Question

我想将两个多个dicts的列表组合成一个新的dicts列表，将新的dicts添加到最终列表中，并将这些视图添加到一起来查看＆＃39;遇到的价值。

a = [{'title': 'Learning How to Program', 'views': 1,'url': '/4XvR', 'slug': 'learning-how-to-program'},
     {'title': 'Mastering Programming', 'views': 3,'url': '/7XqR', 'slug': 'mastering-programming'}]

b = [{'title': 'Learning How to Program', 'views': 7,'url': '/4XvR', 'slug': 'learning-how-to-program'},
     {'title': 'Mastering Programming', 'views': 2,'url': '/7XqR', 'slug': 'mastering-programming'},
     {'title': 'Programming Fundamentals', 'views': 1,'url': '/93hB', 'slug': 'programming-fundamentals'}]

所需的输出是：

c = [{'title': 'Learning How to Program', 'views': 8,'url': '/4XvR', 'slug': 'learning-how-to-program'},
     {'title': 'Mastering Programming', 'views': 5,'url': '/7XqR', 'slug': 'mastering-programming'},
     {'title': 'Programming Fundamentals', 'views': 1,'url': '/93hB', 'slug': 'programming-fundamentals'}]

我发现Is there any pythonic way to combine two dicts (adding values for keys that appear in both)? - 但我不明白如何在我的情况下获得所需的输出，有两个多个dicts列表。

Answer 1

您需要将输入词典转换为(title: count)对，并将其用作Counter中的键和值;然后在求和之后，您可以将它们转换回旧格式：

from collections import Counter

summed = sum((Counter({elem['title']: elem['views']}) for elem in a + b), Counter())
c = [{'title': title, 'views': counts} for title, counts in summed.items()]

演示：

>>> from collections import Counter
>>> a = [{'title': 'Learning How to Program', 'views': 1},
...      {'title': 'Mastering Programming', 'views': 3}]
>>> b = [{'title': 'Learning How to Program', 'views': 7},
...      {'title': 'Mastering Programming', 'views': 2},
...      {'title': 'Programming Fundamentals', 'views': 1}]
>>> summed = sum((Counter({elem['title']: elem['views']}) for elem in a + b), Counter())
>>> summed
Counter({'Learning How to Program': 8, 'Mastering Programming': 5, 'Programming Fundamentals': 1})
>>> [{'title': title, 'views': counts} for title, counts in summed.items()]
[{'views': 8, 'title': 'Learning How to Program'}, {'views': 5, 'title': 'Mastering Programming'}, {'views': 1, 'title': 'Programming Fundamentals'}]

此处的目标是每个计数都有一个唯一标识符。如果您的词典更复杂，您需要将整个字典（减去计数）转换为唯一标识符，或者从字典中选择一个值作为该标识符。然后将每个标识符的视图计数加起来。

从您更新的示例中，该URL将是一个很好的标识符。这样您就可以收集视图计数：

per_url = {}
for entry in a + b:
    key = entry['url']
    if key not in per_url:
        per_url[key] = entry.copy()
    else:
        per_url[key]['views'] += entry['views']

c = per_url.values()  # use list(per_url.values()) on Python 3

这只是使用字典本身（或至少是遇到的第一个字典的副本）来对视图计数求和：

>>> from pprint import pprint
>>> a = [{'title': 'Learning How to Program', 'views': 1,'url': '/4XvR', 'slug': 'learning-how-to-program'},
...      {'title': 'Mastering Programming', 'views': 3,'url': '/7XqR', 'slug': 'mastering-programming'}]
>>> b = [{'title': 'Learning How to Program', 'views': 7,'url': '/4XvR', 'slug': 'learning-how-to-program'},
...      {'title': 'Mastering Programming', 'views': 2,'url': '/7XqR', 'slug': 'mastering-programming'},
...      {'title': 'Programming Fundamentals', 'views': 1,'url': '/93hB', 'slug': 'programming-fundamentals'}]
>>> per_url = {}
>>> for entry in a + b:
...     key = entry['url']
...     if key not in per_url:
...         per_url[key] = entry.copy()
...     else:
...         per_url[key]['views'] += entry['views']
... 
>>> per_url
{'/93hB': {'url': '/93hB', 'title': 'Programming Fundamentals', 'slug': 'programming-fundamentals', 'views': 1}, '/4XvR': {'url': '/4XvR', 'title': 'Learning How to Program', 'slug': 'learning-how-to-program', 'views': 8}, '/7XqR': {'url': '/7XqR', 'title': 'Mastering Programming', 'slug': 'mastering-programming', 'views': 5}}
>>> pprint(per_url.values())
[{'slug': 'programming-fundamentals',
  'title': 'Programming Fundamentals',
  'url': '/93hB',
  'views': 1},
 {'slug': 'learning-how-to-program',
  'title': 'Learning How to Program',
  'url': '/4XvR',
  'views': 8},
 {'slug': 'mastering-programming',
  'title': 'Mastering Programming',
  'url': '/7XqR',
  'views': 5}]

