df[column]=应用（lambda行：row.sort_值（）[1]）行为怪异

Question

问题是： 为什么top_2_的值应该不同于top_2_is，或者换句话说，如果将apply函数分配给某个列，为什么结果是错误的？在

编辑：因为我认为这个问题有点误解，所以我又举了一个例子。 EDIT2：我使用python2.7.12:：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

最好的， 一月

Answer 1

import pandas as pd

d = {'one' : [1., 2., 3., 4.],
     'two' : [2., 3., 4., 5.]}
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()[3], 1)
df52['top_2_should'] = top_2_should
print(df52)

退货：

^{pr2}$