Python 用列表中的项目替换子字符串的多个实例

我有一个字符串，如下所示：

e = "how are you how do you how are they how"

我的预期产出是：

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

我正在尝试以下方法：

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”中的每个“how”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']

以便我可以通过以下方式加入项目：

' '.join(expected_output)

要获得：

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

您可以使用：

输出

how1 are you how2 do you how3 are they how4

---

不需要使代码复杂，只需添加一个计数器并将其添加到每个“how”中即可。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)

---

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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)