Question

实现此解决方案的任何替代方法？当有许多键匹配时，使用str.contains（）并不是很优雅。

df = DataFrame({'A':['Cat had a nap','Dog had puppies','Did you see a Donkey','kitten got angry','puppy was cute']})
    dic = {'Cat':'Cat','kitten':'Cat','Dog':'Dog','puppy':'Dog'}

               A
0   Cat had a nap
1   Dog had puppies
2   Did you see a Donkey
3   kitten got angry
4   puppy was cute


df['Cat'] = (df['A'].astype(str).str.contains('Cat')|df['A'].astype(str).str.contains('kitten')).replace({False:0, True:1})
df['Dog'] = (df['A'].astype(str).str.contains('Dog')|df['A'].astype(str).str.contains('puppy')).replace({False:0, True:1})
df



    A                    Cat    Dog
0   Cat had a nap          1    0
1   Dog had puppies        0    1
2   Did you see a Donkey   0    0
3   kitten got angry       1    0
4   puppy was cute         0    1

Answer 1

将|用于str.contains中的正则表达式or，使用强制转换布尔值来整数astype：

df['Cat'] = df['A'].astype(str).str.contains('Cat|kitten').astype(int)
df['Dog'] = df['A'].astype(str).str.contains('Dog|puppy').astype(int)

类似：

a = df['A'].astype(str)
df['Cat'] = a.str.contains('Cat|kitten').astype(int)
df['Dog'] = a.str.contains('Dog|puppy').astype(int)

print (df)
                      A  Cat  Dog
0         Cat had a nap    1    0
1       Dog had puppies    0    1
2  Did you see a Donkey    0    0
3      kitten got angry    1    0
4        puppy was cute    0    1

使用list s字典的更动态解决方案：

dic = {'Cat':['Cat','kitten'],'Dog':['Dog','puppy']}
for k, v in dic.items():
    df[k] = df['A'].astype(str).str.contains('|'.join(v)).astype(int)
print (df)
                      A  Cat  Dog
0         Cat had a nap    1    0
1       Dog had puppies    0    1
2  Did you see a Donkey    0    0
3      kitten got angry    1    0
4        puppy was cute    0    1

匹配子串的替代解决方案

1 个答案: