在 Pandas 中同时使用 loc 和 iloc
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Use loc and iloc together in pandas
提问by Alex
Say I have the following dataframe, and I want to change the two elements in column cthat correspond to the first two elements in column athat are equal to 1to equal 2.
假设我有以下数据框，我想更改 columnc中对应于 column中前两个元素的两个元素a等于1equal 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?
第二行中的代码不起作用，因为 iloc 设置了副本，因此不会修改原始数据帧。我该怎么做？
采纳答案by jezrael
You can use Index.isin:
您可以使用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 0without duplicates):
如果索引是nice（从0没有重复的开始）：
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:
另一种解决方案：
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
回答by ayhan
A dirty way would be:
一种肮脏的方式是：
df.loc[df[df['a']==1][:2].index, 'c'] = 2
