[python] Set value for particular cell in pandas DataFrame with iloc

I have a question similar to this and this. The difference is that I have to select row by position, as I do not know the index.

I want to do something like df.iloc[0, 'COL_NAME'] = x, but iloc does not allow this kind of access. If I do df.iloc[0]['COL_NAME] = x the warning about chained indexing appears.

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

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

Then use .loc:

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

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.

You can use:

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

set_valye 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)

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.