Альтернативные решения для сопоставления подстрок

Question

Есть ли альтернативные методы для достижения этого решения? Использование str.contains () не очень элегантно, когда нужно сопоставить много ключей.

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

Используйте | для регулярных выражений or в str.contains с приведенным логическим значением к целому числу 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)

Аналогично:

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

Более динамичное решение со словарем list s:

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