use map. Using the original example data -
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
df['col_3'] = list(map(get_sublist,df['col_1'],df['col_2']))
#In Python 2 don't convert above to list
We could pass as many arguments as we wanted into the function this way. The output is what we wanted
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
I'm going to put in a vote for np.vectorize. It allows you to just shoot over x number of columns and not deal with the dataframe in the function, so it's great for functions you don't control or doing something like sending 2 columns and a constant into a function (i.e. col_1, col_2, 'foo').
import numpy as np
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
#df['col_3'] = df[['col_1','col_2']].apply(get_sublist,axis=1)
# expect above to output df as below
df.loc[:,'col_3'] = np.vectorize(get_sublist, otypes=["O"]) (df['col_1'], df['col_2'])
df
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
Here is a faster solution:
def func_1(a,b):
return a + b
df["C"] = func_1(df["A"].to_numpy(),df["B"].to_numpy())
This is 380 times faster than df.apply(f, axis=1) from #Aman and 310 times faster than df['col_3'] = df.apply(lambda x: f(x.col_1, x.col_2), axis=1) from #ajrwhite.
I add some benchmarks too:
Results:
FUNCTIONS TIMINGS GAIN
apply lambda 0.7 x 1
apply 0.56 x 1.25
map 0.3 x 2.3
np.vectorize 0.01 x 70
f3 on Series 0.0026 x 270
f3 on np arrays 0.0018 x 380
f3 numba 0.0018 x 380
In short:
Using apply is slow. We can speed up things very simply, just by using a function that will operate directly on Pandas Series (or better on numpy arrays). And because we will operate on Pandas Series or numpy arrays, we will be able to vectorize the operations. The function will return a Pandas Series or numpy array that we will assign as a new column.
And here is the benchmark code:
import timeit
timeit_setup = """
import pandas as pd
import numpy as np
import numba
np.random.seed(0)
# Create a DataFrame of 10000 rows with 2 columns "A" and "B"
# containing integers between 0 and 100
df = pd.DataFrame(np.random.randint(0,10,size=(10000, 2)), columns=["A", "B"])
def f1(a,b):
# Here a and b are the values of column A and B for a specific row: integers
return a + b
def f2(x):
# Here, x is pandas Series, and corresponds to a specific row of the DataFrame
# 0 and 1 are the indexes of columns A and B
return x[0] + x[1]
def f3(a,b):
# Same as f1 but we will pass parameters that will allow vectorization
# Here, A and B will be Pandas Series or numpy arrays
# with df["C"] = f3(df["A"],df["B"]): Pandas Series
# with df["C"] = f3(df["A"].to_numpy(),df["B"].to_numpy()): numpy arrays
return a + b
#numba.njit('int64[:](int64[:], int64[:])')
def f3_numba_vectorize(a,b):
# Here a and b are 2 numpy arrays with dtype int64
# This function must return a numpy array whith dtype int64
return a + b
"""
test_functions = [
'df["C"] = df.apply(lambda row: f1(row["A"], row["B"]), axis=1)',
'df["C"] = df.apply(f2, axis=1)',
'df["C"] = list(map(f3,df["A"],df["B"]))',
'df["C"] = np.vectorize(f3) (df["A"].to_numpy(),df["B"].to_numpy())',
'df["C"] = f3(df["A"],df["B"])',
'df["C"] = f3(df["A"].to_numpy(),df["B"].to_numpy())',
'df["C"] = f3_numba_vectorize(df["A"].to_numpy(),df["B"].to_numpy())'
]
for test_function in test_functions:
print(min(timeit.repeat(setup=timeit_setup, stmt=test_function, repeat=7, number=10)))
Output:
0.7
0.56
0.3
0.01
0.0026
0.0018
0.0018
Final note: things could be optimzed with Cython and other numba tricks too.
The way you have written f it needs two inputs. If you look at the error message it says you are not providing two inputs to f, just one. The error message is correct.
The mismatch is because df[['col1','col2']] returns a single dataframe with two columns, not two separate columns.
You need to change your f so that it takes a single input, keep the above data frame as input, then break it up into x,y inside the function body. Then do whatever you need and return a single value.
You need this function signature because the syntax is .apply(f)
So f needs to take the single thing = dataframe and not two things which is what your current f expects.
