在 Pandas 中同时使用 loc 和 iloc

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?

Answer 1

一种肮脏的方式是：

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

Answer 2

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