繁体   English   中英

熊猫-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

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

在此处输入图片说明

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM