用列表中的项替换子字符串的多个实例

Question

我有一个如下字符串：

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，我想用'ls'中的项替换，最后使用join函数来获得所需的输出。

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)

要得到：

out = "how1 are you how2 do you how3 are they how4"

Answer 1

你可以使用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)

产量

how1 are you how2 do you how3 are they how4

Answer 2

没有必要让你的代码变得复杂，只需添加一个计数器并将其添加到每个“如何”。 最后制作新的字符串。

e = "how are you how do you how are they how"
e_ok = ""
count = 1
for i in e.split():
    if i == "how":
        i = i+str(count)
        count += 1
    e_ok += i + " "
print(e_ok)