Set value for particular cell in pandas DataFrame with iloc
Solution 1:
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
Solution 2:
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.
Solution 3:
If you know the position, why not just get the index from that?
Then use .loc
:
df.loc[index, 'COL_NAME'] = x
Solution 4:
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
Solution 5:
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