2023年2月16日 07:18:36go评论86阅读模式

英文:

List of dictionaries: Sort elements and insert sum

问题

以下是已经翻译好的代码部分：

我有一个类似这样无序的字典列表:
[{ 'A': 'LabelA', 'B': 'DescriptionA', 'C': '1', 'D': '5', 'sum': '0' },
 { 'A': 'LabelB', 'B': 'DescriptionB', 'C': '2', 'D': '6', 'sum': '0' },
 { 'A': 'LabelB', 'B': 'DescriptionB', 'C': '3', 'D': '7', 'sum': '0' },
 { 'A': 'LabelA', 'B': 'DescriptionA', 'C': '4', 'D': '8', 'sum': '0' }]
我想首先对它进行排序，然后在前面添加一个总和，像这样:
[{ 'A': 'LabelA', 'B': 'DescriptionA', 'C': '5', 'D': '13', 'sum': '1' },
 { 'A': 'LabelA', 'B': 'DescriptionA', 'C': '1', 'D': '5', 'sum': '0' },
 { 'A': 'LabelA', 'B': 'DescriptionA', 'C': '4', 'D': '8', 'sum': '0' },
 { 'A': 'LabelB', 'B': 'DescriptionB', 'C': '5', 'D': '13', 'sum': '1' },
 { 'A': 'LabelB', 'B': 'DescriptionB', 'C': '2', 'D': '6', 'sum': '0' },
 { 'A': 'LabelB', 'B': 'DescriptionB', 'C': '3', 'D': '7', 'sum': '0' }]

希望这可以帮助您理解如何对列表进行排序并添加总和。

英文:

I have a list of dictionaries in an unstructured order like this:

[{&#39;A&#39;: &#39;LabelA&#39;, &#39;B&#39;: &#39;DescriptionA&#39;, &#39;C&#39;: &#39;1&#39;, &#39;D&#39;: &#39;5&#39;, &#39;sum&#39;: &#39;0&#39;},
 {&#39;A&#39;: &#39;LabelB&#39;, &#39;B&#39;: &#39;DescriptionB&#39;, &#39;C&#39;: &#39;2&#39;, &#39;D&#39;: &#39;6&#39;, &#39;sum&#39;: &#39;0&#39;},
 {&#39;A&#39;: &#39;LabelB&#39;, &#39;B&#39;: &#39;DescriptionB&#39;, &#39;C&#39;: &#39;3&#39;, &#39;D&#39;: &#39;7&#39;, &#39;sum&#39;: &#39;0&#39;},
 {&#39;A&#39;: &#39;LabelA&#39;, &#39;B&#39;: &#39;DescriptionA&#39;, &#39;C&#39;: &#39;4&#39;, &#39;D&#39;: &#39;8&#39;, &#39;sum&#39;: &#39;0&#39;}
]

and I want to sort it first and add a sum in front of it like this:

[{&#39;A&#39;: &#39;LabelA&#39;, &#39;B&#39;: &#39;DescriptionA&#39;, &#39;C&#39;: &#39;5&#39;, &#39;D&#39;: &#39;13&#39;, &#39;sum&#39;: &#39;1&#39;},
 {&#39;A&#39;: &#39;LabelA&#39;, &#39;B&#39;: &#39;DescriptionA&#39;, &#39;C&#39;: &#39;1&#39;, &#39;D&#39;: &#39;5&#39;, &#39;sum&#39;: &#39;0&#39;},
 {&#39;A&#39;: &#39;LabelA&#39;, &#39;B&#39;: &#39;DescriptionA&#39;, &#39;C&#39;: &#39;4&#39;, &#39;D&#39;: &#39;8&#39;, &#39;sum&#39;: &#39;0&#39;},
 {&#39;A&#39;: &#39;LabelB&#39;, &#39;B&#39;: &#39;DescriptionB&#39;, &#39;C&#39;: &#39;5&#39;, &#39;D&#39;: &#39;13&#39;, &#39;sum&#39;: &#39;1&#39;},
 {&#39;A&#39;: &#39;LabelB&#39;, &#39;B&#39;: &#39;DescriptionB&#39;, &#39;C&#39;: &#39;2&#39;, &#39;D&#39;: &#39;6&#39;, &#39;sum&#39;: &#39;0&#39;},
 {&#39;A&#39;: &#39;LabelB&#39;, &#39;B&#39;: &#39;DescriptionB&#39;, &#39;C&#39;: &#39;3&#39;, &#39;D&#39;: &#39;7&#39;, &#39;sum&#39;: &#39;0&#39;}
]

I have followed multiple tutorials how to loop through the list to either sort it or to create a sum, but I don't have any clue how to a) do both and b) add the sum at the right position.

