使用iloc为pandas DataFrame中的特定单元格设置值

问题描述

我有一个与和 this .区别在于我必须按位置选择行，因为我不知道索引.

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.

我想做类似df.iloc[0, 'COL_NAME'] = x的操作，但是iloc不允许这种访问.如果执行df.iloc[0]['COL_NAME] = x，则会出现有关链接索引的警告.

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.

推荐答案

对于混合位置和索引，请使用.ix.但是您需要确保索引不是整数，否则会引起混乱.

For mixed position and index, use .ix. BUT you need to make sure that your index is not of integer, otherwise it will cause confusions.

df.ix[0, 'COL_NAME'] = x

更新:

或者，尝试

Update:

Alternatively, try

df.iloc[0, df.columns.get_loc('COL_NAME')] = x

示例:

import pandas as pd
import numpy as np

# your data
# ========================
np.random.seed(0)
df = pd.DataFrame(np.random.randn(10, 2), columns=['col1', 'col2'], index=np.random.randint(1,100,10)).sort_index()

print(df)


      col1    col2
10  1.7641  0.4002
24  0.1440  1.4543
29  0.3131 -0.8541
32  0.9501 -0.1514
33  1.8676 -0.9773
36  0.7610  0.1217
56  1.4941 -0.2052
58  0.9787  2.2409
75 -0.1032  0.4106
76  0.4439  0.3337

# .iloc with get_loc
# ===================================
df.iloc[0, df.columns.get_loc('col2')] = 100

df

      col1      col2
10  1.7641  100.0000
24  0.1440    1.4543
29  0.3131   -0.8541
32  0.9501   -0.1514
33  1.8676   -0.9773
36  0.7610    0.1217
56  1.4941   -0.2052
58  0.9787    2.2409
75 -0.1032    0.4106
76  0.4439    0.3337

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