ITMISS
您可以re.findall使用itertools.groupby:import re, itertools as itdef get_vals(d): r = [(a, list(b)) for a, b in it.groupby(re.findall('\w+\=|[^\s,]+', d), key=lambda x:x[-1] == '=')] return {r[i][-1][0][:-1]:', '.join(r[i+1][-1]) for i in range(0, len(r), 2)}tests = ['key1=value1, key2=value2, key3=value3', 'key1=va, lue1, key2=valu, e2, test, key3=value3']print(list(map(get_vals, tests)))输出:[{'key1': 'value1', 'key2': 'value2', 'key3': 'value3'}, {'key1': 'va, lue1', 'key2': 'valu, e2, test', 'key3': 'value3'}]
梵蒂冈之花
使用@Ajax1234 的示例,re.split()并向前看:import restr="key1=value1, key2=value2, key3=value3, key1=va, lue1, key2=valu, e2, test, key3=value3"re.split(", (?=[^ ]+=)",str)['key1=value1', 'key2=value2', 'key3=value3', 'key1=va, lue1', 'key2=valu, e2, test', 'key3=value3']