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 NaN
s.
Any suggestions?
df['x']['C']
), use df.ix['x','C']
.
dataframe[column (series)] [row (Series index)]
, whereas many people (including myself) are more used to the dataframe[row][column]
order. As a Matlab and R programmer the latter feels more intuitive to me but that apparently is not the way Pandas works..
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)
df['x']['C'] = 10
.
df=df.append(df.sum(numeric_only=True),ignore_index=True)
?
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.
df.loc[df['age']==3, ['age-group']] = 'toddler'
Try using df.loc[row_index,col_indexer] = value
None
, however, Pandas expects the column to already exist; otherwise it may set a wrong value - in my case it was NaN
which led to other problems. If the column doesn't exist, Pandas gives a hidden AttributeError
that I was only able to read during debugging.
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
ix
is deprecated: pandas-docs.github.io/pandas-docs-travis/…
This is the only thing that worked for me!
df.loc['C', 'x'] = 10
Learn more about .loc
here.
.loc
replace .iat/.at
?
at
Similar to loc
, in that both provide label-based lookups. Use at
if you only need to get or set a single value in a DataFrame or Series. From padas doc
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
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
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
you can use .iloc
.
df.iloc[[2], [0]] = 10
df.iloc[[2:8], [0]] = [2,3,4,5,6,7]
which the method df.loc()
does natively.
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.
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
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.
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)
... 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)
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)
... 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'])
Success story sharing
df.x
in the API. What did you mean?'x'
is the name of a column indf
.df.x
returns aSeries
with the values in columnx
. I'll change it todf['x']
since this notation will work with any column name (unlike the dot notation) and I think is clearer.df.x
was some unknown new method alongsidedf.xs, df.ix
df_temp.iat[0, df_temp.columns.get_loc('Cash')] = df_temp['Cash'].iloc[0] + start_val