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 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'm trying to perform dataframe union of thousands of dataframes in a Python list. I'm using two approaches I found. The first one is by means of for loop union and the second one is using functools.reduce. Both them work well for toy examples, however for thousands of dataframes I'm experimenting a severe overhead, probably caused by code out of the JVM, sequentialy appending each dataframe at a time (using both merging approaches).
from functools import reduce # For Python 3.x
from pyspark.sql import DataFrame
# The reduce approach
def unionAll(dfs):
return reduce(DataFrame.unionAll, dfs)
df_list = [td2, td3, td4, td5, td6, td7, td8, td9, td10]
df = unionAll(df_list)
#The loop approach
df = df_list[0].union(df_list[1])
for d in df_list[2:]:
df = df.union(d)
The question is how to perform this multiple dataframe operation efficiently, probably circunventing the overhead caused by merging dataframes one-by-one.
Thank you very much
You are currently joining your DataFrames like this:
(((td1 + td2) + td3) + td4)
At each stage, you are concatenating a huge dataframe with a small dataframe, resulting in a copy at each step and a lot of wasted memory. I would suggest combining them like this:
(td1 + td2) + (td3 + td4)
The idea is to iteratively coalesce pairs of roughly the same size until you are left with a single result. Here is a prototype:
def pairwise_reduce(op, x):
while len(x) > 1:
v = [op(i, j) for i, j in zip(x[::2], x[1::2])]
if len(x) > 1 and len(x) % 2 == 1:
v[-1] = op(v[-1], x[-1])
x = v
return x[0]
result = pairwise_reduce(DataFrame.unionAll, df_list)
You will see how this makes a huge difference for python lists.
from functools import reduce
from operator import add
x = [[1, 2, 3], [4, 5, 6], [7, 8], [9, 10, 11, 12]] * 1000
%timeit sum(x, [])
%timeit reduce(add, x)
%timeit pairwise_reduce(add, x)
64.2 ms ± 606 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
66.3 ms ± 679 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
970 µs ± 9.02 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
sum(x, []) == reduce(add, x) == pairwise_reduce(add, x)
# True
I have a large array with uuids, lets call it labels. Now I need for every different uuid in this array a bool mask which shows me at which positions in the array every uuid is located. I need this for later computations.
I use pandas' get_dummies() function to create a one-hot encoding of the labels array. Each column of the resulting dataframe is then casted to a a boolean array and stored in a dictionary. The key of the entry is the uuid.
The creation of the dataframe with the get_dummies() function is always as fast as I need it. But casting the columns to bools gets really slow:
import pandas as pd
import numpy as np
labels = np.random.randint(0, 10000, 500000)
%timeit -n 1 -r 1 d = pd.get_dummies(labels); d = {key: d[key].astype(bool) for i, key in enumerate(d.columns.values)}
>>52.5 s ± 0 ns per loop (mean ± std. dev. of 1 run, 1 loop each)
#smaller dataset
labels = np.random.randint(0, 10000, 100000)
%timeit -n 1 -r 1 d = pd.get_dummies(labels); d = {key: d[key].astype(bool) for i, key in enumerate(d.columns.values)}
>>3.52 s ± 0 ns per loop (mean ± std. dev. of 1 run, 1 loop each)
#without casting to bool
labels = np.random.randint(0, 10000, 500000)
%timeit -n 1 -r 1 d = pd.get_dummies(labels); d = {key: d[key] for i, key in enumerate(d.columns.values)}
>>1.24 s ± 0 ns per loop (mean ± std. dev. of 1 run, 1 loop each)
How can I make this faster, i.e. how can I get my boolean masks from the one-hot encoding?
In order to convert the df to boolean values you can convert it to a numpy array and compare it to 1 and make a df again:
%timeit pd.DataFrame(d.values==1)
1 loop, best of 3: 281 ms per loop
Its not a good idea to follow my original advice from the comment (a was short one zero when i did the timings there)
%timeit d==1
1 loop, best of 3: 4.83 s per loop
I think pandas is much slower here because its iterating over the columns internally.
edit:
to retain the original index you can do:
e = pd.DataFrame(d.values==1)
e.index = d.index
edit2:
to save another 60 ms its also possible to use pandas eval function
%timeit pd.eval('d==1')
1 loop, best of 3: 220 ms per loop
In the following pandas.DataFframe:
df =
alfa beta ceta
a,b,c c,d,e g,e,h
a,b d,e,f g,h,k
j,k c,k,l f,k,n
How to drop the rows in which the column values for alfa has more than 2 elements? This can be done using the length function, I know but not finding a specific answer.
df = df[['alfa'].str.split(',').map(len) < 3]
You can do that test to each row in turn using pandas.DataFrame.apply()
print(df[df['alfa'].apply(lambda x: len(x.split(',')) < 3)])
Gives:
alfa beta ceta
1 a,b d,e,f g,h,k
2 j,k c,k,l f,k,n
Here is an option that is the easiest to remember and still embracing the DataFrame which is the "bleeding heart" of Pandas:
1) Create a new column in the dataframe with a value for the length:
df['length'] = df.alfa.str.len()
2) Index using the new column:
df = df[df.length < 3]
Then the comparison to the above timings, which are not really relevant in this case as the data is very small, and usually is less important than how likely you're going to remember how to do something and not having to interrupt your workflow:
step 1:
%timeit df['length'] = df.alfa.str.len()
359 µs ± 6.83 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
step 2:
df = df[df.length < 3]
627 µs ± 76.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
The good news is that when the size grows, time does not grow linearly. For example doing the same operation with 30,000 rows of data takes about 3ms (so 10,000x data, 3x speed increase). Pandas DataFrame is like a train, takes energy to get it going (so not great for small things under absolute comparison, but objectively does not matter much does it...as with small data things are fast anyways).
This is the numpy version of #NickilMaveli's answer.
mask = np.core.defchararray.count(df.alfa.values.astype(str), ',') <= 1
pd.DataFrame(df.values[mask], df.index[mask], df.columns)
alfa beta ceta
1 a,b d,e,f g,h,k
2 j,k c,k,l f,k,n
naive timing
How's this?
df = df[df['alpha'].str.split(',', expand=True)[2].isnull()]
Using expand=True creates a new dataframe with one column for each item in the list. If the list has three or more items, then the third column will have a non-null value.
One problem with this approach is that if none of the lists have three or more items, selecting column [2] will cause a KeyError. Based on this, it's safer to use the solution posted by #Stephen Rauch.
There are at-least two ways to subset the given DF:
1) Split on the comma separator and then compute length of the resulting list:
df[df['alfa'].str.split(",").str.len().lt(3)]
2) Count number of commas and add 1 to the result to account for the last character:
df[df['alfa'].str.count(",").add(1).lt(3)]
Both produce: