按 "OTHER" python 重命名频率较低的类别

Question

在我的数据框中，我有一些包含 100 多个不同类别的分类列。我想按最常见的方式对类别进行排名。我保留了前 9 个最频繁的类别，而不太频繁的类别通过以下方式自动重命名：OTHER

示例：

这是我的 df :

print(df)

    Employee_number                 Jobrol
0                 1        Sales Executive
1                 2     Research Scientist
2                 3  Laboratory Technician
3                 4        Sales Executive
4                 5     Research Scientist
5                 6  Laboratory Technician
6                 7        Sales Executive
7                 8     Research Scientist
8                 9  Laboratory Technician
9                10        Sales Executive
10               11     Research Scientist
11               12  Laboratory Technician
12               13        Sales Executive
13               14     Research Scientist
14               15  Laboratory Technician
15               16        Sales Executive
16               17     Research Scientist
17               18     Research Scientist
18               19                Manager
19               20        Human Resources
20               21        Sales Executive


valCount = df['Jobrol'].value_counts()

valCount

Sales Executive          7
Research Scientist       7
Laboratory Technician    5
Manager                  1
Human Resources          1

我保留前 3 个类别，然后将其余类别重命名为 "OTHER"，我应该如何进行？

谢谢。

Answer 1

将您的系列转换为分类，提取计数不在前 3 名的类别，添加一个新类别，例如'Other'，然后替换之前计算的类别：

df['Jobrol'] = df['Jobrol'].astype('category')

others = df['Jobrol'].value_counts().index[3:]
label = 'Other'

df['Jobrol'] = df['Jobrol'].cat.add_categories([label])
df['Jobrol'] = df['Jobrol'].replace(others, label)

注意：很想通过df['Jobrol'].cat.rename_categories(dict.fromkeys(others, label))重命名类别来组合类别，但这行不通将意味着多个相同标签的类别，这是不可能的。

上述解决方案可适用于按 count 过滤。例如，要仅包含计数为 1 的类别，您可以这样定义 others：

counts = df['Jobrol'].value_counts()
others = counts[counts == 1].index

Answer 2

使用value_counts with numpy.where:

need = df['Jobrol'].value_counts().index[:3]
df['Jobrol'] = np.where(df['Jobrol'].isin(need), df['Jobrol'], 'OTHER')

valCount = df['Jobrol'].value_counts()
print (valCount)
Research Scientist       7
Sales Executive          7
Laboratory Technician    5
OTHER                    2
Name: Jobrol, dtype: int64

另一个解决方案：

N = 3
s = df['Jobrol'].value_counts()
valCount = s.iloc[:N].append(pd.Series(s.iloc[N:].sum(), index=['OTHER']))
print (valCount)
Research Scientist       7
Sales Executive          7
Laboratory Technician    5
OTHER                    2
dtype: int64

Answer 3

一行解决方案：

limit = 500
df['Jobrol'] = df['Jobrol'].map({x[0]: x[0] if x[1] > limit else 'other' for x in dict(df['Jobrol'].value_counts()).items()})

按 "OTHER" python 重命名频率较低的类别

Rename the less frequent categories by "OTHER" python

python

pandas

dataframe

counter

categorical-data