Python:结合低频因素/类别计数 [英] Python: Combining Low Frequency Factors/Category Counts

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问题描述

R中有一个很棒的解决方案.

我的df.column看起来像:

Windows
Windows
Mac
Mac
Mac
Linux
Windows
...

我想在此df.column向量中将低频类别替换为其他".例如,我需要df.column看起来像

I want to replace low frequency categories with 'Other' in this df.column vector. For example, I need my df.column to look like

Windows
Windows
Mac
Mac
Mac
Linux -> Other
Windows
...

我想重命名这些稀有类别,以减少回归中的因素数量.这就是为什么我需要原始向量.在python中,运行命令以获取频率表后,我得到:

I would like to rename these rare categories, to reduce the number of factors in my regression. This is why I need the original vector. In python, after running the command to get the frequency table I get:

pd.value_counts(df.column)


Windows          26083
iOS              19711
Android          13077
Macintosh         5799
Chrome OS          347
Linux              285
Windows Phone      167
(not set)           22
BlackBerry          11

我想知道是否有一种方法可以将"Chrome OS","Linux"(低频数据)重命名为另一个类别(例如,其他"类别),并以一种有效的方式进行重命名.

I wonder if there is a method to rename 'Chrome OS', 'Linux' (low frequency data) into another category (for example category 'Other', and do so in an efficient way.

推荐答案

通过查找占用百分比来屏蔽,即:

Mask by finding percentage of occupency i.e :

series = pd.value_counts(df.column)
mask = (series/series.sum() * 100).lt(1)
# To replace df['column'] use np.where I.e 
df['column'] = np.where(df['column'].isin(series[mask].index),'Other',df['column'])

要使用sum更改索引:

To change the index with sum :

new = series[~mask]
new['Other'] = series[mask].sum()

Windows      26083
iOS          19711
Android      13077
Macintosh     5799
Other          832
Name: 1, dtype: int64

如果要替换索引,则:

series.index = np.where(series.index.isin(series[mask].index),'Other',series.index)

Windows      26083
iOS          19711
Android      13077
Macintosh     5799
Other          347
Other          285
Other          167
Other           22
Other           11
Name: 1, dtype: int64

说明

(series/series.sum() * 100) # This will give you the percentage i.e 

Windows          39.820158
iOS              30.092211
Android          19.964276
Macintosh         8.853165
Chrome OS         0.529755
Linux             0.435101
Windows Phone     0.254954
(not set)         0.033587
BlackBerry        0.016793
Name: 1, dtype: float64

.lt(1)等于小于1.这会根据该掩码索引为您提供一个布尔掩码并分配数据

.lt(1) is equivalent to lesser than 1. That gives you a boolean mask, based on that mask index and assign the data

这篇关于Python:结合低频因素/类别计数的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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