Since you haven't provided the body of f I can't help in anymore detail - but this should provide the way out without fundamentally changing your code or using some other methods rather than apply
Another option is df.itertuples() (generally faster and recommended over df.iterrows() by docs and user testing):
import pandas as pd
df = pd.DataFrame([range(4) for _ in range(4)], columns=list("abcd"))
df
a b c d
0 0 1 2 3
1 0 1 2 3
2 0 1 2 3
3 0 1 2 3
df["e"] = [sum(row) for row in df[["b", "d"]].itertuples(index=False)]
df
a b c d e
0 0 1 2 3 4
1 0 1 2 3 4
2 0 1 2 3 4
3 0 1 2 3 4
Since itertuples returns an Iterable of namedtuples, you can access tuple elements both as attributes by column name (aka dot notation) and by index:
b, d = row
b = row.b
d = row[1]
My example to your questions:
def get_sublist(row, col1, col2):
return mylist[row[col1]:row[col2]+1]
df.apply(get_sublist, axis=1, col1='col_1', col2='col_2')
It can be done in two simple ways:
Let's say, we want sum of col1 and col2 in output column named col_sum
Method 1
f = lambda x : x.col1 + x.col2
df['col_sum'] = df.apply(f, axis=1)
Method 2
def f(x):
x['col_sum'] = x.col_1 + col_2
return x
df = df.apply(f, axis=1)
Method 2 should be used when some complex function has to applied to the dataframe. Method 2 can also be used when output in multiple columns is required.
I suppose you don't want to change get_sublist function, and just want to use DataFrame's apply method to do the job. To get the result you want, I've wrote two help functions: get_sublist_list and unlist. As the function name suggest, first get the list of sublist, second extract that sublist from that list. Finally, We need to call apply function to apply those two functions to the df[['col_1','col_2']] DataFrame subsequently.
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
def get_sublist_list(cols):
return [get_sublist(cols[0],cols[1])]
def unlist(list_of_lists):
return list_of_lists[0]
df['col_3'] = df[['col_1','col_2']].apply(get_sublist_list,axis=1).apply(unlist)
df
If you don't use [] to enclose the get_sublist function, then the get_sublist_list function will return a plain list, it'll raise ValueError: could not broadcast input array from shape (3) into shape (2), as #Ted Petrou had mentioned.
If you have a huge data-set, then you can use an easy but faster(execution time) way of doing this using swifter:
import pandas as pd
import swifter
def fnc(m,x,c):
return m*x+c
df = pd.DataFrame({"m": [1,2,3,4,5,6], "c": [1,1,1,1,1,1], "x":[5,3,6,2,6,1]})
df["y"] = df.swifter.apply(lambda x: fnc(x.m, x.x, x.c), axis=1)
Suppose I have a df which has columns of 'ID', 'col_1', 'col_2'. And I define a function :
f = lambda x, y : my_function_expression.
Now I want to apply the f to df's two columns 'col_1', 'col_2' to element-wise calculate a new column 'col_3' , somewhat like :
df['col_3'] = df[['col_1','col_2']].apply(f)
# Pandas gives : TypeError: ('<lambda>() takes exactly 2 arguments (1 given)'
How to do ?
** Add detail sample as below ***
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
#df['col_3'] = df[['col_1','col_2']].apply(get_sublist,axis=1)
# expect above to output df as below
ID col_1 col_2 col_3
0 1 0 1 ['a', 'b']
1 2 2 4 ['c', 'd', 'e']
2 3 3 5 ['d', 'e', 'f']
There is a clean, one-line way of doing this in Pandas:
df['col_3'] = df.apply(lambda x: f(x.col_1, x.col_2), axis=1)
This allows f to be a user-defined function with multiple input values, and uses (safe) column names rather than (unsafe) numeric indices to access the columns.
Example with data (based on original question):
import pandas as pd
df = pd.DataFrame({'ID':['1', '2', '3'], 'col_1': [0, 2, 3], 'col_2':[1, 4, 5]})
mylist = ['a', 'b', 'c', 'd', 'e', 'f']
def get_sublist(sta,end):
return mylist[sta:end+1]
df['col_3'] = df.apply(lambda x: get_sublist(x.col_1, x.col_2), axis=1)
Output of print(df):
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
If your column names contain spaces or share a name with an existing dataframe attribute, you can index with square brackets:
df['col_3'] = df.apply(lambda x: f(x['col 1'], x['col 2']), axis=1)
Here's an example using apply on the dataframe, which I am calling with axis = 1.
