Set value for particular cell in pandas DataFrame with iloc

One thing I would add here is that the at function on a dataframe is much faster particularly if you are doing a lot of assignments of individual (not slice) values.

df.at[index, 'col_name'] = x

In my experience I have gotten a 20x speedup. Here is a write up that is Spanish but still gives an impression of what's going on.

For mixed position and index, use .ix. BUT you need to make sure that your index is not of integer, otherwise it will cause confusions.

df.ix[0, 'COL_NAME'] = x

Update:

Alternatively, try

df.iloc[0, df.columns.get_loc('COL_NAME')] = x

Example:

import pandas as pd
import numpy as np

# your data
# ========================
np.random.seed(0)
df = pd.DataFrame(np.random.randn(10, 2), columns=['col1', 'col2'], index=np.random.randint(1,100,10)).sort_index()

print(df)


      col1    col2
10  1.7641  0.4002
24  0.1440  1.4543
29  0.3131 -0.8541
32  0.9501 -0.1514
33  1.8676 -0.9773
36  0.7610  0.1217
56  1.4941 -0.2052
58  0.9787  2.2409
75 -0.1032  0.4106
76  0.4439  0.3337

# .iloc with get_loc
# ===================================
df.iloc[0, df.columns.get_loc('col2')] = 100

df

      col1      col2
10  1.7641  100.0000
24  0.1440    1.4543
29  0.3131   -0.8541
32  0.9501   -0.1514
33  1.8676   -0.9773
36  0.7610    0.1217
56  1.4941   -0.2052
58  0.9787    2.2409
75 -0.1032    0.4106
76  0.4439    0.3337

If you know the position, why not just get the index from that?

Then use .loc:

df.loc[index, 'COL_NAME'] = x

You can use:

df.set_value('Row_index', 'Column_name', value)

set_value is ~100 times faster than .ix method. It also better then use df['Row_index']['Column_name'] = value.

But since set_value is deprecated now so .iat/.at are good replacements.

For example if we have this 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"] we can do

df.iat[0,0] = 2

or df.at[0,'A'] = 2