测试 a=("ab/cd/ef", "ab/cd/ef/g", "ab/cd/") 输出 {'ab/cd/': 1}#in_set 应是一个有序的字符串集合 def remove_duplicate_string(in_set): if len(in_set) < 2: return in_set duplicate_str_dict = {} same_str_dict = {} old_str = "" same_str = "" for str in in_set: if len(same_str) > 1: if str.find(same_str) >= 0: # same_str 应该是两个字符串的交集,如果能跟same_str匹配,则直接pass continue if len(old_str) > 0: same_str = get_same_string(old_str, str) print(f"{same_str} vs {old_str} vs {str}") if len(same_str) > 1: same_str_dict[same_str] = 1 old_str = str if len(same_str_dict) < 1 or len(same_str_dict) == len(in_set): return in_set if len(same_str_dict) < len(in_set): return remove_duplicate_string(same_str_dict) return same_str_dict
October 7, 2021
python批量提取一个set中,最小相同路径
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