使用 .map() 在 Pandas DataFrame 中有效地创建附加列
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Efficiently creating additional columns in a pandas DataFrame using .map()
提问by Daniel Romero
I am analyzing a data set that is similar in shape to the following example. I have two different types of data (abcdata and xyzdata):
我正在分析一个与以下示例形状相似的数据集。我有两种不同类型的数据（abc数据和xyz数据）：
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
I want to create a function that adds a categorizing column for each abccolumn that exists in the dataframe. Using lists of column names and a category mapping dictionary, I was able to get my desired result.
我想创建一个函数，为数据框中存在的每个abc列添加一个分类列。使用列名列表和类别映射字典，我能够得到我想要的结果。
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
The end result:
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
While the forloop at the end works fine, I feel like I should be using Python's lambdafunction, but can't seem to figure it out.
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。
Is there a more efficient way to map in a dynamic number of abc-type columns?
有没有更有效的方法来映射动态数量的abc类型列？
回答by Andy Hayden
You can use applymapwith the dictionary getmethod:
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
And put this to the specified columns:
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)
Note: you can construct abc_columnsrelatively efficiently using a list comprehension:
注意：您可以abc_columns使用列表推导式相对高效地构建：
abc_columns = [col for col in df.columns if str(col).startswith('abc')]