Note the difference is that instead of trying to pass two values to the function f, rewrite the function to accept a pandas Series object, and then index the Series to get the values needed.
In [49]: df
Out[49]:
0 1
0 1.000000 0.000000
1 -0.494375 0.570994
2 1.000000 0.000000
3 1.876360 -0.229738
4 1.000000 0.000000
In [50]: def f(x):
....: return x[0] + x[1]
....:
In [51]: df.apply(f, axis=1) #passes a Series object, row-wise
Out[51]:
0 1.000000
1 0.076619
2 1.000000
3 1.646622
4 1.000000
Depending on your use case, it is sometimes helpful to create a pandas group object, and then use apply on the group.
A simple solution is:
df['col_3'] = df[['col_1','col_2']].apply(lambda x: f(*x), axis=1)
A interesting question! my answer as below:
import pandas as pd
def sublst(row):
return lst[row['J1']:row['J2']]
df = pd.DataFrame({'ID':['1','2','3'], 'J1': [0,2,3], 'J2':[1,4,5]})
print df
lst = ['a','b','c','d','e','f']
df['J3'] = df.apply(sublst,axis=1)
print df
Output:
ID J1 J2
0 1 0 1
1 2 2 4
2 3 3 5
ID J1 J2 J3
0 1 0 1 [a]
1 2 2 4 [c, d]
2 3 3 5 [d, e]
I changed the column name to ID,J1,J2,J3 to ensure ID < J1 < J2 < J3, so the column display in right sequence.
One more brief version:
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'J1': [0,2,3], 'J2':[1,4,5]})
print df
lst = ['a','b','c','d','e','f']
df['J3'] = df.apply(lambda row:lst[row['J1']:row['J2']],axis=1)
print df
The method you are looking for is Series.combine.
However, it seems some care has to be taken around datatypes.
In your example, you would (as I did when testing the answer) naively call
df['col_3'] = df.col_1.combine(df.col_2, func=get_sublist)
However, this throws the error:
ValueError: setting an array element with a sequence.
My best guess is that it seems to expect the result to be of the same type as the series calling the method (df.col_1 here). However, the following works:
df['col_3'] = df.col_1.astype(object).combine(df.col_2, func=get_sublist)
df
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
Returning a list from apply is a dangerous operation as the resulting object is not guaranteed to be either a Series or a DataFrame. And exceptions might be raised in certain cases. Let's walk through a simple example:
df = pd.DataFrame(data=np.random.randint(0, 5, (5,3)),
columns=['a', 'b', 'c'])
df
a b c
0 4 0 0
1 2 0 1
2 2 2 2
3 1 2 2
4 3 0 0
There are three possible outcomes with returning a list from apply
1) If the length of the returned list is not equal to the number of columns, then a Series of lists is returned.
df.apply(lambda x: list(range(2)), axis=1) # returns a Series
0 [0, 1]
1 [0, 1]
2 [0, 1]
3 [0, 1]
4 [0, 1]
dtype: object
2) When the length of the returned list is equal to the number of
columns then a DataFrame is returned and each column gets the
corresponding value in the list.
df.apply(lambda x: list(range(3)), axis=1) # returns a DataFrame
a b c
0 0 1 2
1 0 1 2
2 0 1 2
3 0 1 2
4 0 1 2
3) If the length of the returned list equals the number of columns for the first row but has at least one row where the list has a different number of elements than number of columns a ValueError is raised.
i = 0
def f(x):
global i
if i == 0:
i += 1
return list(range(3))
return list(range(4))
df.apply(f, axis=1)
ValueError: Shape of passed values is (5, 4), indices imply (5, 3)
Answering the problem without apply
Using apply with axis=1 is very slow. It is possible to get much better performance (especially on larger datasets) with basic iterative methods.
