Python Pandas - 根据索引替换值
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Pandas - Replace values based on index
提问by Colin O'Brien
Apologies if this has been asked before, I can't seem to find an answer.
抱歉，如果之前有人问过这个问题，我似乎找不到答案。
If I create a dataframe like so:
如果我像这样创建一个数据框：
import pandas as pd, numpy as np
df = pd.DataFrame(np.random.randint(0,100,size=(100, 2)), columns=list('AB'))
How would I change the entry in column A to be the number 16 from row 0 -15, for example? In other words, how do I replace cells based purely on index?
例如，如何将 A 列中的条目更改为第 0 -15 行的数字 16？换句话说，如何完全根据索引替换单元格？
回答by jezrael
Use loc:
使用loc：
df.loc[0:15,'A'] = 16
print (df)
A B
0 16 45
1 16 5
2 16 97
3 16 58
4 16 26
5 16 87
6 16 51
7 16 17
8 16 39
9 16 73
10 16 94
11 16 69
12 16 57
13 16 24
14 16 43
15 16 77
16 41 0
17 3 21
18 0 98
19 45 39
20 66 62
21 8 53
22 69 47
23 48 53
Solution with ixis deprecated.
解决方案与ix被弃用。
回答by amandeep1991
One more solution is
另一种解决方案是
df.at[0:15, 'A']=16
print(df.head(20))
OUTPUT:
输出：
A B
0 16 44
1 16 86
2 16 97
3 16 79
4 16 94
5 16 24
6 16 88
7 16 43
8 16 64
9 16 39
10 16 84
11 16 42
12 16 8
13 16 72
14 16 23
15 16 28
16 18 11
17 76 15
18 12 38
19 91 6
回答by nischi
In addition to the other answers, here is what you can do if you have a list of individual indices:
除了其他答案之外，如果您有单个索引列表，您可以执行以下操作：
indices = [0,1,3,6,10,15]
df.loc[indices,'A'] = 16
print(df.head(16))
Output:
输出：
A B
0 16 4
1 16 4
2 4 3
3 16 4
4 1 1
5 3 0
6 16 4
7 2 1
8 4 4
9 3 4
10 16 0
11 3 1
12 4 2
13 2 2
14 2 1
15 16 1
