如何将 pandas isin 用于多列
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how to use pandas isin for multiple columns
提问by Jun Jang
I want to find the values of col1and col2where the col1and col2of the first dataframe are both in the second dataframe.
我想找到的值col1，并col2在col1与col2第一个数据帧的都是在第二个数据帧。
These rows should be in the result dataframe:
这些行应该在结果数据框中：
pizza, boy
pizza, girl
ice cream, boy
披萨，男孩
披萨，女孩
冰淇淋，男孩
because all three rows are in the first and second dataframes.
因为所有三行都在第一个和第二个数据帧中。
How do I possibly accomplish this? I was thinking of using isin, but I am not sure how to use it when I have to consider more than one column.
我怎么可能做到这一点？我正在考虑使用isin，但是当我必须考虑不止一列时，我不确定如何使用它。
回答by unutbu
Perform an inner mergeon col1and col2:
在和上执行内部合并：col1col2
import pandas as pd
df1 = pd.DataFrame({'col1': ['pizza', 'hamburger', 'hamburger', 'pizza', 'ice cream'], 'col2': ['boy', 'boy', 'girl', 'girl', 'boy']}, index=range(1,6))
df2 = pd.DataFrame({'col1': ['pizza', 'pizza', 'chicken', 'cake', 'cake', 'chicken', 'ice cream'], 'col2': ['boy', 'girl', 'girl', 'boy', 'girl', 'boy', 'boy']}, index=range(10,17))
print(pd.merge(df2.reset_index(), df1, how='inner').set_index('index'))
yields
产量
col1 col2
index
10 pizza boy
11 pizza girl
16 ice cream boy
The purpose of the reset_indexand set_indexcalls are to preserve df2's index as in the desired result you posted. If the index is not important, then
reset_index和set_index调用的目的是在df2您发布的所需结果中保留的索引。如果索引不重要，那么
pd.merge(df2, df1, how='inner')
# col1 col2
# 0 pizza boy
# 1 pizza girl
# 2 ice cream boy
would suffice.
就足够了。
Alternatively, you could construct MultiIndexsout of the col1and col2columns, and then call the MultiIndex.isinmethod:
或者，您可以从和列中构造MultiIndexs，然后调用方法：col1col2MultiIndex.isin
index1 = pd.MultiIndex.from_arrays([df1[col] for col in ['col1', 'col2']])
index2 = pd.MultiIndex.from_arrays([df2[col] for col in ['col1', 'col2']])
print(df2.loc[index2.isin(index1)])
yields
产量
col1 col2
10 pizza boy
11 pizza girl
16 ice cream boy
回答by Ningrong Ye
Thank you unutbu! Here is a little update.
谢谢你！这是一个小更新。
import pandas as pd
df1 = pd.DataFrame({'col1': ['pizza', 'hamburger', 'hamburger', 'pizza', 'ice cream'], 'col2': ['boy', 'boy', 'girl', 'girl', 'boy']}, index=range(1,6))
df2 = pd.DataFrame({'col1': ['pizza', 'pizza', 'chicken', 'cake', 'cake', 'chicken', 'ice cream'], 'col2': ['boy', 'girl', 'girl', 'boy', 'girl', 'boy', 'boy']}, index=range(10,17))
df1[df1.set_index(['col1','col2']).index.isin(df2.set_index(['col1','col2']).index)]
return:
返回：
col1 col2
1 pizza boy
4 pizza girl
5 ice cream boy
回答by u9628793
If somehow you must stick to isinor the negate version ~isin.
You may first create a new column, with the concatenation of col1, col2. Then use isinto filter your data. Here is the code:
如果不知何故你必须坚持isin或否定版本~isin。您可以先创建一个新列，并连接col1, col2。然后用于isin过滤您的数据。这是代码：
import pandas as pd
df1 = pd.DataFrame({'col1': ['pizza', 'hamburger', 'hamburger', 'pizza', 'ice cream'], 'col2': ['boy', 'boy', 'girl', 'girl', 'boy']}, index=range(1,6))
df2 = pd.DataFrame({'col1': ['pizza', 'pizza', 'chicken', 'cake', 'cake', 'chicken', 'ice cream'], 'col2': ['boy', 'girl', 'girl', 'boy', 'girl', 'boy', 'boy']}, index=range(10,17))
df1['indicator'] = df1['col1'].str.cat(df1['col2'])
df2['indicator'] = df2['col1'].str.cat(df2['col2'])
df2.loc[df2['indicator'].isin(df1['indicator'])].drop(columns=['indicator'])
which gives
这使
col1 col2
10 pizza boy
11 pizza girl
16 ice cream boy
