Set value for particular cell in pandas DataFrame using index


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?

Asked By: Mitkp



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


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['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['C', 'x'] = 10
100000 loops, best of 3: 9.2 µs per loop
Answered By: unutbu

The recommended way (according to the maintainers) to set a value is:


Using ‘chained indexing’ (df['x']['C']) may lead to problems.


Answered By: Yariv

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)

Answered By: RukTech

Try using df.loc[row_index,col_indexer] = value

Answered By: Yash

This is the only thing that worked for me!

df.loc['C', 'x'] = 10

Learn more about .loc here.

Answered By: Alon Galor

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():
        if (row1[u'src_id'] == row['SRC_ID']) is True:
Answered By: Joshua Magaña

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.

Answered By: Blairg23

you can use .iloc.

df.iloc[[2], [0]] = 10
Answered By: Muge Cevik

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)
Answered By: Kirill Dolmatov

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

Answered By: andrei deusteanu

In my example i just change it in selected cell

    for index, row in result.iterrows():
        if np.isnan(row['weight']):
  [index, 'weight'] = 0.0

‘result’ is a dataField with column ‘weight’

This will change the value of cth row and
xth column.

Answered By: Sujit Singh

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)

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)


# 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):
    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):
    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):
  [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):
    return df

t0 = time.time()    
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))

t0 = time.time()    
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))

t0 = time.time()    
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))

t0 = time.time()    
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))

t0 = time.time()    
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))

t0 = time.time()    
t1 = time.time()
print('Needed {} seconds'.format(t1-t0))
Answered By: gebbissimo

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

>>>[2, "Cars"] = "Chevrolet Corvette"
>>> df
                  Cars  Prices (U$)
0               Audi TT        120.0
1 Lamborghini Aventador        260.0
2    Chevrolet Corvette        240.0


Answered By: ivanleoncz

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 work for both type of data frames, df.iloc only works with row/column integer indices, df.loc and supports for setting values using column names and/or integer indices.

When the specified index does not exist, both df.loc and 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'][2,'y'] = 400

# rows/columns specified does not exist, appends new rows/columns to existing data frame['D','w'] = 9000
df1.loc['E','q'] = 499

# using df[<some_column_name>] == <condition> to retrieve target rows[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[[0,2,"D"],'x'] = 10[:, ['y', 'w']] = 8000

>>> 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
Answered By: Good Will

.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 :

  1. df.iat[0,0] = 2
  2.[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   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   108
1   30
2   30
3   16
4   62
5   126
6   22
Answered By: DINA TAKLIT

To set values, use:[0, 'clm1'] = 0
  • The fastest recommended method for setting variables.
  • set_value, ix have been deprecated.
  • No warning, unlike iloc and loc
Answered By: Miladiouss

I tested and the output is df.set_value is little faster, but the official method 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[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.

Answered By: prosti

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

  1. Replace value of one column for conditional_index
df.loc[conditional_index , [col name]]= <new value>
  1. Replace value of multiple column for conditional_index
df.loc[conditional_index, [col1,col2]]= <new value>
  1. 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.

Answered By: Atta Jutt

Soo, your question to convert NaN at [‘x’,C] to value 10

the answer is..


alternative code is

df.loc['C', 'x']=10
Answered By: Ichsan

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

Answered By: KasperGL

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[0, 0] = '"236"76"'
#  %timeit[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 is the fastest approach.

Answered By: Gonçalo Peres

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
Answered By: Serge Stroobandt
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