Create larger dataframe
df1 = df.sample(100000, replace=True).reset_index(drop=True)
Timings
# apply is slow with axis=1
%timeit df1.apply(lambda x: mylist[x['col_1']: x['col_2']+1], axis=1)
2.59 s ± 76.8 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# zip - similar to #Thomas
%timeit [mylist[v1:v2+1] for v1, v2 in zip(df1.col_1, df1.col_2)]
29.5 ms ± 534 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
#Thomas answer
%timeit list(map(get_sublist, df1['col_1'],df1['col_2']))
34 ms ± 459 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
I'm sure this isn't as fast as the solutions using Pandas or Numpy operations, but if you don't want to rewrite your function you can use map. Using the original example data -
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
df['col_3'] = list(map(get_sublist,df['col_1'],df['col_2']))
#In Python 2 don't convert above to list
We could pass as many arguments as we wanted into the function this way. The output is what we wanted
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
I'm going to put in a vote for np.vectorize. It allows you to just shoot over x number of columns and not deal with the dataframe in the function, so it's great for functions you don't control or doing something like sending 2 columns and a constant into a function (i.e. col_1, col_2, 'foo').
import numpy as np
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
#df['col_3'] = df[['col_1','col_2']].apply(get_sublist,axis=1)
# expect above to output df as below
df.loc[:,'col_3'] = np.vectorize(get_sublist, otypes=["O"]) (df['col_1'], df['col_2'])
df
ID col_1 col_2 col_3
0 1 0 1 [a, b]
1 2 2 4 [c, d, e]
2 3 3 5 [d, e, f]
Here is a faster solution:
def func_1(a,b):
return a + b
df["C"] = func_1(df["A"].to_numpy(),df["B"].to_numpy())
This is 380 times faster than df.apply(f, axis=1) from #Aman and 310 times faster than df['col_3'] = df.apply(lambda x: f(x.col_1, x.col_2), axis=1) from #ajrwhite.
I add some benchmarks too:
Results:
FUNCTIONS TIMINGS GAIN
apply lambda 0.7 x 1
apply 0.56 x 1.25
map 0.3 x 2.3
np.vectorize 0.01 x 70
f3 on Series 0.0026 x 270
f3 on np arrays 0.0018 x 380
f3 numba 0.0018 x 380
In short:
Using apply is slow. We can speed up things very simply, just by using a function that will operate directly on Pandas Series (or better on numpy arrays). And because we will operate on Pandas Series or numpy arrays, we will be able to vectorize the operations. The function will return a Pandas Series or numpy array that we will assign as a new column.
And here is the benchmark code:
import timeit
timeit_setup = """
import pandas as pd
import numpy as np
import numba
np.random.seed(0)
# Create a DataFrame of 10000 rows with 2 columns "A" and "B"
# containing integers between 0 and 100
df = pd.DataFrame(np.random.randint(0,10,size=(10000, 2)), columns=["A", "B"])
def f1(a,b):
# Here a and b are the values of column A and B for a specific row: integers
return a + b
def f2(x):
# Here, x is pandas Series, and corresponds to a specific row of the DataFrame
# 0 and 1 are the indexes of columns A and B
return x[0] + x[1]
def f3(a,b):
# Same as f1 but we will pass parameters that will allow vectorization
# Here, A and B will be Pandas Series or numpy arrays
# with df["C"] = f3(df["A"],df["B"]): Pandas Series
# with df["C"] = f3(df["A"].to_numpy(),df["B"].to_numpy()): numpy arrays
return a + b
#numba.njit('int64[:](int64[:], int64[:])')
def f3_numba_vectorize(a,b):
# Here a and b are 2 numpy arrays with dtype int64
# This function must return a numpy array whith dtype int64
return a + b
"""
test_functions = [
'df["C"] = df.apply(lambda row: f1(row["A"], row["B"]), axis=1)',
'df["C"] = df.apply(f2, axis=1)',
'df["C"] = list(map(f3,df["A"],df["B"]))',
'df["C"] = np.vectorize(f3) (df["A"].to_numpy(),df["B"].to_numpy())',
'df["C"] = f3(df["A"],df["B"])',
'df["C"] = f3(df["A"].to_numpy(),df["B"].to_numpy())',
'df["C"] = f3_numba_vectorize(df["A"].to_numpy(),df["B"].to_numpy())'
]
for test_function in test_functions:
print(min(timeit.repeat(setup=timeit_setup, stmt=test_function, repeat=7, number=10)))
Output:
0.7
0.56
0.3
0.01
0.0026
0.0018
0.0018
Final note: things could be optimzed with Cython and other numba tricks too.
The way you have written f it needs two inputs. If you look at the error message it says you are not providing two inputs to f, just one. The error message is correct.
