Use loc and iloc together in pandas

Question

Say I have the following dataframe, and I want to change the two elements in column c that correspond to the first two elements in column a that are equal to 1 to equal 2.

>>> df = pd.DataFrame({"a" : [1,1,1,1,2,2,2,2], "b" : [2,3,1,4,5,6,7,2], "c" : [1,2,3,4,5,6,7,8]})
>>> df.loc[df["a"] == 1, "c"].iloc[0:2] = 2
>>> df
   a  b  c
0  1  2  1
1  1  3  2
2  1  1  3
3  1  4  4
4  2  5  5
5  2  6  6
6  2  7  7
7  2  2  8

The code in the second line doesn't work because iloc sets a copy, so the original dataframe is not modified. How would I do this?

Does this answer your question? [How to set a value in a pandas DataFrame by mixed iloc and loc](https://stackoverflow.com/questions/32258817/how-to-set-a-value-in-a-pandas-dataframe-by-mixed-iloc-and-loc) — Serge Stroobandt, Aug 16 '21 at 08:25

ayhan · Answer 1 · 2021-07-28T11:07:10.980

6

A dirty way would be:

df.loc[df[df['a'] == 1][:2].index, 'c'] = 2

edited Jul 28 '21 at 11:07

answered Jun 19 '16 at 19:53

ayhan

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2

I don't think this is dirty! This is an example of a class of slicing where it's positional along one axis, and by index on the other. – piRSquared Jun 20 '16 at 00:21
4

I was able to save one char: `df.loc[df.index[df.a == 1][:2], 'c'] = 2` – piRSquared Jun 20 '16 at 00:27

jezrael · Accepted Answer · 2016-06-19T21:20:37.247

You can use Index.isin:

import pandas as pd

df = pd.DataFrame({"a" : [1,1,1,1,2,2,2,2], 
                   "b" : [2,3,1,4,5,6,7,2],
                   "c" : [1,2,3,4,5,6,7,8]})

#more general index                       
df.index = df.index + 10
print (df)
    a  b  c
10  1  2  1
11  1  3  2
12  1  1  3
13  1  4  4
14  2  5  5
15  2  6  6
16  2  7  7
17  2  2  8

print (df.index.isin(df.index[:2]))
[ True  True False False False False False False]

df.loc[(df["a"] == 1) & (df.index.isin(df.index[:2])), "c"] = 2
print (df)
    a  b  c
10  1  2  2
11  1  3  2
12  1  1  3
13  1  4  4
14  2  5  5
15  2  6  6
16  2  7  7
17  2  2  8

If index is nice (starts from 0 without duplicates):

df.loc[(df["a"] == 1) & (df.index < 2), "c"] = 2
print (df)
   a  b  c
0  1  2  2
1  1  3  2
2  1  1  3
3  1  4  4
4  2  5  5
5  2  6  6
6  2  7  7
7  2  2  8

Another solution:

mask = df["a"] == 1
mask = mask & (mask.cumsum() < 3)

df.loc[mask.index[:2], "c"] = 2
print (df)
   a  b  c
0  1  2  2
1  1  3  2
2  1  1  3
3  1  4  4
4  2  5  5
5  2  6  6
6  2  7  7
7  2  2  8

Use loc and iloc together in pandas

2 Answers2