Answer 2

首先，您需要将输入转换为dicts，例如

b = {'Learning How to Program': 7,
     'Mastering Programming': 2,
     'Programming Fundamentals': 1}

之后，应用找到的解决方案，然后将其转换回dicts列表。

Answer 3

这是一个简单的问题。浏览所有条目，在第一次遇到条目时复制条目，并在后续遭遇中添加视图：

summary = {}    
for entry in a + b:
    key = entry['url']
    if key not in summary:
        summary[key] = entry.copy()
    else:
        summary[key]['views'] += entry['views']
c = list(summary.values())

Answer 4

它可能不是最pythonic的解决方案：

def coalesce(d1,d2):
    combined = [i for i in d1]
    for d in d2:
        found = False
        for itr in combined:          
            if itr['title'] == d['title']:
                itr['views'] += d['views']
                found = True
                break
        if not found:
             combined.append(d)
     return combined

Answer 5

非最佳，但有效：

>>> from collections import Counter
>>> from pprint import pprint
>>> a = [{'title': 'Learning How to Program', 'views': 1,'url': '/4XvR', 'slug': 'learning-how-to-program'},
...      {'title': 'Mastering Programming', 'views': 3,'url': '/7XqR', 'slug': 'mastering-programming'}]
>>> b = [{'title': 'Learning How to Program', 'views': 7,'url': '/4XvR', 'slug': 'learning-how-to-program'},
...      {'title': 'Mastering Programming', 'views': 2,'url': '/7XqR', 'slug': 'mastering-programming'},
...      {'title': 'Programming Fundamentals', 'views': 1,'url': '/93hB', 'slug': 'programming-fundamentals'}]
>>> summed = sum((Counter({x['slug']: x['views']}) for x in a+b), Counter())
>>> c = dict()
>>> _ = [c.update({x['slug']: x}) for x in a + b]
>>> _ = [c[x].update({'views': summed[x]}) for x in c.keys()]
>>> pprint(c.values())
[{'slug': 'mastering-programming',
  'title': 'Mastering Programming',
  'url': '/7XqR',
  'views': 5},
 {'slug': 'programming-fundamentals',
  'title': 'Programming Fundamentals',
  'url': '/93hB',
  'views': 1},
 {'slug': 'learning-how-to-program',
  'title': 'Learning How to Program',
  'url': '/4XvR',
  'views': 8}]

基于Martijn的Counter理念，需要更多迭代才能更新计数器值与其他属性，假设它们不会改变。

请注意，有一些＆＃34;加密＆＃34;在发电机中循环......

Answer 6

一个简单的函数，可以为任何给定数量的列表执行所需的操作：

import itertools
from collections import Counter, OrderedDict

def sum_views(*lists):
    views = Counter()
    docs = OrderedDict()  # to preserve input order
    for doc in itertools.chain(*lists):
        slug = doc['slug']
        views[slug] += doc['views']
        docs[slug] = dict(doc)   # shallow copy of original dict
        docs[slug]['views'] = views[slug]
    return docs.values()

Answer 7

假设您不想将其标题为＆＃34; title＆＃34;和＆＃34;观点＆＃34;。更专业的方式就是这样写：

  def combing(x):
     result = {}
     for i in x:
        h = i.values()
        result[h[0]] = result.get(h[0],0)+ h[1]
     return result

combing([{'item': 'item1', 'amount': 400}, {'item': 'item2', 'amount': 
300}, {'item': 'item1', 'amount': 750}])

结合两个dicts列表，将值相加

7 个答案: