如何合并两个具有多个键值对的字典

Question

我有两个字典，如下所示。 我使用的是Python 2.7。

entries_per_day = [ {"time": "October 1", "entries": "5" }, 
                {"time": "October 2", "entries": "3" }, 
                {"time": "October 3", "entries": "1" }, 
                {"time": "October 4", "entries": "0" }, 
                {"time": "October 5", "entries": "23" }]

views_per_day = [ {"time": "October 1", "views": "9" }, 
              {"time": "October 2", "views": "3" }, 
              {"time": "October 3", "views": "5" }, 
              {"time": "October 4", "views": "6" }, 
              {"time": "October 5", "views": "32" }]   

如何将两个字典合并为第三个字典，以使输出看起来像这样：

area_chart_data = [ {"time": "October 1", "entries": "5", "views": "9" }, 
                {"time": "October 2", "entries": "3", "views": "3" }, 
                {"time": "October 3", "entries": "1", "views": "5" }, 
                {"time": "October 4", "entries": "0", "views": "6" }, 
                {"time": "October 5", "entries": "23", "views": "32" }]

我希望“条目”和“视图”键值对与它们最初使用的日期位于同一数据段中。

Answer 1

由于字典条目似乎匹配，因此只需zip两个列表并用第二个字典更新一个字典，然后插入列表中。

area_chart_data = []

for e,v in zip(entries_per_day,views_per_day):
    e.update(v)
    area_chart_data.append(e)

print(area_chart_data)

结果：

[{'views': '9', 'time': 'October 1', 'entries': '5'}, {'views': '3', 'time': 'October 2', 'entries': '3'}, {'views': '5', 'time': 'October 3', 'entries': '1'}, {'views': '6', 'time': 'October 4', 'entries': '0'}, {'views': '32', 'time': 'October 5', 'entries': '23'}]

它更改了第一个列表。 如果您不希望这样做，则必须在更新之前执行e = e.copy()

编辑：如本问答中所述，使用“ dict加法”的单线：

area_chart_data = [dict(e, **v) for e,v in zip(entries_per_day,views_per_day)]

Answer 2

以最简单的形式，您可以遍历一本字典并在第二本字典中搜索相同的键。 找到后，将第一个词典entry_per_day复制到新字典，这样您的新词典将包含键“ time”，“ entries”及其值。 然后用键“ view”更新新字典，它的值来自第二个字典views_per_day 。 现在，将其附加到列表area_chart_data

>>> area_chart_data = []
>>> for d in entries_per_day:
...     for f in views_per_day:
...         if d["time"] == f["time"] :
...             m = dict(d)
...             m["views"] = f["views"]
...             area_chart_data.append(m)

结果：

>>> area_chart_data
[{'time': 'October 1', 'entries': '5', 'views': '9'}, 
 {'time': 'October 2', 'entries': '3', 'views': '3'}, 
 {'time': 'October 3', 'entries': '1', 'views': '5'}, 
 {'time': 'October 4', 'entries': '0', 'views': '6'}, 
 {'time': 'October 5', 'entries': '23', 'views': '32'}]

Answer 3

使用dict.update方法的“原始”解决方案：

area_chart_data = []
for entry in entries_per_day:
    for view in views_per_day:
        if entry['time'] == view['time']:
            d = entry.copy()
            d.update(view)
            area_chart_data.append(d)

print area_chart_data

输出：

[{'time': 'October 1', 'views': '9', 'entries': '5'}, {'time': 'October 2', 'views': '3', 'entries': '3'}, {'time': 'October 3', 'views': '5', 'entries': '1'}, {'time': 'October 4', 'views': '6', 'entries': '0'}, {'time': 'October 5', 'views': '32', 'entries': '23'}]

---

您还可以使用单个列表理解：

area_chart_data = [dict(entry, **view) for entry in entries_per_day 
                   for view in views_per_day if entry['time'] == view['time']]

Answer 4

只需尝试使用zip并用dict-2更新dict-1

lst1 = [ {"time": "October 1", "entries": "5" }, 
         {"time": "October 2", "entries": "3" }, 
       ]

lst2 = [ {"time": "October 1", "views": "9" }, 
         {"time": "October 2", "views": "3" }, ]


for x,y in zip(lst1,lst2):
    x.update(y)

print lst1

输出：

[{'views': '9', 'entries': '5', 'time': 'October 1'}, {'views': '3', 'entries': '3', 'time': 'October 2'}]