Related
I have created a Pandas DataFrame
df = DataFrame(index=['A','B','C'], columns=['x','y'])
and have got this
x y
A NaN NaN
B NaN NaN
C NaN NaN
Now, I would like to assign a value to particular cell, for example to row C and column x.
I would expect to get this result:
x y
A NaN NaN
B NaN NaN
C 10 NaN
with this code:
df.xs('C')['x'] = 10
However, the contents of df has not changed. The dataframe contains yet again only NaNs.
Any suggestions?
RukTech's answer, df.set_value('C', 'x', 10), is far and away faster than the options I've suggested below. However, it has been slated for deprecation.
Going forward, the recommended method is .iat/.at.
Why df.xs('C')['x']=10 does not work:
df.xs('C') by default, returns a new dataframe with a copy of the data, so
df.xs('C')['x']=10
modifies this new dataframe only.
df['x'] returns a view of the df dataframe, so
df['x']['C'] = 10
modifies df itself.
Warning: It is sometimes difficult to predict if an operation returns a copy or a view. For this reason the docs recommend avoiding assignments with "chained indexing".
So the recommended alternative is
df.at['C', 'x'] = 10
which does modify df.
In [18]: %timeit df.set_value('C', 'x', 10)
100000 loops, best of 3: 2.9 µs per loop
In [20]: %timeit df['x']['C'] = 10
100000 loops, best of 3: 6.31 µs per loop
In [81]: %timeit df.at['C', 'x'] = 10
100000 loops, best of 3: 9.2 µs per loop
Update: The .set_value method is going to be deprecated. .iat/.at are good replacements, unfortunately pandas provides little documentation
The fastest way to do this is using set_value. This method is ~100 times faster than .ix method. For example:
df.set_value('C', 'x', 10)
You can also use a conditional lookup using .loc as seen here:
df.loc[df[<some_column_name>] == <condition>, [<another_column_name>]] = <value_to_add>
where <some_column_name is the column you want to check the <condition> variable against and <another_column_name> is the column you want to add to (can be a new column or one that already exists). <value_to_add> is the value you want to add to that column/row.
This example doesn't work precisely with the question at hand, but it might be useful for someone wants to add a specific value based on a condition.
Try using df.loc[row_index,col_indexer] = value
The recommended way (according to the maintainers) to set a value is:
df.ix['x','C']=10
Using 'chained indexing' (df['x']['C']) may lead to problems.
See:
https://stackoverflow.com/a/21287235/1579844
http://pandas.pydata.org/pandas-docs/dev/indexing.html#indexing-view-versus-copy
https://github.com/pydata/pandas/pull/6031
This is the only thing that worked for me!
df.loc['C', 'x'] = 10
Learn more about .loc here.
To set values, use:
df.at[0, 'clm1'] = 0
The fastest recommended method for setting variables.
set_value, ix have been deprecated.
No warning, unlike iloc and loc
.iat/.at is the good solution.
Supposing you have this simple data_frame:
A B C
0 1 8 4
1 3 9 6
2 22 33 52
if we want to modify the value of the cell [0,"A"] u can use one of those solution :
df.iat[0,0] = 2
df.at[0,'A'] = 2
And here is a complete example how to use iat to get and set a value of cell :
def prepossessing(df):
for index in range(0,len(df)):
df.iat[index,0] = df.iat[index,0] * 2
return df
y_train before :
0
0 54
1 15
2 15
3 8
4 31
5 63
6 11
y_train after calling prepossessing function that iat to change to multiply the value of each cell by 2:
0
0 108
1 30
2 30
3 16
4 62
5 126
6 22
I would suggest:
df.loc[index_position, "column_name"] = some_value
To modifiy multiple cells at the same time:
df.loc[start_idx_pos: End_idx_pos, "column_name"] = some_value
Avoid Assignment with Chained Indexing
You are dealing with an assignment with chained indexing which will result in a SettingWithCopy warning. This should be avoided by all means.
Your assignment will have to resort to one single .loc[] or .iloc[] slice, as explained here. Hence, in your case:
df.loc['C', 'x'] = 10
In my example i just change it in selected cell
for index, row in result.iterrows():
if np.isnan(row['weight']):
result.at[index, 'weight'] = 0.0
'result' is a dataField with column 'weight'
Here is a summary of the valid solutions provided by all users, for data frames indexed by integer and string.
df.iloc, df.loc and df.at work for both type of data frames, df.iloc only works with row/column integer indices, df.loc and df.at supports for setting values using column names and/or integer indices.
When the specified index does not exist, both df.loc and df.at would append the newly inserted rows/columns to the existing data frame, but df.iloc would raise "IndexError: positional indexers are out-of-bounds". A working example tested in Python 2.7 and 3.7 is as follows:
import numpy as np, pandas as pd
df1 = pd.DataFrame(index=np.arange(3), columns=['x','y','z'])
df1['x'] = ['A','B','C']
df1.at[2,'y'] = 400
# rows/columns specified does not exist, appends new rows/columns to existing data frame
df1.at['D','w'] = 9000
df1.loc['E','q'] = 499
# using df[<some_column_name>] == <condition> to retrieve target rows
df1.at[df1['x']=='B', 'y'] = 10000
df1.loc[df1['x']=='B', ['z','w']] = 10000
# using a list of index to setup values
df1.iloc[[1,2,4], 2] = 9999
df1.loc[[0,'D','E'],'w'] = 7500
df1.at[[0,2,"D"],'x'] = 10
df1.at[:, ['y', 'w']] = 8000
df1
>>> df1
x y z w q
0 10 8000 NaN 8000 NaN
1 B 8000 9999 8000 NaN
2 10 8000 9999 8000 NaN
D 10 8000 NaN 8000 NaN
E NaN 8000 9999 8000 499.0
you can use .iloc.
df.iloc[[2], [0]] = 10
set_value() is deprecated.
Starting from the release 0.23.4, Pandas "announces the future"...
>>> df
Cars Prices (U$)
0 Audi TT 120.0
1 Lamborghini Aventador 245.0
2 Chevrolet Malibu 190.0
>>> df.set_value(2, 'Prices (U$)', 240.0)
__main__:1: FutureWarning: set_value is deprecated and will be removed in a future release.
Please use .at[] or .iat[] accessors instead
Cars Prices (U$)
0 Audi TT 120.0
1 Lamborghini Aventador 245.0
2 Chevrolet Malibu 240.0
Considering this advice, here's a demonstration of how to use them:
by row/column integer positions
>>> df.iat[1, 1] = 260.0
>>> df
Cars Prices (U$)
0 Audi TT 120.0
1 Lamborghini Aventador 260.0
2 Chevrolet Malibu 240.0
by row/column labels
>>> df.at[2, "Cars"] = "Chevrolet Corvette"
>>> df
Cars Prices (U$)
0 Audi TT 120.0
1 Lamborghini Aventador 260.0
2 Chevrolet Corvette 240.0
References:
pandas.DataFrame.iat
pandas.DataFrame.at
One way to use index with condition is first get the index of all the rows that satisfy your condition and then simply use those row indexes in a multiple of ways
conditional_index = df.loc[ df['col name'] <condition> ].index
Example condition is like
==5, >10 , =="Any string", >= DateTime
Then you can use these row indexes in variety of ways like
Replace value of one column for conditional_index
df.loc[conditional_index , [col name]]= <new value>
Replace value of multiple column for conditional_index
df.loc[conditional_index, [col1,col2]]= <new value>
One benefit with saving the conditional_index is that you can assign value of one column to another column with same row index
df.loc[conditional_index, [col1,col2]]= df.loc[conditional_index,'col name']
This is all possible because .index returns a array of index which .loc can use with direct addressing so it avoids traversals again and again.
I tested and the output is df.set_value is little faster, but the official method df.at looks like the fastest non deprecated way to do it.
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.rand(100, 100))
%timeit df.iat[50,50]=50 # ✓
%timeit df.at[50,50]=50 # ✔
%timeit df.set_value(50,50,50) # will deprecate
%timeit df.iloc[50,50]=50
%timeit df.loc[50,50]=50
7.06 µs ± 118 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
5.52 µs ± 64.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
3.68 µs ± 80.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
98.7 µs ± 1.07 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
109 µs ± 1.42 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Note this is setting the value for a single cell. For the vectors loc and iloc should be better options since they are vectorized.
If one wants to change the cell in the position (0,0) of the df to a string such as '"236"76"', the following options will do the work:
df[0][0] = '"236"76"'
# %timeit df[0][0] = '"236"76"'
# 938 µs ± 83.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Or using pandas.DataFrame.at
df.at[0, 0] = '"236"76"'
# %timeit df.at[0, 0] = '"236"76"'
#15 µs ± 2.09 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Or using pandas.DataFrame.iat
df.iat[0, 0] = '"236"76"'
# %timeit df.iat[0, 0] = '"236"76"'
# 41.1 µs ± 3.09 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Or using pandas.DataFrame.loc
df.loc[0, 0] = '"236"76"'
# %timeit df.loc[0, 0] = '"236"76"'
# 5.21 ms ± 401 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Or using pandas.DataFrame.iloc
df.iloc[0, 0] = '"236"76"'
# %timeit df.iloc[0, 0] = '"236"76"'
# 5.12 ms ± 300 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
If time is of relevance, using pandas.DataFrame.at is the fastest approach.
Soo, your question to convert NaN at ['x',C] to value 10
the answer is..
df['x'].loc['C':]=10
df
alternative code is
df.loc['C', 'x']=10
df
df.loc['c','x']=10
This will change the value of cth row and
xth column.
If you want to change values not for whole row, but only for some columns:
x = pd.DataFrame({'A': [1, 2, 3], 'B': [4, 5, 6]})
x.iloc[1] = dict(A=10, B=-10)
From version 0.21.1 you can also use .at method. There are some differences compared to .loc as mentioned here - pandas .at versus .loc, but it's faster on single value replacement
In addition to the answers above, here is a benchmark comparing different ways to add rows of data to an already existing dataframe. It shows that using at or set-value is the most efficient way for large dataframes (at least for these test conditions).
Create new dataframe for each row and...
... append it (13.0 s)
... concatenate it (13.1 s)
Store all new rows in another container first, convert to new dataframe once and append...
container = lists of lists (2.0 s)
container = dictionary of lists (1.9 s)
Preallocate whole dataframe, iterate over new rows and all columns and fill using
... at (0.6 s)
... set_value (0.4 s)
For the test, an existing dataframe comprising 100,000 rows and 1,000 columns and random numpy values was used. To this dataframe, 100 new rows were added.
Code see below:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Nov 21 16:38:46 2018
#author: gebbissimo
"""
import pandas as pd
import numpy as np
import time
NUM_ROWS = 100000
NUM_COLS = 1000
data = np.random.rand(NUM_ROWS,NUM_COLS)
df = pd.DataFrame(data)
NUM_ROWS_NEW = 100
data_tot = np.random.rand(NUM_ROWS + NUM_ROWS_NEW,NUM_COLS)
df_tot = pd.DataFrame(data_tot)
DATA_NEW = np.random.rand(1,NUM_COLS)
#%% FUNCTIONS
# create and append
def create_and_append(df):
for i in range(NUM_ROWS_NEW):
df_new = pd.DataFrame(DATA_NEW)
df = df.append(df_new)
return df
# create and concatenate
def create_and_concat(df):
for i in range(NUM_ROWS_NEW):
df_new = pd.DataFrame(DATA_NEW)
df = pd.concat((df, df_new))
return df
# store as dict and
def store_as_list(df):
lst = [[] for i in range(NUM_ROWS_NEW)]
for i in range(NUM_ROWS_NEW):
for j in range(NUM_COLS):
lst[i].append(DATA_NEW[0,j])
df_new = pd.DataFrame(lst)
df_tot = df.append(df_new)
return df_tot
# store as dict and
def store_as_dict(df):
dct = {}
for j in range(NUM_COLS):
dct[j] = []
for i in range(NUM_ROWS_NEW):
dct[j].append(DATA_NEW[0,j])
df_new = pd.DataFrame(dct)
df_tot = df.append(df_new)
return df_tot
# preallocate and fill using .at
def fill_using_at(df):
for i in range(NUM_ROWS_NEW):
for j in range(NUM_COLS):
#print("i,j={},{}".format(i,j))
df.at[NUM_ROWS+i,j] = DATA_NEW[0,j]
return df
# preallocate and fill using .at
def fill_using_set(df):
for i in range(NUM_ROWS_NEW):
for j in range(NUM_COLS):
#print("i,j={},{}".format(i,j))
df.set_value(NUM_ROWS+i,j,DATA_NEW[0,j])
return df
#%% TESTS
t0 = time.time()
create_and_append(df)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
t0 = time.time()
create_and_concat(df)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
t0 = time.time()
store_as_list(df)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
t0 = time.time()
store_as_dict(df)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
t0 = time.time()
fill_using_at(df_tot)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
t0 = time.time()
fill_using_set(df_tot)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
I too was searching for this topic and I put together a way to iterate through a DataFrame and update it with lookup values from a second DataFrame. Here is my code.
src_df = pd.read_sql_query(src_sql,src_connection)
for index1, row1 in src_df.iterrows():
for index, row in vertical_df.iterrows():
src_df.set_value(index=index1,col=u'etl_load_key',value=etl_load_key)
if (row1[u'src_id'] == row['SRC_ID']) is True:
src_df.set_value(index=index1,col=u'vertical',value=row['VERTICAL'])
I am trying to compare two columns and then return a third value from one of the two adjacent columns. I have read that using iterrows is not the correct way to accomplish this so I tried making writing my own function. The trouble is figuring out the correct syntax to apply it to the df.
import pandas as pd
d = {'a':[1,2,3], 'b':[4,1,6], 'c':[6,7,8], 'd':[8,9,0]}
df = pd.DataFrame(d)
print(df)
def area_name_final(ms1, ms2, an1, an2):
if ms1 >= ms2:
return an1
else:
return an2
df['e'] = df.apply(area_name_final(df.a, df.b, df.c, df.d), axis=1)
Error:
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
Desired Output:
a b c d e
0 1 4 6 8 8
1 2 1 7 9 7
2 3 6 8 0 0
You can try np.where
import numpy as np
df['e'] = np.where(df['a'] >= df['b'], df['c'], df['d'])
print(df)
a b c d e
0 1 4 6 8 8
1 2 1 7 9 7
2 3 6 8 0 0
To fix your code, you need to pass row not the column to apply function
def area_name_final(row):
if row['a'] >= row['b']:
return row['c']
else:
return row['d']
df['e'] = df.apply(area_name_final, axis=1)
You can use a simple where condition that will be much more efficient (vectorized) than your custom function:
df['e'] = df['c'].where(df['a'].ge(df['b']), df['d'])
output:
a b c d e
0 1 4 6 8 8
1 2 1 7 9 7
2 3 6 8 0 0
Using np.where is definitely a good option. There's another way to do it fancily without calling numpy library.
ddf = pd.MultiIndex.from_frame(df)
result = [i[2] if i[0] >= i[1] else i[3] for i in ddf]
df['e'] = result
df
Out[9]:
a b c d e
0 1 4 6 8 8
1 2 1 7 9 7
2 3 6 8 0 0
As pandas' Multiindex helps to turn all your data in dataframe into rows of tuples, you can easily then compare components in a list/tuple.
Extra
However, of course, np.where will give you the result faster.
def solution_1(df):
df['e'] = np.where(df['a'] >= df['b'], df['c'], df['d'])
def solution_2(df):
ddf = pd.MultiIndex.from_frame(df)
df['e'] = [i[2] if i[0] >= i[1] else i[3] for i in ddf]
%timeit solution_1(df)
268 µs ± 2.59 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit solution_2(df)
1.6 ms ± 185 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
If you create a global pd.MultIndex dataframe, on the other hand, the solution will be faster.
def solution_3(df):
df['e'] = [i[2] if i[0] >= i[1] else i[3] for i in ddf]
%timeit solution_3(df)
60.5 µs ± 722 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Another option, probably the closest to what you already had, would be:
df['e'] = df.apply(lambda x: area_name_final(x.a, x.b, x.c, x.d), axis=1)
I have created a Pandas DataFrame
df = DataFrame(index=['A','B','C'], columns=['x','y'])
and have got this
x y
A NaN NaN
B NaN NaN
C NaN NaN
Now, I would like to assign a value to particular cell, for example to row C and column x.
I would expect to get this result:
x y
A NaN NaN
B NaN NaN
C 10 NaN
with this code:
df.xs('C')['x'] = 10
However, the contents of df has not changed. The dataframe contains yet again only NaNs.
Any suggestions?
RukTech's answer, df.set_value('C', 'x', 10), is far and away faster than the options I've suggested below. However, it has been slated for deprecation.
Going forward, the recommended method is .iat/.at.
Why df.xs('C')['x']=10 does not work:
df.xs('C') by default, returns a new dataframe with a copy of the data, so
df.xs('C')['x']=10
modifies this new dataframe only.
df['x'] returns a view of the df dataframe, so
df['x']['C'] = 10
modifies df itself.
Warning: It is sometimes difficult to predict if an operation returns a copy or a view. For this reason the docs recommend avoiding assignments with "chained indexing".
So the recommended alternative is
df.at['C', 'x'] = 10
which does modify df.
In [18]: %timeit df.set_value('C', 'x', 10)
100000 loops, best of 3: 2.9 µs per loop
In [20]: %timeit df['x']['C'] = 10
100000 loops, best of 3: 6.31 µs per loop
In [81]: %timeit df.at['C', 'x'] = 10
100000 loops, best of 3: 9.2 µs per loop
Update: The .set_value method is going to be deprecated. .iat/.at are good replacements, unfortunately pandas provides little documentation
The fastest way to do this is using set_value. This method is ~100 times faster than .ix method. For example:
df.set_value('C', 'x', 10)
You can also use a conditional lookup using .loc as seen here:
df.loc[df[<some_column_name>] == <condition>, [<another_column_name>]] = <value_to_add>
where <some_column_name is the column you want to check the <condition> variable against and <another_column_name> is the column you want to add to (can be a new column or one that already exists). <value_to_add> is the value you want to add to that column/row.
This example doesn't work precisely with the question at hand, but it might be useful for someone wants to add a specific value based on a condition.
Try using df.loc[row_index,col_indexer] = value
The recommended way (according to the maintainers) to set a value is:
df.ix['x','C']=10
Using 'chained indexing' (df['x']['C']) may lead to problems.
See:
https://stackoverflow.com/a/21287235/1579844
http://pandas.pydata.org/pandas-docs/dev/indexing.html#indexing-view-versus-copy
https://github.com/pydata/pandas/pull/6031
This is the only thing that worked for me!
df.loc['C', 'x'] = 10
Learn more about .loc here.
To set values, use:
df.at[0, 'clm1'] = 0
The fastest recommended method for setting variables.
set_value, ix have been deprecated.
No warning, unlike iloc and loc
.iat/.at is the good solution.
Supposing you have this simple data_frame:
A B C
0 1 8 4
1 3 9 6
2 22 33 52
if we want to modify the value of the cell [0,"A"] u can use one of those solution :
df.iat[0,0] = 2
df.at[0,'A'] = 2
And here is a complete example how to use iat to get and set a value of cell :
def prepossessing(df):
for index in range(0,len(df)):
df.iat[index,0] = df.iat[index,0] * 2
return df
y_train before :
0
0 54
1 15
2 15
3 8
4 31
5 63
6 11
y_train after calling prepossessing function that iat to change to multiply the value of each cell by 2:
0
0 108
1 30
2 30
3 16
4 62
5 126
6 22
I would suggest:
df.loc[index_position, "column_name"] = some_value
To modifiy multiple cells at the same time:
df.loc[start_idx_pos: End_idx_pos, "column_name"] = some_value
Avoid Assignment with Chained Indexing
You are dealing with an assignment with chained indexing which will result in a SettingWithCopy warning. This should be avoided by all means.
Your assignment will have to resort to one single .loc[] or .iloc[] slice, as explained here. Hence, in your case:
df.loc['C', 'x'] = 10
In my example i just change it in selected cell
for index, row in result.iterrows():
if np.isnan(row['weight']):
result.at[index, 'weight'] = 0.0
'result' is a dataField with column 'weight'
Here is a summary of the valid solutions provided by all users, for data frames indexed by integer and string.
df.iloc, df.loc and df.at work for both type of data frames, df.iloc only works with row/column integer indices, df.loc and df.at supports for setting values using column names and/or integer indices.
When the specified index does not exist, both df.loc and df.at would append the newly inserted rows/columns to the existing data frame, but df.iloc would raise "IndexError: positional indexers are out-of-bounds". A working example tested in Python 2.7 and 3.7 is as follows:
import numpy as np, pandas as pd
df1 = pd.DataFrame(index=np.arange(3), columns=['x','y','z'])
df1['x'] = ['A','B','C']
df1.at[2,'y'] = 400
# rows/columns specified does not exist, appends new rows/columns to existing data frame
df1.at['D','w'] = 9000
df1.loc['E','q'] = 499
# using df[<some_column_name>] == <condition> to retrieve target rows
df1.at[df1['x']=='B', 'y'] = 10000
df1.loc[df1['x']=='B', ['z','w']] = 10000
# using a list of index to setup values
df1.iloc[[1,2,4], 2] = 9999
df1.loc[[0,'D','E'],'w'] = 7500
df1.at[[0,2,"D"],'x'] = 10
df1.at[:, ['y', 'w']] = 8000
df1
>>> df1
x y z w q
0 10 8000 NaN 8000 NaN
1 B 8000 9999 8000 NaN
2 10 8000 9999 8000 NaN
D 10 8000 NaN 8000 NaN
E NaN 8000 9999 8000 499.0
you can use .iloc.
df.iloc[[2], [0]] = 10
set_value() is deprecated.
Starting from the release 0.23.4, Pandas "announces the future"...
>>> df
Cars Prices (U$)
0 Audi TT 120.0
1 Lamborghini Aventador 245.0
2 Chevrolet Malibu 190.0
>>> df.set_value(2, 'Prices (U$)', 240.0)
__main__:1: FutureWarning: set_value is deprecated and will be removed in a future release.
Please use .at[] or .iat[] accessors instead
Cars Prices (U$)
0 Audi TT 120.0
1 Lamborghini Aventador 245.0
2 Chevrolet Malibu 240.0
Considering this advice, here's a demonstration of how to use them:
by row/column integer positions
>>> df.iat[1, 1] = 260.0
>>> df
Cars Prices (U$)
0 Audi TT 120.0
1 Lamborghini Aventador 260.0
2 Chevrolet Malibu 240.0
by row/column labels
>>> df.at[2, "Cars"] = "Chevrolet Corvette"
>>> df
Cars Prices (U$)
0 Audi TT 120.0
1 Lamborghini Aventador 260.0
2 Chevrolet Corvette 240.0
References:
pandas.DataFrame.iat
pandas.DataFrame.at
One way to use index with condition is first get the index of all the rows that satisfy your condition and then simply use those row indexes in a multiple of ways
conditional_index = df.loc[ df['col name'] <condition> ].index
Example condition is like
==5, >10 , =="Any string", >= DateTime
Then you can use these row indexes in variety of ways like
Replace value of one column for conditional_index
df.loc[conditional_index , [col name]]= <new value>
Replace value of multiple column for conditional_index
df.loc[conditional_index, [col1,col2]]= <new value>
One benefit with saving the conditional_index is that you can assign value of one column to another column with same row index
df.loc[conditional_index, [col1,col2]]= df.loc[conditional_index,'col name']
This is all possible because .index returns a array of index which .loc can use with direct addressing so it avoids traversals again and again.
I tested and the output is df.set_value is little faster, but the official method df.at looks like the fastest non deprecated way to do it.
import numpy as np
import pandas as pd
df = pd.DataFrame(np.random.rand(100, 100))
%timeit df.iat[50,50]=50 # ✓
%timeit df.at[50,50]=50 # ✔
%timeit df.set_value(50,50,50) # will deprecate
%timeit df.iloc[50,50]=50
%timeit df.loc[50,50]=50
7.06 µs ± 118 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
5.52 µs ± 64.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
3.68 µs ± 80.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
98.7 µs ± 1.07 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
109 µs ± 1.42 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Note this is setting the value for a single cell. For the vectors loc and iloc should be better options since they are vectorized.
If one wants to change the cell in the position (0,0) of the df to a string such as '"236"76"', the following options will do the work:
df[0][0] = '"236"76"'
# %timeit df[0][0] = '"236"76"'
# 938 µs ± 83.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Or using pandas.DataFrame.at
df.at[0, 0] = '"236"76"'
# %timeit df.at[0, 0] = '"236"76"'
#15 µs ± 2.09 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Or using pandas.DataFrame.iat
df.iat[0, 0] = '"236"76"'
# %timeit df.iat[0, 0] = '"236"76"'
# 41.1 µs ± 3.09 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Or using pandas.DataFrame.loc
df.loc[0, 0] = '"236"76"'
# %timeit df.loc[0, 0] = '"236"76"'
# 5.21 ms ± 401 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Or using pandas.DataFrame.iloc
df.iloc[0, 0] = '"236"76"'
# %timeit df.iloc[0, 0] = '"236"76"'
# 5.12 ms ± 300 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
If time is of relevance, using pandas.DataFrame.at is the fastest approach.
Soo, your question to convert NaN at ['x',C] to value 10
the answer is..
df['x'].loc['C':]=10
df
alternative code is
df.loc['C', 'x']=10
df
df.loc['c','x']=10
This will change the value of cth row and
xth column.
If you want to change values not for whole row, but only for some columns:
x = pd.DataFrame({'A': [1, 2, 3], 'B': [4, 5, 6]})
x.iloc[1] = dict(A=10, B=-10)
From version 0.21.1 you can also use .at method. There are some differences compared to .loc as mentioned here - pandas .at versus .loc, but it's faster on single value replacement
In addition to the answers above, here is a benchmark comparing different ways to add rows of data to an already existing dataframe. It shows that using at or set-value is the most efficient way for large dataframes (at least for these test conditions).
Create new dataframe for each row and...
... append it (13.0 s)
... concatenate it (13.1 s)
Store all new rows in another container first, convert to new dataframe once and append...
container = lists of lists (2.0 s)
container = dictionary of lists (1.9 s)
Preallocate whole dataframe, iterate over new rows and all columns and fill using
... at (0.6 s)
... set_value (0.4 s)
For the test, an existing dataframe comprising 100,000 rows and 1,000 columns and random numpy values was used. To this dataframe, 100 new rows were added.
Code see below:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Wed Nov 21 16:38:46 2018
#author: gebbissimo
"""
import pandas as pd
import numpy as np
import time
NUM_ROWS = 100000
NUM_COLS = 1000
data = np.random.rand(NUM_ROWS,NUM_COLS)
df = pd.DataFrame(data)
NUM_ROWS_NEW = 100
data_tot = np.random.rand(NUM_ROWS + NUM_ROWS_NEW,NUM_COLS)
df_tot = pd.DataFrame(data_tot)
DATA_NEW = np.random.rand(1,NUM_COLS)
#%% FUNCTIONS
# create and append
def create_and_append(df):
for i in range(NUM_ROWS_NEW):
df_new = pd.DataFrame(DATA_NEW)
df = df.append(df_new)
return df
# create and concatenate
def create_and_concat(df):
for i in range(NUM_ROWS_NEW):
df_new = pd.DataFrame(DATA_NEW)
df = pd.concat((df, df_new))
return df
# store as dict and
def store_as_list(df):
lst = [[] for i in range(NUM_ROWS_NEW)]
for i in range(NUM_ROWS_NEW):
for j in range(NUM_COLS):
lst[i].append(DATA_NEW[0,j])
df_new = pd.DataFrame(lst)
df_tot = df.append(df_new)
return df_tot
# store as dict and
def store_as_dict(df):
dct = {}
for j in range(NUM_COLS):
dct[j] = []
for i in range(NUM_ROWS_NEW):
dct[j].append(DATA_NEW[0,j])
df_new = pd.DataFrame(dct)
df_tot = df.append(df_new)
return df_tot
# preallocate and fill using .at
def fill_using_at(df):
for i in range(NUM_ROWS_NEW):
for j in range(NUM_COLS):
#print("i,j={},{}".format(i,j))
df.at[NUM_ROWS+i,j] = DATA_NEW[0,j]
return df
# preallocate and fill using .at
def fill_using_set(df):
for i in range(NUM_ROWS_NEW):
for j in range(NUM_COLS):
#print("i,j={},{}".format(i,j))
df.set_value(NUM_ROWS+i,j,DATA_NEW[0,j])
return df
#%% TESTS
t0 = time.time()
create_and_append(df)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
t0 = time.time()
create_and_concat(df)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
t0 = time.time()
store_as_list(df)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
t0 = time.time()
store_as_dict(df)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
t0 = time.time()
fill_using_at(df_tot)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
t0 = time.time()
fill_using_set(df_tot)
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
I too was searching for this topic and I put together a way to iterate through a DataFrame and update it with lookup values from a second DataFrame. Here is my code.
src_df = pd.read_sql_query(src_sql,src_connection)
for index1, row1 in src_df.iterrows():
for index, row in vertical_df.iterrows():
src_df.set_value(index=index1,col=u'etl_load_key',value=etl_load_key)
if (row1[u'src_id'] == row['SRC_ID']) is True:
src_df.set_value(index=index1,col=u'vertical',value=row['VERTICAL'])
I have a table below. I would like to return in the haves column for each row in the table, column names where row values equals one, using python and pandas.
