Question

新手问题。

假设我有一个嵌套词典列表。

a = [{"value1": 1234, "value2": 23423423421, "value3": norway, "value4": charlie},
     {"value1": 1398, "value2": 23423412221, "value3": england, "value4": alpha},
     {"value1": 1234, "value2": 23234231221, "value3": norway, "value4": charlie},
     {"value1": 1398, "value2": 23423213121, "value3": england, "value4": alpha}]

我想要的是移动每个副本的singularis条目，其中value1，value3和value4匹配。结果应该是这样的：

b = [{"value1": 1398, "value2": 23423412221, "value3": england, "value4": alpha},
     {"value1": 1234, "value2": 23234231221, "value3": norway, "value4": charlie}]

原始列表a应该保留在原始状态中。

Answer 1

最近有一个类似的问题。试试这个entry。

事实上，你问过这个问题：“假设存在多个条目，其中value3和value4与其他嵌套字典相同。我怎样才能快速轻松地找到并删除那些重复的字典。”

这听起来像是一回事，对吧？

编辑：自由地窃取Alex的代码，它看起来像这样：

import itertools
import pprint
import operator

alpha, charlie, norway, england = range(4)

a = [{"value1": 1234, "value2": 23423423421, "value3": norway, "value4": charlie},
     {"value1": 1398, "value2": 23423412221, "value3": england, "value4": alpha}, 
     {"value1": 1234, "value2": 23234231221, "value3": norway, "value4": charlie}, 
     {"value1": 1398, "value2": 23423213121, "value3": england, "value4": alpha}]


getvals = operator.itemgetter('value1', 'value3', 'value4')

a.sort(key=getvals)

b = [g.next() for _, g in itertools.groupby(a, getvals)]
pprint.pprint(b)

结果是：

[{'value1': 1234, 'value2': 23423423421L, 'value3': 2, 'value4': 1},
 {'value1': 1398, 'value2': 23423412221L, 'value3': 3, 'value4': 0}]

重复词典列表将单个词条复制到另一个列表

1 个答案: