检查 Pandas 列是否包含来自另一列的值
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Check if Pandas column contains value from another column
提问by ChrisArmstrong
if df['col']='a','b','c'and df2['col']='a123','b456','d789'how do I create df2['is_contained']='a','b','no_match'where if values from df['col']are found within values from df2['col']the df['col']value is returned and if no match is found, 'no_match' is returned? Also I don't expect there to be multiple matches, but in the unlikely case there are, I'd want to return a string like 'Multiple Matches'.
如果df['col']='a','b','c'和df2['col']='a123','b456','d789'我如何创建df2['is_contained']='a','b','no_match'，如果从值，其中df['col']来自值范围内被发现df2['col']的df['col']返回值，如果没有发现匹配，“NO_MATCH”返回？此外，我不希望有多个匹配项，但在不太可能的情况下，我想返回一个像“多个匹配项”这样的字符串。
回答by hernamesbarbara
With this toy data set, we want to add a new column to df2which will contain no_matchfor the first three rows, and the last row will contain the value 'd'due to the fact that that row's colvalue (the letter 'a') appears in df1.
有了这个玩具数据集，我们想添加一个新列，df2其中将包含no_match前三行，最后一行将包含值，'d'因为该行的col值（字母'a'）出现在 df1 中。
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
df1 = pd.DataFrame({'col': ['a', 'b', 'c', 'd']})
df2 = pd.DataFrame({'col': ['a123','b456','d789', 'a']})
In other words, values from df1should be used to populate this new column in df2only when a row's df2['col']value appears somewhere in df1['col'].
换句话说，从值df1应采用填充这个新列df2只有当行的df2['col']值出现在某处df1['col']。
In [2]: df1
Out[2]:
col
0 a
1 b
2 c
3 d
In [3]: df2
Out[3]:
col
0 a123
1 b456
2 d789
3 a
If this is the right way to understand your question, then you can do this with pandas isin:
如果这是理解您的问题的正确方法，那么您可以使用 pandas 执行此操作isin：
In [4]: df2.col.isin(df1.col)
Out[4]:
0 False
1 False
2 False
3 True
Name: col, dtype: bool
This evaluates to Trueonly when a value in df2.colis also in df1.col.
True仅当 in 中的值df2.col也在 in时才计算为df1.col。
Then you can use np.wherewhich is more or less the same as ifelsein R if you are familiar with R at all.
然后，如果您完全熟悉 R，则可以使用与 R 中np.where大致相同的 which ifelse。
In [5]: np.where(df2.col.isin(df1.col), df1.col, 'NO_MATCH')
Out[5]:
0 NO_MATCH
1 NO_MATCH
2 NO_MATCH
3 d
Name: col, dtype: object
For rows where a df2.colvalue appears in df1.col, the value from df1.colwill be returned for the given row index. In cases where the df2.colvalue is not a member of df1.col, the default 'NO_MATCH'value will be used.
对于df2.col出现在df1.col中的值的行，df1.col将为给定的行索引返回值 from 。如果该df2.col值不是 的成员df1.col，'NO_MATCH'则将使用默认值。
回答by Andy Hayden
In 0.13, you can use str.extract:
在 0.13 中，您可以使用str.extract：
In [11]: df1 = pd.DataFrame({'col': ['a', 'b', 'c']})
In [12]: df2 = pd.DataFrame({'col': ['d23','b456','a789']})
In [13]: df2.col.str.extract('(%s)' % '|'.join(df1.col))
Out[13]:
0 NaN
1 b
2 a
Name: col, dtype: object
