a = input()  # you don't need to wrap these in str() since in python3 input always returns a string
b = input()
output = list()

for i in a.split(' '):  # split input a by spaces
    output.append(b[:len(i)])  # split input b
    b = b[len(i):]  # update b

print(output)

输出：

['aaab', 'bbb']

你可以这样做：

a = input()
b = input()

splitted_b = []
idx = 0
for word in a.split():
    w_len = len(word)
    splitted_b.append(b[idx:idx+w_len])
    idx += w_len

print(splitted_b)

其思想是从b中取出a上每个单词长度的连续子串。

您可以不使用索引，而是迭代a的每个字符。如果字符不是空格，请将b的下一个字符添加到b_word。如果是空格，请将b_word添加到b_array

b_iter = iter(b) # Create an iterator from b so we can get the next character when needed

b_word = []
b_array = []

for char in a:
    # If char is a space, and b_word isn't empty, append it to the result
    if char == " " and b_word:
        b_array.append("".join(b_word))
        b_word = []
    else:
        b_word.append(next(b_iter)) # Append the next character from b to b_word

if b_word: # If anything left over in b_word, append it to the result
    b_array.append("".join(b_word))

得到b_array = ['aaab', 'bbb']
注意，我将b_word更改为每次添加字符时.append都会使用的 list，这样可以防止每次添加字符时重新创建整个字符串，然后在添加到b_array之前使用"".join(b_word)连接所有字符。

所以为了适应输入中任意数量的空格，它变得有点复杂，因为字母的索引会随着每个空格的增加而改变。所以为了收集字符串中的所有空格，我创建了这个循环，它将考虑多个空格，并随着初始单词中的每个新空格改变索引。

indexs = []
new = ''
for i in range(len(a)):
    if len(indexs) > 0:
        if a[i] == ' ':
            indexs.append(i-len(indexs))
    else:
        if a[i] == ' ':
            indexs.append(i)

然后我们简单地将它们连接在一起，创建一个在预定索引处包含空格的新字符串。

for i in range(len(b)):
    if i in indexs:
        print(i)
        new += " "
        new += b[i]
    else:
        new += b[i]
print(new)

希望这个有用。

代码

sone = input()                 
stwo = 'zzzzzxxxyyyyy'
nwz = []
wrd = ''
cnt = 0
idx = 0

spc = sone.split(' ')      #split by whitespace
a = [len(i) for i in spc]  #list word lengths w/out ws

for i in stwo:
  if cnt == a[idx]:        #if current iter eq. word length w/out ws
    nwz.append(wrd)        #append the word
    wrd = ''               #clear old word
    wrd = wrd + i          #start new word
    idx = idx + 1            
    cnt = 0
    
  else:
    wrd = wrd + i          #building word
  cnt = cnt + 1
    
nwz.append(wrd)            #append remaining word
print(nwz)

结果

>'split and match'
['zzzzz', 'xxx', 'yyyyy']

考虑基于迭代器和itertools.islice方法的解决方案：

import itertools

def split_by_space(s1, s2):
    chunks = s1.split()
    it = iter(s2)   # second string as iterator
    return [''.join(itertools.islice(it, len(c))) for c in chunks]

print(split_by_space('head eat or', 'aaaabbbcc'))  # ['aaaa', 'bbb', 'cc']