pandas -检查一个数据帧中的字符串列是否包含来自另一个数据帧的一对字符串 [英] Pandas - check if a string column in one dataframe contains a pair of strings from another dataframe
问题描述
This question is based on another question I asked, where I didn't cover the problem entirely: Pandas - check if a string column contains a pair of strings
这是问题的修改版本.
This is a modified version of the question.
我有两个数据框:
df1 = pd.DataFrame({'consumption':['squirrel ate apple', 'monkey likes apple',
'monkey banana gets', 'badger gets banana', 'giraffe eats grass', 'badger apple loves', 'elephant is huge', 'elephant eats banana tree', 'squirrel digs in grass']})
df2 = pd.DataFrame({'food':['apple', 'apple', 'banana', 'banana'],
'creature':['squirrel', 'badger', 'monkey', 'elephant']})
目标是测试df1.consumptions中是否存在df.food:df.creature对.
The goal is to test if df.food:df.creature pairs are present in df1.consumptions.
在上面的示例中，此测试的预期答案为:
The expected answer for this test in the above example would be :
['True', 'False', 'True', 'False', 'False', 'True', 'False', 'True', 'False']
模式是:
松鼠吃了苹果= True，因为松鼠和苹果是一对. 猴子喜欢apple = False，因为我们不要找猴子和苹果.
squirrel ate apple = True since squirrel and apple is a pair. monkey likes apple = False since monkey and apple is not a pair we are looking for.
我当时正在考虑构建一个具有成对值的数据帧的字典，其中每个数据帧都将用于一个生物，例如松鼠，猴子等，然后使用np.where创建一个布尔表达式并执行一个str.contains.
I was thinking of constructing a dictionary of dataframes of the pair-values where each dataframe would be for one creature for e.g.squirrel, monkey etc. and then using np.where to create a boolean expression and perform a str.contains.
不确定这是否是最简单的方法.
Not sure if that is the easiest way.
推荐答案
请考虑以下矢量化方法:
Consider this vectorized approach:
from sklearn.feature_extraction.text import CountVectorizer
vect = CountVectorizer()
X = vect.fit_transform(df1.consumption)
Y = vect.transform(df2.creature + ' ' + df2.food)
res = np.ravel(np.any((X.dot(Y.T) > 1).todense(), axis=1))
结果:
In [67]: res
Out[67]: array([ True, False, True, False, False, True, False, True, False], dtype=bool)
说明:
In [68]: pd.DataFrame(X.toarray(), columns=vect.get_feature_names())
Out[68]:
apple ate badger banana digs eats elephant gets giraffe grass huge in is likes loves monkey squirrel tree
0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0
2 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0
3 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0
4 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0
5 1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
6 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0
7 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 1
8 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 1 0
In [69]: pd.DataFrame(Y.toarray(), columns=vect.get_feature_names())
Out[69]:
apple ate badger banana digs eats elephant gets giraffe grass huge in is likes loves monkey squirrel tree
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0
3 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0
更新:
In [92]: df1['match'] = np.ravel(np.any((X.dot(Y.T) > 1).todense(), axis=1))
In [93]: df1
Out[93]:
consumption match
0 squirrel ate apple True
1 monkey likes apple False
2 monkey banana gets True
3 badger gets banana False
4 giraffe eats grass False
5 badger apple loves True
6 elephant is huge False
7 elephant eats banana tree True
8 squirrel digs in grass False
9 squirrel.eats/apple True # <----- NOTE
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