我有一个如下所示的字符串:
e = "how are you how do you how are they how"
我的预期输出是:
out = "how1 are you how2 do you how3 are they how4"
我正在尝试以下方式:
def givs(y,x):
tt = []
va = [i+1 for i in list(range(y.count(x)))]
for i in va:
tt.append(x+str(i))
return tt
ls = givs(e, 'how')
ls = ['how1', 'how2', 'how3', 'how4']
fg = []
for i in e.split(' '):
fg.append(i)
fg = ['how', 'are', 'you', 'how', 'do', 'you', 'how', 'are', 'they', 'how']
对于“fg”中“how”的每个实例，我想替换为“ls”中的项目，最后使用 join 函数来获取所需的输出。
expected_output = ['how1', 'are', 'you', 'how2', 'do', 'you', 'how3', 'are',
'they', 'how4']
这样我就可以通过以下方式加入项目:
' '.join(expected_output)
获取:
out = "how1 are you how2 do you how3 are they how4"
最佳答案
您可以使用itertools.count :
from itertools import count
counter = count(1)
e = "how are you how do you how are they how"
result = ' '.join([w if w != "how" else w + str(next(counter)) for w in e.split()])
print(result)
输出
how1 are you how2 do you how3 are they how4
关于python - 将子字符串的多个实例替换为列表中的项目，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/53874578/