Question

如果公共密钥具有相同的值，我正在尝试组合n个表（包含dicts的列表）。因此我使用* arg和另外两个表。我的代码工作正常，有两个表但是由于某些未知原因* args什么都不返回。我的代码：

def join(tbl1, tbl2, *args):
    _dic = {}
    _list = []
    dictsOutOfArgs = []
    for d in args:
        dictsOutOfArgs.append(d)
    for t1, t2 , arg in zip(tbl1, tbl2, dictsOutOfArgs):
        for i , j , z in zip(t1, t2, arg):
            if i == j == z and t1.get(i) == t2.get(j) == arg.get(z):
                _dic = t1.copy()
                _dic.update(t2)
                _dic.update(arg)
                _list.append(_dic)

    return _list

if __name__ == '__main__':
  import pprint
  print(test_answer())
  print(test_answer1())
  series = [
  {'s_id': 'bb', 'title': 'Breaking Bad'},
  {'s_id': 'bcs', 'title': 'Better Call Saul'}]
  characters = [
  {'c_id': 'ww', 'family_name': 'White', 'given_name': 'Walter'},
  {'c_id': 'sw', 'family_name': 'White', 'given_name': 'Skyler'},
  {'c_id': 'sg', 'family_name': 'Goodman', 'given_name': 'Saul'}]
  series_characters = [
  {'c_id': 'ww', 's_id': 'bb'},
  {'c_id': 'sw', 's_id': 'bb'},
  {'c_id': 'sg', 's_id': 'bb'},
  {'c_id': 'sg', 's_id': 'bcs'}]
  x = join(series_characters, series, characters)

  pprint.pprint(x)

预期产出：

[{'c_id': 'ww',
    'family_name': 'White',
    'given_name': 'Walter',
    's_id': 'bb',
    'title': 'Breaking Bad'},
    {'c_id': 'sw',
    'family_name': 'White',
    'given_name': 'Skyler',
    's_id': 'bb',
    'title': 'Breaking Bad'},
    {'c_id': 'sg',
    'family_name': 'Goodman',
    'given_name': 'Saul',
    's_id': 'bb',
    'title': 'Breaking Bad'},
    {'c_id': 'sg',
    'family_name': 'Goodman',
    'given_name': 'Saul',
    's_id': 'bcs',
    'title': 'Better Call Saul'}]

返回是：[] 我认为原因是argss没有任何值，所以if条件返回false。

任何想法？

Answer 1

也许这样的事情（注意它对表的顺序有点挑剔）：

def join(tbl1, *args):
    ret_list = []
    for obj in tbl1:
        consolidated = obj.copy()
        for join_tbl in args:
            keys = consolidated.viewkeys() & join_tbl[0].viewkeys()
            match_obj = next(o for o in join_tbl if all(consolidated.get(k) == o.get(k) for k in keys))
            if match_obj:
                consolidated.update(match_obj)
        ret_list.append(consolidated)
    return ret_list

join(series_characters, series, characters)返回

{'c_id': 'ww', 'given_name': 'Walter', 'title': 'Breaking Bad', 's_id': 'bb', 'family_name': 'White'}
{'c_id': 'sw', 'given_name': 'Skyler', 'title': 'Breaking Bad', 's_id': 'bb', 'family_name': 'White'}
{'c_id': 'sg', 'given_name': 'Saul', 'title': 'Breaking Bad', 's_id': 'bb', 'family_name': 'Goodman'}
{'c_id': 'sg', 'given_name': 'Saul', 'title': 'Better Call Saul', 's_id': 'bcs', 'family_name': 'Goodman'}

将n个词组合在一个词典列表中

1 个答案: