使用 iloc 为 PandasDataFrame 中的特定单元格设置值

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

Then use .loc:

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

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

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.

Another way is to get the row index and then use df.loc or df.at.

# get row index 'label' from row number 'irow'
label = df.index.values[irow]
df.at[label, 'COL_NAME'] = x

another way is, you assign a column value for a given row based on the index position of a row, the index position always starts with zero, and the last index position is the length of the dataframe:

df["COL_NAME"].iloc[0]=x

Extending Jianxun's answer, using set_value mehtod in pandas. It sets value for a column at given index.

From pandas documentations:

DataFrame.set_value(index, col, value)

To set value at particular index for a column, do:

df.set_value(index, 'COL_NAME', x)

Hope it helps.

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

To modify the value in a cell at the intersection of row "r" (in column "A") and column "C"

1. retrieve the index of the row "r" in column "A"

       i = df[ df['A']=='r' ].index.values[0]
   

2. modify the value in the desired column "C"

       df.loc[i,"C"]="newValue"
   

Note: before, be sure to reset the index of rows ...to have a nice index list!

        df=df.reset_index(drop=True)