The mismatch is because df[['col1','col2']] returns a single dataframe with two columns, not two separate columns.
You need to change your f so that it takes a single input, keep the above data frame as input, then break it up into x,y inside the function body. Then do whatever you need and return a single value.
You need this function signature because the syntax is .apply(f)
So f needs to take the single thing = dataframe and not two things which is what your current f expects.
Since you haven't provided the body of f I can't help in anymore detail - but this should provide the way out without fundamentally changing your code or using some other methods rather than apply
Another option is df.itertuples() (generally faster and recommended over df.iterrows() by docs and user testing):
import pandas as pd
df = pd.DataFrame([range(4) for _ in range(4)], columns=list("abcd"))
df
a b c d
0 0 1 2 3
1 0 1 2 3
2 0 1 2 3
3 0 1 2 3
df["e"] = [sum(row) for row in df[["b", "d"]].itertuples(index=False)]
df
a b c d e
0 0 1 2 3 4
1 0 1 2 3 4
2 0 1 2 3 4
3 0 1 2 3 4
Since itertuples returns an Iterable of namedtuples, you can access tuple elements both as attributes by column name (aka dot notation) and by index:
b, d = row
b = row.b
d = row[1]
My example to your questions:
def get_sublist(row, col1, col2):
return mylist[row[col1]:row[col2]+1]
df.apply(get_sublist, axis=1, col1='col_1', col2='col_2')
It can be done in two simple ways:
Let's say, we want sum of col1 and col2 in output column named col_sum
Method 1
f = lambda x : x.col1 + x.col2
df['col_sum'] = df.apply(f, axis=1)
Method 2
def f(x):
x['col_sum'] = x.col_1 + col_2
return x
df = df.apply(f, axis=1)
Method 2 should be used when some complex function has to applied to the dataframe. Method 2 can also be used when output in multiple columns is required.
I suppose you don't want to change get_sublist function, and just want to use DataFrame's apply method to do the job. To get the result you want, I've wrote two help functions: get_sublist_list and unlist. As the function name suggest, first get the list of sublist, second extract that sublist from that list. Finally, We need to call apply function to apply those two functions to the df[['col_1','col_2']] DataFrame subsequently.
import pandas as pd
df = pd.DataFrame({'ID':['1','2','3'], 'col_1': [0,2,3], 'col_2':[1,4,5]})
mylist = ['a','b','c','d','e','f']
def get_sublist(sta,end):
return mylist[sta:end+1]
def get_sublist_list(cols):
return [get_sublist(cols[0],cols[1])]
def unlist(list_of_lists):
return list_of_lists[0]
df['col_3'] = df[['col_1','col_2']].apply(get_sublist_list,axis=1).apply(unlist)
df
If you don't use [] to enclose the get_sublist function, then the get_sublist_list function will return a plain list, it'll raise ValueError: could not broadcast input array from shape (3) into shape (2), as #Ted Petrou had mentioned.
If you have a huge data-set, then you can use an easy but faster(execution time) way of doing this using swifter:
import pandas as pd
import swifter
def fnc(m,x,c):
return m*x+c
df = pd.DataFrame({"m": [1,2,3,4,5,6], "c": [1,1,1,1,1,1], "x":[5,3,6,2,6,1]})
df["y"] = df.swifter.apply(lambda x: fnc(x.m, x.x, x.c), axis=1)
I have a simple looking problem. I have a dataframe df with two columns. For each of the strings that occurs in either of these columns I would like to count the number of rows which has the symbol in either column.
E.g.
g k
a h
c i
j e
d i
i h
b b
d d
i a
d h
The following code works but is very inefficient.
for elem in set(df.values.flat):
print elem, len(df.loc[(df[0] == elem) | (df[1] == elem)])
a 2
c 1
b 1
e 1
d 3
g 1
i 4
h 3
k 1
j 1
This is however very inefficient and my dataframe is large. The inefficiency comes from calling df.loc[(df[0] == elem) | (df[1] == elem)] separately for every distinct symbol in df.
Is there a fast way of doing this?
You can use loc to filter out row level matches from 'col2', append the filtered 'col2' values to 'col1', and then call value_counts:
counts = df['col1'].append(df.loc[df['col1'] != df['col2'], 'col2']).value_counts()
The resulting output:
i 4
d 3
h 3
a 2
j 1
k 1
c 1
g 1
b 1
e 1
Note: You can add .sort_index() to the end of the counting code if you want the output to appear in alphabetical order.
