如何在分隔符处拆分字符串，除非该分隔符后跟某个模式？

Question

我有一个Python字符串

string = aaa1bbb1ccc1ddd

我希望像这样分开它

re.split('[split at all occurrences of "1", unless the 1 is followed by a c]', string)

所以结果是

['aaa', 'bbb1ccc', 'ddd']

我该怎么做？

Answer 1

对正则表达式和re模块使用负向前瞻：

>>> string = 'aaa1bbb1ccc1ddd'
>>> import re
>>> re.split(r"1(?!c)", string)
['aaa', 'bbb1ccc', 'ddd']

Answer 2

def split_by_delim_except(s, delim, bar):
    escape = '\b'
    find = delim + bar
    return map(lambda s: s.replace(escape, find),
               s.replace(find, escape).split(delim))

split_by_delim_except('aaa1bbb1ccc1ddd', '1', 'c')

Answer 3

虽然不如正则表达式那么漂亮，但我的以下代码返回相同的结果：

string = 'aaa1bbb1ccc1ddd'

在＆＃39; 1＆＃39;

的所有实例中拆分字符串

p1 = string.split('1')

创建一个新的空列表，以便我们可以将所需的项目附加到

new_result = []

count = 0
for j in p1:

    if j.startswith('c'):

        # This removes the previous element from the list and stores it in a variable.
        prev_element = new_result.pop(count-1)

        prev_one_plus_j = prev_element + '1' + j

        new_result.append(prev_one_plus_j)

    else:
        new_result.append(j)

    count += 1

print (new_result)

输出：

[＆＃39; aaa＆＃39;，＆＃39; bbb1ccc＆＃39;，＆＃39; ddd＆＃39;]