Question

我有两个熊猫系列的文本列，我怎么能得到它们的交集？

print(df)

0  {this, is, good}
1  {this, is, not, good}

print(df1)

0  {this, is}
1  {good, bad}

我正在寻找类似下面的输出。

print(df2)

0  {this, is}
1  {good}

我已经尝试过了，但是它返回了

df.apply(lambda x: x.intersection(df1))
TypeError: unhashable type: 'set'

Answer 1

这种方法对我有用

import pandas as pd
import numpy as np

data = np.array([{'this', 'is', 'good'},{'this', 'is', 'not', 'good'}])
data1 = np.array([{'this', 'is'},{'good', 'bad'}])
df = pd.Series(data)
df1 = pd.Series(data1)

df2 = pd.Series([df[i] & df1[i] for i in xrange(df.size)])
print(df2)

Answer 2

看起来很简单：

s1 = pd.Series([{'this', 'is', 'good'}, {'this', 'is', 'not', 'good'}])
s2 = pd.Series([{'this', 'is'}, {'good', 'bad'}])
s1 - (s1 - s2)  
#Out[122]: 
#0    {this, is}
#1        {good}
#dtype: object

Answer 3

我很感谢上面的回答。如果您有 DataFrame ，这是一个简单的示例，可以解决相同的问题（据我猜测，在调查了df和df1之类的变量名之后，您曾要求使用< strong> DataFrame 。）。

此df.apply(lambda row: row[0].intersection(df1.loc[row.name][0]), axis=1)将执行此操作。让我们看看我如何到达解决方案。

https://stackoverflow.com/questions/266582...的答案对我很有帮助。

>>> import pandas as pd

>>> 
>>> df = pd.DataFrame({
...     "set": [{"this", "is", "good"}, {"this", "is", "not", "good"}]
... })
>>> 
>>> df
                     set
0       {this, is, good}
1  {not, this, is, good}
>>> 
>>> df1 = pd.DataFrame({
...     "set": [{"this", "is"}, {"good", "bad"}]
... })
>>> 
>>> df1
           set
0   {this, is}
1  {bad, good}
>>>
>>> df.apply(lambda row: row[0].intersection(df1.loc[row.name][0]), axis=1)
0    {this, is}
1        {good}
dtype: object
>>>

我如何到达上述解决方案？

>>> df.apply(lambda x: print(x.name), axis=1)
0
1
0    None
1    None
dtype: object
>>> 
>>> df.loc[0]
set    {this, is, good}
Name: 0, dtype: object
>>> 
>>> df.apply(lambda row: print(row[0]), axis=1)
{'this', 'is', 'good'}
{'not', 'this', 'is', 'good'}
0    None
1    None
dtype: object
>>> 
>>> df.apply(lambda row: print(type(row[0])), axis=1)
<class 'set'>
<class 'set'>
0    None
1    None
dtype: object
>>> df.apply(lambda row: print(type(row[0]), df1.loc[row.name]), axis=1)
<class 'set'> set    {this, is}
Name: 0, dtype: object
<class 'set'> set    {good}
Name: 1, dtype: object
0    None
1    None
dtype: object
>>> df.apply(lambda row: print(type(row[0]), type(df1.loc[row.name])), axis=1)
<class 'set'> <class 'pandas.core.series.Series'>
<class 'set'> <class 'pandas.core.series.Series'>
0    None
1    None
dtype: object
>>> df.apply(lambda row: print(type(row[0]), type(df1.loc[row.name][0])), axis=1)
<class 'set'> <class 'set'>
<class 'set'> <class 'set'>
0    None
1    None
dtype: object
>>>

Answer 4

类似于上面，除非您要将所有内容保留在一个数据框中

Current df:
df = pd.DataFrame({0: np.array([{'this', 'is', 'good'},{'this', 'is', 'not', 'good'}]), 1: np.array([{'this', 'is'},{'good', 'bad'}])})

Intersection of series 0 & 1
df[2] = df.apply(lambda x: x[0] & x[1], axis=1)

如何获得两个熊猫系列文本列的交集？

4 个答案:

我如何到达上述解决方案？