Python pandas - 为每列添加lambda

Question

我有一个如下所示的数据框：

DayOfWeek  Sunday  Monday  Tuesday  Wednesday  Thursday  Friday  Saturday
00            0.0     0.0      0.0       19.0       0.0     4.0       0.0
01            0.0     0.0      0.0        0.0       0.0     7.0       0.0
07            0.0     0.0      3.0        5.0       3.0     0.0       1.0
08            0.0    17.0     16.0        8.0      10.0     1.0       0.0
09           10.0    48.0     30.0       86.0      12.0     3.0       0.0
10           70.0    58.0      3.0       36.0      52.0    70.0       0.0
11           32.0    26.0      0.0       20.0      38.0    42.0       0.0
12           21.0     9.0     83.0       32.0     129.0    57.0       0.0
13           53.0    51.0     55.0       36.0      18.0    32.0       0.0
14           64.0    62.0     24.0       21.0      53.0    61.0       0.0
15           46.0   121.0     37.0       31.0      58.0    54.0       0.0
16           95.0   139.0     86.0       58.0      79.0    11.0       0.0
17          113.0    56.0     73.0      146.0      78.0    17.0       0.0

我想把它作为百分比，所以我想总结每一列，并在每个单元格中我想要除以列的总和，所以我做了这个代码：

df_day = df_day.apply(lambda x: round(100 * x / df_day.groupby('DayOfWeek').size().sum()))

但它不起作用......

有什么想法吗？

Answer 1

我觉得你需要分div通过总结列sum ，然后由多个mul ，如有必要round ：

print (df_day.sum())
Sunday       504.0
Monday       587.0
Tuesday      410.0
Wednesday    498.0
Thursday     530.0
Friday       359.0
Saturday       1.0
dtype: float64

print (df_day.div(df_day.sum(), axis=1).mul(100).round(0))
           Sunday  Monday  Tuesday  Wednesday  Thursday  Friday  Saturday
DayOfWeek                                                                
0             0.0     0.0      0.0        4.0       0.0     1.0       0.0
1             0.0     0.0      0.0        0.0       0.0     2.0       0.0
7             0.0     0.0      1.0        1.0       1.0     0.0     100.0
8             0.0     3.0      4.0        2.0       2.0     0.0       0.0
9             2.0     8.0      7.0       17.0       2.0     1.0       0.0
10           14.0    10.0      1.0        7.0      10.0    19.0       0.0
11            6.0     4.0      0.0        4.0       7.0    12.0       0.0
12            4.0     2.0     20.0        6.0      24.0    16.0       0.0
13           11.0     9.0     13.0        7.0       3.0     9.0       0.0
14           13.0    11.0      6.0        4.0      10.0    17.0       0.0
15            9.0    21.0      9.0        6.0      11.0    15.0       0.0
16           19.0    24.0     21.0       12.0      15.0     3.0       0.0
17           22.0    10.0     18.0       29.0      15.0     5.0       0.0

apply慢的解决方案：

print (df_day.apply(lambda x: round(100 * x / df_day.sum()), axis=1))

时间 ：

In [171]: %timeit (df_day.div(df_day.sum(), axis=1).mul(100).round(0))
1000 loops, best of 3: 1.89 ms per loop

In [172]: %timeit (df_day.apply(lambda x: round(100 * x / df_day.sum()), axis=1))
100 loops, best of 3: 5.18 ms per loop