熊猫:如何根据其他列值的条件对列进行求和?

Sha*_*ang 10 python conditional dataframe pandas

我有以下pandas DataFrame.

import pandas as pd
df = pd.read_csv('filename.csv')

print(df)

     dog      A         B           C
0     dog1    0.787575  0.159330    0.053095
1     dog10   0.770698  0.169487    0.059815
2     dog11   0.792689  0.152043    0.055268
3     dog12   0.785066  0.160361    0.054573
4     dog13   0.795455  0.150464    0.054081
5     dog14   0.794873  0.150700    0.054426
..    ....
8     dog19   0.811585  0.140207    0.048208
9     dog2    0.797202  0.152033    0.050765
10    dog20   0.801607  0.145137    0.053256
11    dog21   0.792689  0.152043    0.055268
    ....
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我总结列上创建一个新列"A","B","C"如下:

df['total_ABC'] = df[["A", "B", "B"]].sum(axis=1)
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现在我想基于条件执行此操作,即如果"A" < 0.78然后创建新的求和列df['smallA_sum'] = df[["A", "B", "B"]].sum(axis=1).否则,该值应为零.

如何创建这样的条件语句?

我的想法是使用

df['smallA_sum'] = df1.apply(lambda row: (row['A']+row['B']+row['C']) if row['A'] < 0.78))
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但是,这不起作用,我无法指定轴.

如何根据其他列的值创建列?

您也可以为每个df['dog'] == 'dog2'创建列dog2_sum,即

 df['dog2_sum'] = df1.apply(lambda row: (row['A']+row['B']+row['C']) if df['dog'] == 'dog2'))
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但我的方法不正确.

`

EdC*_*ica 11

以下应该可以工作,这里我们屏蔽满足条件的df,这将设置为不满足条件NaN的行,所以我们调用fillna新的col:

In [67]:
df = pd.DataFrame(np.random.randn(5,3), columns=list('ABC'))
df

Out[67]:
          A         B         C
0  0.197334  0.707852 -0.443475
1 -1.063765 -0.914877  1.585882
2  0.899477  1.064308  1.426789
3 -0.556486 -0.150080 -0.149494
4 -0.035858  0.777523 -0.453747

In [73]:    
df['total'] = df.loc[df['A'] > 0,['A','B']].sum(axis=1)
df['total'].fillna(0, inplace=True)
df

Out[73]:
          A         B         C     total
0  0.197334  0.707852 -0.443475  0.905186
1 -1.063765 -0.914877  1.585882  0.000000
2  0.899477  1.064308  1.426789  1.963785
3 -0.556486 -0.150080 -0.149494  0.000000
4 -0.035858  0.777523 -0.453747  0.000000
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另一种方法是调用wheresum结果,这需要一个值PARAM当条件不满足返回:

In [75]:
df['total'] = df[['A','B']].sum(axis=1).where(df['A'] > 0, 0)
df

Out[75]:
          A         B         C     total
0  0.197334  0.707852 -0.443475  0.905186
1 -1.063765 -0.914877  1.585882  0.000000
2  0.899477  1.064308  1.426789  1.963785
3 -0.556486 -0.150080 -0.149494  0.000000
4 -0.035858  0.777523 -0.453747  0.000000
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  • `.where()` 解决方案是完美的!谢谢 (2认同)