Question

我有一个熊猫数据框A，列keywords为：-

 keywords
 ['loans','mercedez','bugatti','a4']
 ['trump','usa','election','president']
 ['galaxy','7s','canon','macbook']
 ['beiber','spiderman','marvels','ironmen']
 .........................................
 .........................................
 .........................................

我还有另一个熊猫数据框B，其中列category和words是逗号分隔的字符串，如：-

category              words
audi                  audi a4,audi a6
bugatti               bugatti veyron, bugatti chiron
mercedez              mercedez s-class, mercedez e-class
dslr                  canon, nikon
apple                 iphone 7s,iphone 6s,iphone 5
finance               sales,loans,sales price
politics              donald trump, election, votes
entertainment         spiderman,captain america, ironmen
music                 justin beiber, rihana,drake
........              ..............
.........             .........

所有我想将dataframe A列keywords与dataframe B列words映射并分配一个对应的category。 keywords列的映射应与列word的字符串中的每个单词匹配。例如：-关键字a4应该与列audi a4中的字符串words中的两个单词匹配。预期结果将是：-

  keywords                                       matched_category
  ['loans','mercedez','bugatti','a4']            ['finance','mercedez','mercedez','bugatti','bugatti','audi']                                    
  ['trump','usa','election','president']         ['politics','politics']                                           
  ['galaxy','7s','canon','macbook']              ['apple','dslr']
  ['beiber','spiderman','marvels','ironmen']     ['music','entertaiment','entertainment','entertainment']

Answer 1

一种方法是使用pandas.transform：

import pandas as pd

A = pd.DataFrame({'keywords': [['loans','mercedez','bugatti','a4'],
                           ['trump','usa','election','president']]})
B = pd.DataFrame({'category': ['audi', 'finance'],
                  'words': ['audi a4,audi a6', 'sales,loans,sales price']})

def match_category_to_keywords(kws):
    ret = []
    for kw in kws:
        m = B['words'].transform(lambda words: any([kw in w for w in words.split(',')]))
        ret.extend(B['category'].loc[m].tolist())
    return pd.np.unique(ret)

A['matched_category'] = A['keywords'].transform(lambda kws: match_category_to_keywords(kws))
print(A)

输出：

                            keywords matched_category
0     [loans, mercedez, bugatti, a4]  [audi, finance]
1  [trump, usa, election, president]               []

Answer 2

希望您可以使用：

from operator import itemgetter
from itertools import groupby

d = [['4027221', 'MX', '0.4', 3], 
     ['4027221', 'MX', '30', 1], 
     ['4027222', 'MX', '0.4', 3], 
     ['4027222', 'MX', '30', 1]]


d.sort()
d = [min(g, key=lambda s: s[-2]) for _, g in groupby(d, key=lambda s: s[:-2])]
[['4027221', 'MX', '0.4', 3], ['4027222', 'MX', '0.4', 3]]

#create dictionary by split comma and whitespaces
d = df2.set_index('category')['words'].str.split(',\s*|\s+').to_dict()
#flatten lists to dictionary
d1 = {k: oldk for oldk, oldv in d.items() for k in oldv}
print (d1)
{'audi': 'audi', 'a4': 'audi', 'a6': 'audi', 'bugatti': 'bugatti', 
 'veyron': 'bugatti', 'chiron': 'bugatti', 'mercedez': 'mercedez', 
 's-class': 'mercedez', 'e-class': 'mercedez', 'canon': 'dslr', 
 'nikon': 'dslr', 'iphone': 'apple', '7s': 'apple', '6s': 'apple',
 '5': 'apple', 'sales': 'finance', 'loans': 'finance', 'price': 'finance', 
 'donald': 'politics', 'trump': 'politics', 'election': 'politics', 
 'votes': 'politics', 'spiderman': 'entertainment', 'captain': 'entertainment',
 'america': 'entertainment', 'ironmen': 'entertainment', 'justin': 'music', 
 'beiber': 'music', 'rihana': 'music', 'drake': 'music'}

将pandas列的元素与另一个pandas数据框的列匹配

2 个答案: