Pandas 将一列列表转换为哑元
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Pandas convert a column of list to dummies
提问by user2900369
I have a dataframe where one column is a list of groups each of my users belongs to. Something like:
我有一个数据框，其中一列是我的每个用户所属的组列表。就像是：
index groups
0 ['a','b','c']
1 ['c']
2 ['b','c','e']
3 ['a','c']
4 ['b','e']
And what I would like to do is create a series of dummy columns to identify which groups each user belongs to in order to run some analyses
我想做的是创建一系列虚拟列来标识每个用户属于哪些组，以便运行一些分析
index a b c d e
0 1 1 1 0 0
1 0 0 1 0 0
2 0 1 1 0 1
3 1 0 1 0 0
4 0 1 0 0 0
pd.get_dummies(df['groups'])
won't work because that just returns a column for each different list in my column.
将不起作用，因为这只会为我的列中的每个不同列表返回一列。
The solution needs to be efficient as the dataframe will contain 500,000+ rows. Any advice would be appreciated!
该解决方案需要高效，因为数据帧将包含 500,000 多行。任何意见，将不胜感激！
回答by joris
Using sfor your df['groups']:
使用s您的df['groups']：
In [21]: s = pd.Series({0: ['a', 'b', 'c'], 1:['c'], 2: ['b', 'c', 'e'], 3: ['a', 'c'], 4: ['b', 'e'] })
In [22]: s
Out[22]:
0 [a, b, c]
1 [c]
2 [b, c, e]
3 [a, c]
4 [b, e]
dtype: object
This is a possible solution:
这是一个可能的解决方案：
In [23]: pd.get_dummies(s.apply(pd.Series).stack()).sum(level=0)
Out[23]:
a b c e
0 1 1 1 0
1 0 0 1 0
2 0 1 1 1
3 1 0 1 0
4 0 1 0 1
The logic of this is:
这样做的逻辑是：
.apply(Series)converts the series of lists to a dataframe.stack()puts everything in one column again (creating a multi-level index)pd.get_dummies( )creating the dummies.sum(level=0) for remerging the different rows that should be one row (by summing up the second level, only keeping the original level (level=0))
.apply(Series)将一系列列表转换为数据框.stack()再次将所有内容放在一列中（创建多级索引）pd.get_dummies( )创建假人.sum(level=0) 用于重新合并应该是一行的不同行（通过总结第二级，只保留原始级别 (level=0)）
An slight equivalent is pd.get_dummies(s.apply(pd.Series), prefix='', prefix_sep='').sum(level=0, axis=1)
一个轻微的等价物是 pd.get_dummies(s.apply(pd.Series), prefix='', prefix_sep='').sum(level=0, axis=1)
If this will be efficient enough, I don't know, but in any case, if performance is important, storing lists in a dataframe is not a very good idea.
我不知道这是否足够有效，但无论如何，如果性能很重要，将列表存储在数据框中并不是一个好主意。
回答by Teoretic
Very fast solution in case you have a large dataframe
非常快速的解决方案，以防您有大型数据框
Using sklearn.preprocessing.MultiLabelBinarizer
使用sklearn.preprocessing.MultiLabelBinarizer
import pandas as pd
from sklearn.preprocessing import MultiLabelBinarizer
df = pd.DataFrame(
{'groups':
[['a','b','c'],
['c'],
['b','c','e'],
['a','c'],
['b','e']]
}, columns=['groups'])
s = df['groups']
mlb = MultiLabelBinarizer()
pd.DataFrame(mlb.fit_transform(s),columns=mlb.classes_, index=df.index)
Result:
结果：
a b c e
0 1 1 1 0
1 0 0 1 0
2 0 1 1 1
3 1 0 1 0
4 0 1 0 1
回答by Paulo Alves
Even though this quest was answered, I have a faster solution:
即使这个任务得到了回答，我有一个更快的解决方案：
df.groups.apply(lambda x: pd.Series([1] * len(x), index=x)).fillna(0, downcast='infer')
And, in case you have empty groups or NaN, you could just:
而且，如果您有空组或NaN，您可以：
df.loc[df.groups.str.len() > 0].apply(lambda x: pd.Series([1] * len(x), index=x)).fillna(0, downcast='infer')
How it works
这个怎么运作
Inside the lambda, xis your list, for example ['a', 'b', 'c']. So pd.Serieswill be as follows:
在 lambda 中，x是您的列表，例如['a', 'b', 'c']. 所以pd.Series将如下：
In [2]: pd.Series([1, 1, 1], index=['a', 'b', 'c'])
Out[2]:
a 1
b 1
c 1
dtype: int64
When all pd.Seriescomes together, they become pd.DataFrameand their indexbecome columns; missing indexbecame a columnwith NaNas you can see next:
当所有pd.Series走到一起，他们变得pd.DataFrame和他们index成为columns; 丢失index变成了一个column，NaN你可以在下面看到：
In [4]: a = pd.Series([1, 1, 1], index=['a', 'b', 'c'])
In [5]: b = pd.Series([1, 1, 1], index=['a', 'b', 'd'])
In [6]: pd.DataFrame([a, b])
Out[6]:
a b c d
0 1.0 1.0 1.0 NaN
1 1.0 1.0 NaN 1.0
Now fillnafills those NaNwith 0:
现在，fillna填充那些NaN有0：
In [7]: pd.DataFrame([a, b]).fillna(0)
Out[7]:
a b c d
0 1.0 1.0 1.0 0.0
1 1.0 1.0 0.0 1.0
And downcast='infer'is to downcast from floatto int:
并且downcast='infer'是从float到int：
In [11]: pd.DataFrame([a, b]).fillna(0, downcast='infer')
Out[11]:
a b c d
0 1 1 1 0
1 1 1 0 1
PS.: It's not required the use of .fillna(0, downcast='infer').
PS.: 不需要使用.fillna(0, downcast='infer').