答案1

得分: 0

以下是您要的代码部分的翻译：

那么，看起来您想要按 `A` 列的值进行分组。所以，只需使用常规的分组方法：
data = [{'A': 'LabelA', 'B': 'DescriptionA', 'C': '1', 'D': '5', 'sum': '0'},
        {'A': 'LabelB', 'B': 'DescriptionB', 'C': '2', 'D': '6', 'sum': '0'},
        {'A': 'LabelB', 'B': 'DescriptionB', 'C': '3', 'D': '7', 'sum': '0'},
        {'A': 'LabelA', 'B': 'DescriptionA', 'C': '4', 'D': '8', 'sum': '0'}]
grouper = {}
for row in data:
    grouper.setdefault(row['A'], []).append(row)

这将给您：

from pprint import pprint
pprint(grouper)
{'LabelA': [{'A': 'LabelA',
             'B': 'DescriptionA',
             'C': '1',
             'D': '5',
             'sum': '0'},
            {'A': 'LabelA',
             'B': 'DescriptionA',
             'C': '4',
             'D': '8',
             'sum': '0'}],
 'LabelB': [{'A': 'LabelB',
             'B': 'DescriptionB',
             'C': '2',
             'D': '6',
             'sum': '0'},
            {'A': 'LabelB',
             'B': 'DescriptionB',
             'C': '3',
             'D': '7',
             'sum': '0'}]}

然后，遍历分组以填充最终列表。首先计算“sum”，追加该值，然后追加分组。

首先，我会定义一个小的辅助函数：

def _add_rows(row1, row2):
    row1['C'] = str(int(row1['C']) + int(row2['C']))
    row1['D'] = str(int(row1['D']) + int(row2['D']))

然后，记得将 sum 设置为 '1'。所以类似这样：

final = []
for (first, *rest) in grouper.values():
    row_sum = first.copy()
    row_sum['sum'] = '1'
    for row in rest:
        _add_rows(row_sum, row)
    final.append(row_sum)
    final.append(first)
    final.extend(rest)

这将得到：

[{'A': 'LabelA', 'B': 'DescriptionA', 'C': '5', 'D': '13', 'sum': '1'},
 {'A': 'LabelA', 'B': 'DescriptionA', 'C': '1', 'D': '5', 'sum': '0'},
 {'A': 'LabelA', 'B': 'DescriptionA', 'C': '4', 'D': '8', 'sum': '0'},
 {'A': 'LabelB', 'B': 'DescriptionB', 'C': '5', 'D': '13', 'sum': '1'},
 {'A': 'LabelB', 'B': 'DescriptionB', 'C': '2', 'D': '6', 'sum': '0'},
 {'A': 'LabelB', 'B': 'DescriptionB', 'C': '3', 'D': '7', 'sum': '0'}]

请注意，我将键 C、D 和 sum 保持为字符串，但它们实际上应该是 int 对象。

英文:

So, it looks like you want to group the values by 'A'. So just use the regular grouping idiom:

&gt;&gt;&gt; data = [{&#39;A&#39;: &#39;LabelA&#39;, &#39;B&#39;: &#39;DescriptionA&#39;, &#39;C&#39;: &#39;1&#39;, &#39;D&#39;: &#39;5&#39;, &#39;sum&#39;: &#39;0&#39;},
...  {&#39;A&#39;: &#39;LabelB&#39;, &#39;B&#39;: &#39;DescriptionB&#39;, &#39;C&#39;: &#39;2&#39;, &#39;D&#39;: &#39;6&#39;, &#39;sum&#39;: &#39;0&#39;},
...  {&#39;A&#39;: &#39;LabelB&#39;, &#39;B&#39;: &#39;DescriptionB&#39;, &#39;C&#39;: &#39;3&#39;, &#39;D&#39;: &#39;7&#39;, &#39;sum&#39;: &#39;0&#39;},
...  {&#39;A&#39;: &#39;LabelA&#39;, &#39;B&#39;: &#39;DescriptionA&#39;, &#39;C&#39;: &#39;4&#39;, &#39;D&#39;: &#39;8&#39;, &#39;sum&#39;: &#39;0&#39;}
... ]
&gt;&gt;&gt; grouper = {}
&gt;&gt;&gt; for row in data:
...     grouper.setdefault(row[&#39;A&#39;],[]).append(row)
...

