Pandas

Question

这个问题是基于我问的另一个问题，我没有完全涵盖这个问题：

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']})

目标是测试 df.food:df.creature 对是否存在于 df1.consumptions.

中

上例中此测试的预期答案为：

['True', 'False', 'True', 'False', 'False', 'True', 'False', 'True', 'False']

模式是：

squirrel ate apple = True 因为松鼠和苹果是一对。 monkey likes apple = False 因为 monkey 和 apple 不是我们要找的一对。

我正在考虑构建一对值的数据帧字典，其中每个数据帧将用于 e.g.squirrel、猴子等的一个生物，然后使用 np.where 创建一个布尔表达式并执行 str.contains。

不确定这是否是最简单的方法。

Answer 1

考虑这种矢量化方法：

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)

解释：

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

Answer 2

这是我使用理解和 zip
的答案请注意，这会检查 df1

中的子字符串

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。尊重他所做的……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

简单测试

Pandas - 检查一个数据框中的字符串列是否包含来自另一个数据框中的一对字符串

Pandas - check if a string column in one dataframe contains a pair of strings from another dataframe

python

boolean-expression

string-matching