Location House car Toys haves
x 1 1 3 House, Car
y 2 1 1 Car, toys
First compare values by eq (==) with dot product with columns names and last remove last separator values by rstrip if performance is important
df['haves'] = df.eq(1).dot(df.columns + ', ').str.rstrip(', ')
#solution with omiting first column
#df['haves'] = df.iloc[:, 1:].eq(1).dot(df.columns[1:] + ', ').str.rstrip(', ')
print (df)
Location House car Toys haves
0 x 1 1 3 House, car
1 y 2 1 1 car, Toys
Details:
print (df.eq(1))
Location House car Toys
0 False True True False
1 False False True True
print (df.eq(1).dot(df.columns + ', '))
0 House, car,
1 car, Toys,
dtype: object
Performance: depends of number of 1 values, number of columns and rows, but because dot is vectorized it is faster like loop solutions:
#2k rows
df = pd.concat([df] * 1000, ignore_index=True)
In [183]: %timeit df['haves'] = df.eq(1).dot(df.columns + ', ').str.rstrip(', ')
2.65 ms ± 34.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
#working if no missing values
In [184]: %timeit df['haves'] = [x.rstrip(', ') for x in df.eq(1).dot(df.columns + ', ')]
2.43 ms ± 38.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
#jpp answer
In [185]: %timeit df['haves'] = [', '.join(df.columns[1:][idx]) for idx in df.iloc[:, 1:].eq(1).values]
86.5 ms ± 4.32 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
#Naga Kiran removed answer
In [186]: %timeit df['have'] = df.apply(lambda x: ','.join(x[x.eq(1)].index),1)
813 ms ± 8.66 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Assuming you need to create the haves series, you can use a list comprehension:
df['haves'] = [', '.join(df.columns[1:][idx]) for idx in df.iloc[:, 1:].eq(1).values]
print(df)
Location House car Toys haves
0 x 1 1 3 House, car
1 y 2 1 1 car, Toys
I don't believe this task is easily vectorisable since you can have a variable number of values satisfying your condition, and your result will be an object dtype series.
Here is a simple way which is only a little slower than the dot method and may be easier to understand. It does use numpy to create the cols array which speeds things up considerably vs. just using df.columns as a list.
import numpy as np
# numpy array of dataframe column names
cols = np.array(df.columns)
# boolean array to mark where dataframe values equal 1
b = (df.values == 1)
# list comprehension to join column names for each boolean row result
df['haves'] = [', '.join(cols[(row_index)]) for row_index in b]
If you have a Pandas dataframe like this, filtering this way works:
df = pd.DataFrame({'name1': ['apple','pear','applepie','APPLE'],
'name2': ['APPLE','PEAR','apple','APPLE']
})
df[df['name1'] != df['name2']] # works
But how do you filter rows, when you want to compare the upper values of the columns?
df[df['name1'].upper() != df['name2'].upper()] # does not work
You need to use pandas.Series.str.upper() as df['name1'] is a series of strings and hence we use .str string accessor for vectorized string manipulation.
df[df['name1'].str.upper() != df['name2'].str.upper()]
Output:
name1 name2
2 applepie apple
For ASCII-only, check above :)
Just as an observation, following this very good answer from #Veedrac, if you want to compare case-insensitive for lots of rows in many languages, you might want to normalize and casefold the values first
df.col.str.normalize('NFKD').transform(str.casefold)
Example
df=pd.DataFrame({'t':['a','b','A', 'ê', 'ê', 'Ê', 'ß', 'ss']})
df.t.duplicated()
0 False
1 False
2 False
3 False
4 False
5 False
6 False
7 False
and
df.t.str.lower().duplicated()
0 False
1 False
2 True
3 False
4 False
5 True
6 False
7 False
But
df.t.str.normalize('NFKD').transform(str.casefold).duplicated(keep=False)
0 True
1 False
2 True
3 True
4 True
5 True
6 True
7 True
Often times it can be faster to use list comprehensions when dealing with strings in pandas.
pd.DataFrame(
[[i, j] for i, j in zip(df.name1, df.name2) if i.upper() != j.upper()],
columns=df.columns
)
name1 name2
0 applepie apple
Some timings:
In [159]: df = pd.concat([df]*10000)
In [160]: %%timeit
...: pd.DataFrame(
...: [[i, j] for i, j in zip(df.name1, df.name2) if i.upper() != j.upper()]
...: ,
...: columns=df.columns
...: )
...:
14.2 ms ± 68.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [161]: %timeit df[df['name1'].str.upper() != df['name2'].str.upper()]
35.6 ms ± 160 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)