Question

我有一个这样的清单：

table=[{'device': 'a', 'iface': 'i1'}, {'device': 'a', 'iface': 'i2'}, {'device': 'a', 'iface': 'i4'}, {'device': 'b', 'iface': 'i5'}, {'device': 'a', 'iface': 'i3'}, {'device': 'b', 'iface': 'i7'}]

我想将该列表分组到这样的新列表中：

newtable=[{'device':'a','iface':['i1','i2','i4','i3']},{'device':'b','iface':['i5','i7']}]

我曾尝试使用collections模块中的defaultdict，但结果并非我的预期。请帮我。提前致谢

这里我做了什么：

from collections import defaultdict
table=[{'device': 'a', 'iface': 'i1'}, {'device': 'a', 'iface': 'i2'}, {'device': 'a', 'iface': 'i4'}, {'device': 'b', 'iface': 'i5'}, {'device': 'a', 'iface': 'i3'}, {'device': 'b', 'iface': 'i7'}]
newtable=defaultdict(list)
for row in table:
    newtable[row['device']].append(row['iface'])
newtable=list(newtable)

我从这里迷路了。对不起我的英语。

Answer 1

这是一种方法 - 只使用defaultdict可能还不够。您可能需要进一步处理它（只需确保使用更好的变量名:)）：

>>> from collections import defaultdict
>>> xx = defaultdict(list)
>>> table=[{'device': 'a', 'iface': 'i1'}, {'device': 'a', 'iface': 'i2'}, {'device': 'a', 'iface': 'i4'}, {'device': 'b', 'iface': 'i5'}, {'device': 'a', 'iface': 'i3'}, {'device': 'b', 'iface': 'i7'}]
>>> for tab in table:
...   xx[tab['device']].append(tab['iface'])
... 
>>> xx
defaultdict(<type 'list'>, {'a': ['i1', 'i2', 'i4', 'i3'], 'b': ['i5', 'i7']})
>>> yy = dict(xx)
>>> yy
{'a': ['i1', 'i2', 'i4', 'i3'], 'b': ['i5', 'i7']}

>>> new_table = []
>>> for a in yy:
...   new_table.append({'device': a, 'iface': yy[a]})
... 
>>> new_table
[{'device': 'a', 'iface': ['i1', 'i2', 'i4', 'i3']}, {'device': 'b', 'iface': ['i5', 'i7']}]

Answer 2

我选择将列表table中的项目按设备值收集到地图中，并在迭代同一设备的新项目时继续扩展iface列表：

table=[{'device': 'a', 'iface': 'i1'}, {'device': 'a', 'iface': 'i2'}, {'device': 'a', 'iface': 'i4'}, {'device': 'b', 'iface': 'i5'}, {'device': 'a', 'iface': 'i3'}, {'device': 'b', 'iface': 'i7'}]
new_table = []
tmp_map = {}
for itm in table:
    map_itm = tmp_map.get(itm['device'])
    if not map_itm:
        tmp_map[itm['device']] = {'device':itm['device'] ,'iface': [itm['iface']]}
    else:
        tmp_map[itm['device']]['iface'] = [itm['iface']]+ map_itm['iface']
print tmp_map.values()

<强>输出：

[{'device': 'a', 'iface': ['i3', 'i4', 'i2', 'i1']}, {'device': 'b', 'iface': ['i7', 'i5']}]

Answer 3

你做了一本字典。然后，您不能在列表中的键和值之间建立链接。如果你想要类似于字典的东西，你可以做两个循环，一个用于键，另一个用于值，然后将其包含在列表中。您可以要求System.out.println（键+“：”+值）

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