Question

PYTHON：我的字典是：

abc = {[
    ["buy", "contract_type"], ["apple", "product"], ["from", "o"],
    ["Alex", "trader_name"], ["heeb", "trader_name"], ["of", "o"],
    ["APPLE", "counter_party"], ["INC", "counter_party"]]}

我想要的是，基本上在字典中隐含相似键的值

new_abc = {[
    ["buy", "contract_type"], ["apple", "product"], ["from", "o"],
    ["Alex heeb", "trader_name"], ["of", "o"],
    ["APPLE INC", "counter_party"]]}

Answer 1

尝试一下：

d = {}
for a, b in abc:
    d.setdefault(b, []).append(a)
new_abc = [ [ ' '.join(d.pop(b)), b ] for a, b in abc if b in d ]

但是正如@Sayse已经指出的那样，这还将结合'o'值'from'和'of'：

[['buy', 'contract_type'], ['apple', 'product'], ['from of', 'o'],
 ['Alex heeb', 'trader_name'], ['APPLE INC', 'counter_party']]

Answer 2

尝试一下：

from itertools import groupby    

abc = [
["buy", "contract_type"], ["apple", "product"], ["from", "o"],
["Alex", "trader_name"], ["heeb", "trader_name"], ["of", "o"],
["APPLE", "counter_party"], ["INC", "counter_party"]]

new= [] 
for k, g in groupby(sorted(abc, key=lambda x:x[1]),key =lambda x:x[1]): 
    new.append(['',k]) 
    for i in g: 
        new[-1][0] += i[0]+ ' '

new将是您的预期输出：

[['buy ', 'contract_type'],
 ['APPLE INC ', 'counter_party'],
 ['from of ', 'o'],
 ['apple ', 'product'],
 ['Alex heeb ', 'trader_name']]

PYTHON：如果在字典中发现相似的值，则将值压缩

2 个答案: