使用.map()在pandas DataFrame中有效创建其他列 [英] Efficiently creating additional columns in a pandas DataFrame using .map()

问题描述

我正在分析形状类似于以下示例的数据集.我有两种不同类型的数据( abc 数据和 xyz 数据):

I am analyzing a data set that is similar in shape to the following example. I have two different types of data (abc data and xyz data):

   abc1  abc2  abc3  xyz1  xyz2  xyz3
0     1     2     2     2     1     2
1     2     1     1     2     1     1
2     2     2     1     2     2     2
3     1     2     1     1     1     1
4     1     1     2     1     2     1

我想创建一个为数据框中存在的每个 abc 列添加一个分类列的函数.使用列名列表和类别映射字典，我可以获得所需的结果.

I want to create a function that adds a categorizing column for each abc column that exists in the dataframe. Using lists of column names and a category mapping dictionary, I was able to get my desired result.

abc_columns = ['abc1', 'abc2', 'abc3']
xyz_columns = ['xyz1', 'xyz2', 'xyz3']
abc_category_columns = ['abc1_category', 'abc2_category', 'abc3_category']
categories = {1: 'Good', 2: 'Bad', 3: 'Ugly'}

for i in range(len(abc_category_columns)):
    df3[abc_category_columns[i]] = df3[abc_columns[i]].map(categories)

print df3

最终结果:

   abc1  abc2  abc3  xyz1  xyz2  xyz3 abc1_category abc2_category abc3_category
0     1     2     2     2     1     2          Good           Bad           Bad
1     2     1     1     2     1     1           Bad          Good          Good
2     2     2     1     2     2     2           Bad           Bad          Good
3     1     2     1     1     1     1          Good           Bad          Good
4     1     1     2     1     2     1          Good          Good           Bad

虽然最后的for循环工作正常，但我觉得我应该使用Python的lambda函数，但似乎无法弄清楚.

While the for loop at the end works fine, I feel like I should be using Python's lambda function, but can't seem to figure it out.

是否有更有效的方法来映射动态数量的 abc 类型的列?

Is there a more efficient way to map in a dynamic number of abc-type columns?

推荐答案

您可以使用 get 方法:

You can use applymap with the dictionary get method:

In [11]: df[abc_columns].applymap(categories.get)
Out[11]:
   abc1  abc2  abc3
0  Good   Bad   Bad
1   Bad  Good  Good
2   Bad   Bad  Good
3  Good   Bad  Good
4  Good  Good   Bad

并将其放入指定的列:

In [12]: abc_categories = map(lambda x: x + '_category', abc_columns)

In [13]: abc_categories
Out[13]: ['abc1_category', 'abc2_category', 'abc3_category']

In [14]: df[abc_categories] = df[abc_columns].applymap(categories.get)

注意:您可以使用列表理解来相对有效地构建abc_columns:

Note: you can construct abc_columns relatively efficiently using a list comprehension:

abc_columns = [col for col in df.columns if str(col).startswith('abc')]

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