This gives you:

&gt;&gt;&gt; from pprint import pprint
&gt;&gt;&gt; pprint(grouper)
{&#39;LabelA&#39;: [{&#39;A&#39;: &#39;LabelA&#39;,
             &#39;B&#39;: &#39;DescriptionA&#39;,
             &#39;C&#39;: &#39;1&#39;,
             &#39;D&#39;: &#39;5&#39;,
             &#39;sum&#39;: &#39;0&#39;},
            {&#39;A&#39;: &#39;LabelA&#39;,
             &#39;B&#39;: &#39;DescriptionA&#39;,
             &#39;C&#39;: &#39;4&#39;,
             &#39;D&#39;: &#39;8&#39;,
             &#39;sum&#39;: &#39;0&#39;}],
 &#39;LabelB&#39;: [{&#39;A&#39;: &#39;LabelB&#39;,
             &#39;B&#39;: &#39;DescriptionB&#39;,
             &#39;C&#39;: &#39;2&#39;,
             &#39;D&#39;: &#39;6&#39;,
             &#39;sum&#39;: &#39;0&#39;},
            {&#39;A&#39;: &#39;LabelB&#39;,
             &#39;B&#39;: &#39;DescriptionB&#39;,
             &#39;C&#39;: &#39;3&#39;,
             &#39;D&#39;: &#39;7&#39;,
             &#39;sum&#39;: &#39;0&#39;}]}

Then, iterate over the groups to populate your final list. First calculating the "sum", appending that, then appending the group.

First, I would define a little helper function:

&gt;&gt;&gt; def _add_rows(row1, row2):
...     row1[&#39;C&#39;] = str(int(row1[&#39;C&#39;]) + int(row2[&#39;C&#39;]))
...     row1[&#39;D&#39;] = str(int(row1[&#39;D&#39;]) + int(row2[&#39;D&#39;]))
...

Then remember to set 'sum' to '1'. So something like:

&gt;&gt;&gt; final = []
&gt;&gt;&gt; for (first, *rest) in grouper.values():
...     row_sum = first.copy()
...     row_sum[&#39;sum&#39;] = &#39;1&#39;
...     for row in rest:
...         _add_rows(row_sum, row)
...     final.append(row_sum)
...     final.append(first)
...     final.extend(rest)
...
&gt;&gt;&gt; pprint(final)
[{&#39;A&#39;: &#39;LabelA&#39;, &#39;B&#39;: &#39;DescriptionA&#39;, &#39;C&#39;: &#39;5&#39;, &#39;D&#39;: &#39;13&#39;, &#39;sum&#39;: &#39;1&#39;},
 {&#39;A&#39;: &#39;LabelA&#39;, &#39;B&#39;: &#39;DescriptionA&#39;, &#39;C&#39;: &#39;1&#39;, &#39;D&#39;: &#39;5&#39;, &#39;sum&#39;: &#39;0&#39;},
 {&#39;A&#39;: &#39;LabelA&#39;, &#39;B&#39;: &#39;DescriptionA&#39;, &#39;C&#39;: &#39;4&#39;, &#39;D&#39;: &#39;8&#39;, &#39;sum&#39;: &#39;0&#39;},
 {&#39;A&#39;: &#39;LabelB&#39;, &#39;B&#39;: &#39;DescriptionB&#39;, &#39;C&#39;: &#39;5&#39;, &#39;D&#39;: &#39;13&#39;, &#39;sum&#39;: &#39;1&#39;},
 {&#39;A&#39;: &#39;LabelB&#39;, &#39;B&#39;: &#39;DescriptionB&#39;, &#39;C&#39;: &#39;2&#39;, &#39;D&#39;: &#39;6&#39;, &#39;sum&#39;: &#39;0&#39;},
 {&#39;A&#39;: &#39;LabelB&#39;, &#39;B&#39;: &#39;DescriptionB&#39;, &#39;C&#39;: &#39;3&#39;, &#39;D&#39;: &#39;7&#39;, &#39;sum&#39;: &#39;0&#39;}]

Note, I kept keys 'C', 'D', and 'sum' as strings, but they really should be int objects.

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