我有两个要基于列合并的DataFrame。但是,由于其他拼写方式,空格数量不同,不存在变音符,只要它们彼此相似,我希望能够合并。
任何相似性算法都可以使用(soundex,Levenshtein,difflib)。
假设一个DataFrame具有以下数据:
df1 = DataFrame([[1],[2],[3],[4],[5]], index=['one','two','three','four','five'], columns=['number']) number one 1 two 2 three 3 four 4 five 5 df2 = DataFrame([['a'],['b'],['c'],['d'],['e']], index=['one','too','three','fours','five'], columns=['letter']) letter one a too b three c fours d five e
然后我想得到结果DataFrame
number letter one 1 a two 2 b three 3 c four 4 d five 5 e
类似@locojay建议,你可以申请difflib的get_close_matches到df2的指标,然后应用join:
@locojay
difflib
get_close_matches
join
In [23]: import difflib In [24]: difflib.get_close_matches Out[24]: <function difflib.get_close_matches> In [25]: df2.index = df2.index.map(lambda x: difflib.get_close_matches(x, df1.index)[0]) In [26]: df2 Out[26]: letter one a two b three c four d five e In [31]: df1.join(df2) Out[31]: number letter one 1 a two 2 b three 3 c four 4 d five 5 e
如果这些是列,则可以按照相同的方式应用于该列,然后merge:
df1 = DataFrame([[1,'one'],[2,'two'],[3,'three'],[4,'four'],[5,'five']], columns=['number', 'name']) df2 = DataFrame([['a','one'],['b','too'],['c','three'],['d','fours'],['e','five']], columns=['letter', 'name']) df2['name'] = df2['name'].apply(lambda x: difflib.get_close_matches(x, df1['name'])[0]) df1.merge(df2)