scikit-learn中跨多列的标签编码

问题描述

我正在尝试使用scikit-learn的LabelEncoder对字符串标签的熊猫DataFrame进行编码.由于数据框有许多(50+)列，因此我想避免为每一列创建一个LabelEncoder对象.我宁愿只有一个大的LabelEncoder对象，这些对象可以在 all 我的所有数据列中使用.

I'm trying to use scikit-learn's LabelEncoder to encode a pandas DataFrame of string labels. As the dataframe has many (50+) columns, I want to avoid creating a LabelEncoder object for each column; I'd rather just have one big LabelEncoder objects that works across all my columns of data.

将整个DataFrame扔到LabelEncoder中会产生以下错误.请记住，我在这里使用伪数据.实际上，我正在处理大约50列的字符串标记数据，因此需要一种不按名称引用任何列的解决方案.

Throwing the entire DataFrame into LabelEncoder creates the below error. Please bear in mind that I'm using dummy data here; in actuality I'm dealing with about 50 columns of string labeled data, so need a solution that doesn't reference any columns by name.

import pandas
from sklearn import preprocessing

df = pandas.DataFrame({
    'pets': ['cat', 'dog', 'cat', 'monkey', 'dog', 'dog'],
    'owner': ['Champ', 'Ron', 'Brick', 'Champ', 'Veronica', 'Ron'],
    'location': ['San_Diego', 'New_York', 'New_York', 'San_Diego', 'San_Diego',
                 'New_York']
})

le = preprocessing.LabelEncoder()

le.fit(df)

关于如何解决此问题的任何想法?

Any thoughts on how to get around this problem?

推荐答案

您可以轻松地做到这一点，

You can easily do this though,

df.apply(LabelEncoder().fit_transform)

在scikit-learn 0.20中，推荐的方法是

In scikit-learn 0.20, the recommended way is

OneHotEncoder().fit_transform(df)

，因为OneHotEncoder现在支持字符串输入.ColumnTransformer可以将OneHotEncoder仅应用于某些列.

as the OneHotEncoder now supports string input.Applying OneHotEncoder only to certain columns is possible with the ColumnTransformer.

由于这个答案是一年多以前的，并且引起了很多反对(包括赏金)，所以我可能应该进一步扩大这个范围.

Since this answer is over a year ago, and generated many upvotes (including a bounty), I should probably extend this further.

对于inverse_transform和transform，您必须做一点点改动.

For inverse_transform and transform, you have to do a little bit of hack.

from collections import defaultdict
d = defaultdict(LabelEncoder)

现在，您将所有LabelEncoder列保留为字典.

With this, you now retain all columns LabelEncoder as dictionary.

# Encoding the variable
fit = df.apply(lambda x: d[x.name].fit_transform(x))

# Inverse the encoded
fit.apply(lambda x: d[x.name].inverse_transform(x))

# Using the dictionary to label future data
df.apply(lambda x: d[x.name].transform(x))

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