python - df[column] = apply(lambda row : row. sort_values()[1]) 行为奇怪

Question

问题是:为什么 top_2_should 的值与 top_2_is 不同 - 或者换句话说 - 如果将 apply 函数分配给列，为什么它的结果是错误的？

编辑:因为我认为这个问题有点被误解，所以我创建了另一个例子。EDIT2:我使用Python 2.7.12::Anaconda 4.0.0(64位)::Pandas 0.18.0

import pandas as pd

d = {'one' : [1., 2., 3., 4.],
     'two' : [4., 3., 2., 1.]}
df52 = pd.DataFrame(d)

top_1_should = df52.apply(lambda row: row.sort_values()[0], 1)
top_2_should = df52.apply(lambda row: row.sort_values()[1], 1)
df52['top_1_is'] = df52.apply(lambda row: row.sort_values()[0], 1)
df52['top_1_should'] = top_1_should
df52['top_2_is'] = df52.apply(lambda row: row.sort_values()[1], 1)
df52['top_2_should'] = top_2_should
print df52

   one  two  top_1_is  top_1_should  top_2_is  top_2_should
0  1.0  4.0       1.0           1.0       1.0           4.0
1  2.0  3.0       2.0           2.0       2.0           3.0
2  3.0  2.0       2.0           2.0       2.0           3.0
3  4.0  1.0       1.0           1.0       1.0           4.0

最好，一月

Answer 1

我认为你可以使用Series.sort_values与 values对于中断对齐行:

print (df52.apply(lambda row: row.sort_values().values, axis=1))
   one  two
0  1.0  4.0
1  2.0  3.0
2  2.0  3.0
3  1.0  4.0

或者:

print (pd.DataFrame(np.sort(df52.values), df52.index, df52.columns))
   one  two
0  1.0  4.0
1  2.0  3.0
2  2.0  3.0
3  1.0  4.0

<小时/>

如果使用print，您将获得排序的输出 - 如果之前添加新列，需要更改Series中所选行的位置，即DataFrame<中的列:

top_1_should = df52.apply(lambda row: row.sort_values()[0], 1)
top_2_should = df52.apply(lambda row: row.sort_values()[1], 1)
df52['top_1_is'] = df52.apply(lambda row: row.sort_values()[0], 1)
df52['top_1_should'] = top_1_should
df52['top_2_is'] = df52.apply(lambda row: row.sort_values()[1], 1)
df52['top_2_is'] = df52.apply(lambda row: print(row.sort_values()), 1)
one             1.0
top_1_is        1.0
top_1_should    1.0
top_2_is        1.0
two             4.0
Name: 0, dtype: float64
one             2.0
top_1_is        2.0
top_1_should    2.0
top_2_is        2.0
two             3.0
Name: 1, dtype: float64
two             2.0
top_1_is        2.0
top_1_should    2.0
top_2_is        2.0
one             3.0
Name: 2, dtype: float64
two             1.0
top_1_is        1.0
top_1_should    1.0
top_2_is        1.0
one             4.0
Name: 3, dtype: float64