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Python 中是否有一个函数可以完成 R fct_lump 函数所做的事情（即将所有太小的组分组到一个“其他”组中）？

下面的例子：

library(dplyr)
library(forcats)

> x <- factor(rep(LETTERS[1:9], times = c(40, 10, 5, 27, 1, 1, 1, 1, 1)))

> x
 [1] A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A A B B B B B B B B
[49] B B C C C C C D D D D D D D D D D D D D D D D D D D D D D D D D D D E F G H I
Levels: A B C D E F G H I

> x %>% fct_lump_n(3)
 [1] A     A     A     A     A     A     A     A     A     A     A     A     A     A     A     A    
[17] A     A     A     A     A     A     A     A     A     A     A     A     A     A     A     A    
[33] A     A     A     A     A     A     A     A     B     B     B     B     B     B     B     B    
[49] B     B     Other Other Other Other Other D     D     D     D     D     D     D     D     D    
[65] D     D     D     D     D     D     D     D     D     D     D     D     D     D     D     D    
[81] D     D     Other Other Other Other Other
Levels: A B D Other
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2 回答 2

2 
pip install siuba 
#( in python or anaconda prompth shell)

#use library as:
from siuba.dply.forcats import fct_lump, fct_reorder 

#just like fct_lump of R :

df['Your_column'] = fct_lump(df['Your_column'], n= 10)

df['Your_column'].value_counts() # check your levels

#it reduces the level to 10, lumps all the others as 'Other'
于 2020-11-09T03:34:04.723 回答
0

您可能还想尝试datar：

>>> from datar.all import factor, rep, LETTERS, c, fct_lump_n, fct_count
>>> 
>>> x = factor(rep(LETTERS[:9], times=c(40, 10, 5, 27, 1, 1, 1, 1, 1)))
>>> x >> fct_count()
           f       n
  <category> <int64>
0          A      40
1          B      10
2          C       5
3          D      27
4          E       1
5          F       1
6          G       1
7          H       1
8          I       1

>>> x >> fct_lump_n(3) >> fct_count()
           f       n
  <category> <int64>
0          A      40
1          B      10
2          D      27
3      Other      10

免责声明：我是datar包的作者。

于 2021-09-06T22:17:27.840 回答