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[英]Check if a column of strings from one dataframe contains a substring from a column in another dataframe, and output its mapped data
[英]Pandas - check if a string column in one dataframe contains a pair of strings from another dataframe
这个问题是基于我问的另一个问题,我没有完全解决这个问题: Pandas-检查字符串列是否包含一对字符串
这是问题的修改版本。
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']})
目的是测试df1.consumptions中是否存在df.food:df.creature对。
在上面的示例中,此测试的预期答案为:
['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。 猴子喜欢苹果=错误,因为猴子和苹果不是我们要寻找的一对。
我正在考虑构建一个对值的数据帧字典,其中每个数据帧将用于一个生物,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。
不知道这是否是最简单的方法。
考虑这种向量化方法:
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
这是我使用理解和zip
答案
注意,这会检查df1
子字符串
c = df1.consumption.values.tolist()
f = df2.food.values.tolist()
a = df2.creature.values.tolist()
check = np.array([[fd in cs and cr in cs for fd, cr in zip(f, a)] for cs in c])
check.any(1)
array([ True, False, True, False, False, True, False, True, False], dtype=bool)
这是@MaxU所做的pandas
版本。 尊重他的所作所为...太棒了!
X = df1.consumption.str.get_dummies(' ')
Y = (df2.creature + ' ' + df2.food).str.get_dummies(' ') \
.reindex_axis(X.columns, 1, fill_value=0)
# This is where you can see which rows from `df2` (columns)
# matched with which rows from `df1` (rows)
XY = X.dot(Y.T)
print(XY)
0 1 2 3
0 2 1 0 0
1 1 1 1 0
2 0 0 2 1
3 0 1 1 1
4 0 0 0 0
5 1 2 0 0
6 0 0 0 1
7 0 0 1 2
8 1 0 0 0
# return the desired `True`s and `False`s
XY.gt(1).any(1)
0 True
1 False
2 True
3 False
4 False
5 True
6 False
7 True
8 False
dtype: bool
幼稚的测试
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