Question

这个问题是基于我提出的另一个问题，我没有完全解决这个问题：Pandas - check if a string column contains a pair of strings

这是问题的修改版本。

我有两个数据帧：

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.food中是否存在df.food:df.creature对。

上述示例中此测试的预期答案是：

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

模式是：

squirrel ate apple =真的，因为松鼠和苹果是一对。猴子喜欢苹果=假，因为猴子和苹果不是我们正在寻找的一对。

我正在考虑构建一对数据帧的字典，其中每个数据帧将用于例如松鼠，猴子等的一个生物，然后使用np.where来创建布尔表达式并执行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：

    const appReducer = combineReducers({
      gameSettings: ...,
      gameStatus: ...,
    });

  const store = createStore(
    appReducer,
    enhancer,
  );

结果：

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))

说明：

In [67]: res
Out[67]: array([ True, False,  True, False, False,  True, False,  True, False], dtype=bool)

<强>更新

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

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

天真的测试

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

2 个答案: