使用setdefault重构Python字典列表 [英] Re-structuring a list of Python Dicts using setdefault

问题描述

我试图通过分组"(可能不是正确的表达式，但是将其用作基于SQL的代理)基于(非唯一)值的字典来重新构建Python字典列表.我已经接近了，但是我遇到了最后的障碍，因为我无法解决如何将每个值重新分配给一个名称(即，我得到的看起来像是元组而不是字典).

I am trying to re-structure a list of Python dictionaries by 'grouping' (that's probably not the correct expression, but using it as a proxy based on SQL) the dictionaries based on a (non-unique) value. I have got close, however I'm falling at the final hurdle, in that I couldn't work out how to re-assign each value to a name (i.e. I end up with what looks like a tuple rather than a dict).

另外，我还有一个多余的列表(即，我的输出最终以[[{...}]]而不是[{......]]来显示.

Additionally I have a superfluous list (i.e. my output ends up as [[{...}]] rather than [{...}].

我在这里使用了示例:

>如何将此分组同一个月的字典列表?

这让我非常接近我想要的东西，但是我被困在最后阶段！

Which gets me quite close to what I want, however I'm stuck at the final stage!

market = [
    {'selection_id': 1099, 'value': '11', 'value_name': 'a'},
    {'selection_id': 1099, 'value': '78', 'value_name': 'p'},
    {'selection_id': 1097, 'value': '39', 'value_name': 'b'},
    {'selection_id': 1097, 'value': '52', 'value_name': 'f'},
    {'selection_id': 1098, 'value': '98', 'value_name': 'd'},
    {'selection_id': 1099, 'value': '13', 'value_name': 'y'},
    {'selection_id': 1098, 'value': '4', 'value_name': 'r'},
]

new_structure = {}
new_structure2 = []

for z in market:
        new_structure.setdefault(z['selection_id'], []).append((z['value'], z['value_name']))
        t = [{'selection_id': m, 'value_dict': n} for m, n in new_structure.items()]
new_structure2.append(t)

print(new_structure2)

输出为:

[[{'selection_id': 1099, 'value_dict': [('11', 'a'), ('78', 'p'), ('13',  
 'y')]}, {'selection_id': 1097, 'value_dict': [('39', 'b'), ('52', 'f')]},  
 {'selection_id': 1098, 'value_dict': [('98', 'd'), ('4', 'r')]}]]

哪个位置很近，但是我的目标是:

Which is very close, however what I'm aiming for is:

[{'selection_id': 1099,
  'value_dict': [{'value': '11', 'value_name': 'a'},
                 {'value': '78', 'value_name': 'p'},
                 {'value': '13', 'value_name': 'y'}]},
 {'selection_id': 1097,
  'value_dict': [{'value': '39', 'value_name': 'b'},
                 {'value': '52', 'value_name': 'f'}]},
 {'selection_id': 1098,
  'value_dict': [{'value': '98', 'value_name': 'd'},
                 {'value': '4', 'value_name': 'r'}]}]

我很感激它可能是一个非常简单的修复程序，但此刻它使我逃脱了，因此任何指导将不胜感激！

I appreciate its probably a really simple fix, but its escaping me at the moment, so any guidance would be greatly appreciated!

推荐答案

以下是一些提示:

第一件事是按 selection_id 进行排序:

The first thing is to sort by selection_id:

by_selection_id = operator.itemgetter('selection_id')
market.sort(key=by_selection_id)

然后您可以按 selection_id 分组:

for selection_id, group in itertools.groupby(market, key=by_selection_id):
    print(selection_id, list(group))

您得到:

(1097, [{'value_name': 'b', 'value': '39', 'selection_id': 1097},
        {'value_name': 'f', 'value': '52', 'selection_id': 1097}])
(1098, [{'value_name': 'd', 'value': '98', 'selection_id': 1098},
        {'value_name': 'r', 'value': '4', 'selection_id': 1098}])
(1099, [{'value_name': 'a', 'value': '11', 'selection_id': 1099},
        {'value_name': 'p', 'value': '78', 'selection_id': 1099},
        {'value_name': 'y', 'value': '13', 'selection_id': 1099}])

然后很容易建立最终列表.

Then it is easy to build the final list.

这是使用理解列表/字典的解决方案:

Here is a solution using comprehension list/dict:

new_structure = [{'selection_id': selection_id,
                  'value_dict': [{'value': item['value'],
                                  'value_name': item['value_name']} for item in group]}
                 for selection_id, group in itertools.groupby(market, key=by_selection_id)]

或在append中使用经典列表:

new_structure = []
for selection_id, group in itertools.groupby(market, key=by_selection_id):
    value_dict = [{'value': item['value'], 'value_name': item['value_name']} for item in group]
    new_structure.append({'selection_id': selection_id,
                          'value_dict': value_dict})

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