我正在尝试用另一个列表中的字符串替换一个列表中的字符串。
strlist = ['D-astroid 3-cyclone', 'DL-astroid 3-cyclone', 'DL-astroid', 'D-comment', 'satellite']
to_match = ['astroid 3-cyclone', 'D-comment', 'D-astroid']
预期输出:
str_list = ['astroid 3-cyclone', 'astroid 3-cyclone', 'D-astroid', 'D-comment', 'satellite']
并且还输出包含映射的字典
dict =
{'astroid 3-cyclone':['astroid 3-cyclone', 'astroid 3-cyclone'],
'D-comment':'D-comment',
'D-astroid':'DL-astroid',
}
我正在尝试使用 difflib
以以下方式针对测试用例实现它,
from difflib import SequenceMatcher
from pprint import pprint
def similar(a, b):
return SequenceMatcher(None, a, b).ratio()
strlist = ['D-astroid 3-cyclone', 'DL-astroid 3-cyclone', 'DL-astroid', 'D-comment']
to_match = ['astroid 3-cyclone', 'D-comment', 'D-astroid']
similarity = similar('DL-astroid', 'astroid 3-cyclone')
pprint(similarity)
基本上,如果相似度匹配高于 0.9 或 0.85,则 strlist
中的字符串必须替换为 to_match
列表中的字符串。可以使用两个 for
循环来检查 strlist
中的项目与 to_match
中的项目是否具有较高的相似度(>0.9)。我不确定这是否是一种有效的实现方式。
有什么建议吗?
编辑:我的尝试,但我不确定如何创建字典。
from difflib import SequenceMatcher
from pprint import pprint
def similar(a, to_match):
percent_similarity = [SequenceMatcher(None, a, b).ratio() for b in to_match]
max_value_index = [i for i, j in enumerate(percent_similarity) if j == max(percent_similarity)][0]
map = [to_match[max_value_index] if max(percent_similarity) > 0.9 else a][0]
return map
strlist = ['D-saturn 6-pluto', 'D-astroid 3-cyclone', 'DL-astroid 3-cyclone', 'DL-astroid', 'D-comment', 'literal']
to_match = ['saturn 6-pluto', 'pluto', 'astroid 3-cyclone', 'D-comment', 'D-astroid']
map = [similar(item, to_match) for item in strlist]
pprint(map)
最佳答案
您可以从第二个列表创建字典并将其应用到第一个列表:
strlist = ['D-astroid 3-cyclone', 'DL-astroid 3-cyclone', 'DL-astroid', 'D-comment', 'satellite']
to_match = ['astroid 3-cyclone', 'D-comment', 'D-astroid']
d1 = {i.split('-')[-1]:i for i in to_match}
result1 = [d1.get(i.split('-')[-1], i) for i in strlist]
result2 = {b:[i for i in strlist if i.endswith(a)] for a, b in d1.items()}
result2 = {a:b if len(b) != 1 else b[0] for a, b in result2.items()}
输出:
['astroid 3-cyclone', 'astroid 3-cyclone', 'D-astroid', 'D-comment', 'satellite']
{'astroid 3-cyclone': ['D-astroid 3-cyclone', 'DL-astroid 3-cyclone'], 'D-comment': 'D-comment', 'D-astroid': 'DL-astroid'}
关于python - 根据相似性替换字符串,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57296078/