Timings
Using the following setup to produce a larger sample dataset:
from string import ascii_lowercase
n = 10**5
data = np.random.choice(list(ascii_lowercase), size=(n,2))
df = pd.DataFrame(data, columns=['col1', 'col2'])
def edchum(df):
vals = np.unique(df.values)
count = np.maximum(df['col1'].str.get_dummies().reindex_axis(vals, axis=1).fillna(0), df['col2'].str.get_dummies().reindex_axis(vals, axis=1).fillna(0)).sum()
return count
I get the following timings:
%timeit df['col1'].append(df.loc[df['col1'] != df['col2'], 'col2']).value_counts()
10 loops, best of 3: 19.7 ms per loop
%timeit edchum(df)
1 loop, best of 3: 3.81 s per loop
OK this is much trickier than I thought, not sure how this will scale but if you have a lot of repeating values then it will be more efficient than your current method, basically we can use str.get_dummies and reindex the columns from that result to generate a dummies df for all unique values, we can then use np.maximal on the 2 dfs and sum these:
In [77]:
t="""col1 col2
g k
a h
c i
j e
d i
i h
b b
d d
i a
d h"""
df = pd.read_csv(io.StringIO(t), delim_whitespace=True)
np.maximum(df['col1'].str.get_dummies().reindex_axis(vals, axis=1).fillna(0), df['col2'].str.get_dummies().reindex_axis(vals, axis=1).fillna(0)).sum()
Out[77]:
a 2
b 1
c 1
d 3
e 1
g 1
h 3
i 4
j 1
k 1
dtype: float64
vals here is just the unique values:
In [80]:
vals = np.unique(df.values)
vals
Out[80]:
array(['a', 'b', 'c', 'd', 'e', 'g', 'h', 'i', 'j', 'k'], dtype=object)
Is there a faster way to drop columns that only contain one distinct value than the code below?
cols=df.columns.tolist()
for col in cols:
if len(set(df[col].tolist()))<2:
df=df.drop(col, axis=1)
This is really quite slow for large dataframes. Logically, this counts the number of values in each column when in fact it could just stop counting after reaching 2 different values.
You can use Series.unique() method to find out all the unique elements in a column, and for columns whose .unique() returns only 1 element, you can drop that. Example -
for col in df.columns:
if len(df[col].unique()) == 1:
df.drop(col,inplace=True,axis=1)
A method that does not do inplace dropping -
res = df
for col in df.columns:
if len(df[col].unique()) == 1:
res = res.drop(col,axis=1)
Demo -
In [154]: df = pd.DataFrame([[1,2,3],[1,3,3],[1,2,3]])
In [155]: for col in df.columns:
.....: if len(df[col].unique()) == 1:
.....: df.drop(col,inplace=True,axis=1)
.....:
In [156]: df
Out[156]:
1
0 2
1 3
2 2
Timing results -
In [166]: %paste
def func1(df):
res = df
for col in df.columns:
if len(df[col].unique()) == 1:
res = res.drop(col,axis=1)
return res
## -- End pasted text --
In [172]: df = pd.DataFrame({'a':1, 'b':np.arange(5), 'c':[0,0,2,2,2]})
In [178]: %timeit func1(df)
1000 loops, best of 3: 1.05 ms per loop
In [180]: %timeit df[df.apply(pd.Series.value_counts).dropna(thresh=2, axis=1).columns]
100 loops, best of 3: 8.81 ms per loop
In [181]: %timeit df.apply(pd.Series.value_counts).dropna(thresh=2, axis=1)
100 loops, best of 3: 5.81 ms per loop
The fastest method still seems to be the method using unique and looping through the columns.
