比较 pandas 数据框中的两列以创建第三列 [英] Comparing two columns in pandas dataframe to create a third one

问题描述

我有以下数据框:

In [25]: df1
Out[25]: 
          a         b
0  0.752072  0.813426
1  0.868841  0.354665
2  0.944651  0.745505
3  0.485834  0.163747
4  0.001487  0.820176
5  0.904039  0.136355
6  0.572265  0.250570
7  0.514955  0.868373
8  0.195440  0.484160
9  0.506443  0.523912

现在，我想创建另一列df1['c']，其值在df1['a']和df1['b']中最大.因此，我希望将其作为输出:

Now I want to create another column df1['c'] whose values would be maximum among df1['a'] and df1['b']. Thus, I would like to have this as an output:

In [25]: df1
Out[25]: 
          a         b        c
0  0.752072  0.813426 0.813426
1  0.868841  0.354665 0.868841
2  0.944651  0.745505 0.944651
3  0.485834  0.163747 0.485834
4  0.001487  0.820176 0.820176

我尝试过:

In [23]: df1['c'] = np.where(max(df1['a'], df1['b'], df1['a'], df1['b'])

但是，这会引发语法错误.我看不出有什么方法可以在大熊猫中做到这一点.我的实际数据框太复杂了，因此我想为此提供一个通用的解决方案.有什么想法吗?

However, this throws a syntax error. I don't see any way in which I can do this in pandas. My actual dataframe is way too complex and so I would like to have a generic solution for this. Any ideas?

推荐答案

您可以使用

You can use Series.where:

df['c'] = df.b.where(df.a < df.b, df.a)
print (df)
          a         b         c
0  0.752072  0.813426  0.813426
1  0.868841  0.354665  0.868841
2  0.944651  0.745505  0.944651
3  0.485834  0.163747  0.485834
4  0.001487  0.820176  0.820176
5  0.904039  0.136355  0.904039
6  0.572265  0.250570  0.572265
7  0.514955  0.868373  0.868373
8  0.195440  0.484160  0.484160
9  0.506443  0.523912  0.523912

使用 numpy.where :

df['c'] = np.where(df['a'] > df['b'], df['a'], df['b'])
print (df)
          a         b         c
0  0.752072  0.813426  0.813426
1  0.868841  0.354665  0.868841
2  0.944651  0.745505  0.944651
3  0.485834  0.163747  0.485834
4  0.001487  0.820176  0.820176
5  0.904039  0.136355  0.904039
6  0.572265  0.250570  0.572265
7  0.514955  0.868373  0.868373
8  0.195440  0.484160  0.484160
9  0.506443  0.523912  0.523912

或者找到更简单的 max :

Or simplier is find max:

df['c'] = df[['a','b']].max(axis=1)
print (df)
          a         b         c
0  0.752072  0.813426  0.813426
1  0.868841  0.354665  0.868841
2  0.944651  0.745505  0.944651
3  0.485834  0.163747  0.485834
4  0.001487  0.820176  0.820176
5  0.904039  0.136355  0.904039
6  0.572265  0.250570  0.572265
7  0.514955  0.868373  0.868373
8  0.195440  0.484160  0.484160
9  0.506443  0.523912  0.523912

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