One step:
df = df[[c for c
in list(df)
if len(df[c].unique()) > 1]]
Two steps:
Create a list of column names that have more than 1 distinct value.
keep = [c for c
in list(df)
if len(df[c].unique()) > 1]
Drop the columns that are not in 'keep'
df = df[keep]
Note: this step can also be done using a list of columns to drop:
drop_cols = [c for c
in list(df)
if df[c].nunique() <= 1]
df = df.drop(columns=drop_cols)
df.loc[:,df.apply(pd.Series.nunique) != 1]
For example
In:
df = pd.DataFrame({'A': [10, 20, np.nan, 30], 'B': [10, np.nan, 10, 10]})
df.loc[:,df.apply(pd.Series.nunique) != 1]
Out:
A
0 10
1 20
2 NaN
3 30
Two simple one-liners for either returning a view (shorter version of jz0410's answer)
df.loc[:,df.nunique()!=1]
or dropping inplace (via drop())
df.drop(columns=df.columns[df.nunique()==1], inplace=True)
You can create a mask of your df by calling apply and call value_counts, this will produce NaN for all rows except one, you can then call dropna column-wise and pass param thresh=2 so that there must be 2 or more non-NaN values:
In [329]:
df = pd.DataFrame({'a':1, 'b':np.arange(5), 'c':[0,0,2,2,2]})
df
Out[329]:
a b c
0 1 0 0
1 1 1 0
2 1 2 2
3 1 3 2
4 1 4 2
In [342]:
df[df.apply(pd.Series.value_counts).dropna(thresh=2, axis=1).columns]
Out[342]:
b c
0 0 0
1 1 0
2 2 2
3 3 2
4 4 2
Output from the boolean conditions:
In [344]:
df.apply(pd.Series.value_counts)
Out[344]:
a b c
0 NaN 1 2
1 5 1 NaN
2 NaN 1 3
3 NaN 1 NaN
4 NaN 1 NaN
In [345]:
df.apply(pd.Series.value_counts).dropna(thresh=2, axis=1)
Out[345]:
b c
0 1 2
1 1 NaN
2 1 3
3 1 NaN
4 1 NaN
Many examples in thread and this thread does not worked for my df. Those worked:
# from: https://stackoverflow.com/questions/33144813/quickly-drop-dataframe-columns-with-only-one-distinct-value
# from: https://stackoverflow.com/questions/20209600/pandas-dataframe-remove-constant-column
import pandas as pd
import numpy as np
data = {'var1': [1,2,3,4,5,np.nan,7,8,9],
'var2':['Order',np.nan,'Inv','Order','Order','Shp','Order', 'Order','Inv'],
'var3':[101,101,101,102,102,102,103,103,np.nan],
'var4':[np.nan,1,1,1,1,1,1,1,1],
'var5':[1,1,1,1,1,1,1,1,1],
'var6':[np.nan,np.nan,np.nan,np.nan,np.nan,np.nan,np.nan,np.nan,np.nan],
'var7':["a","a","a","a","a","a","a","a","a"],
'var8': [1,2,3,4,5,6,7,8,9]}
df = pd.DataFrame(data)
df_original = df.copy()
#-------------------------------------------------------------------------------------------------
df2 = df[[c for c
in list(df)
if len(df[c].unique()) > 1]]
#-------------------------------------------------------------------------------------------------
keep = [c for c
in list(df)
if len(df[c].unique()) > 1]
df3 = df[keep]
#-------------------------------------------------------------------------------------------------
keep_columns = [col for col in df.columns if len(df[col].unique()) > 1]
df5 = df[keep_columns].copy()
#-------------------------------------------------------------------------------------------------
for col in df.columns:
if len(df[col].unique()) == 1:
df.drop(col,inplace=True,axis=1)
I would like to throw in:
pandas 1.0.3
ids = df.nunique().values>1
df.loc[:,ids]
not that slow:
2.81 ms ± 115 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
df=df.loc[:,df.nunique()!=Numberofvalues]
None of the solutions worked in my use-case because I got this error: (my dataframe contains list item).
TypeError: unhashable type: 'list'
The solution that worked for me is this:
ndf = df.describe(include="all").T
new_cols = set(df.columns) - set(ndf[ndf.unique == 1].index)
df = df[list(new_cols)]
One line
df=df[[i for i in df if len(set(df[i]))>1]]
One of the solutions with pipe (convenient if used often):
def drop_unique_value_col(df):
return df.loc[:,df.apply(pd.Series.nunique) != 1]
df.pipe(drop_unique_value_col)
This will drop all the columns with only one distinct value.
for col in Dataframe.columns:
if len(Dataframe[col].value_counts()) == 1:
Dataframe.drop([col], axis=1, inplace=True)
Most 'pythonic' way of doing it I could find:
df = df.loc[:, (df != df.iloc[0]).any()]