从字典中删除重复值而不删除键 [英] Remove duplicate value from dictionary without removing key

问题描述

我是Python的新手，不得不通过试用和错误的学习，但是我无法找到我遇到的问题的解决方案。

I'm new to Python and am having to learn by trial and error, but I haven't been able to find a solution to the problem I'm having.

我有一本类似这样的字典：

I have a dictionary that looks something like this:

myDict = {'key1': ['item1', 'item2', 'item3'], 'key2': ['item4', 'item5', 'item6'],  
'key3': 'item7', 'key4': 'item8', 'key5': ['item1', 'item2', 'item3'], 'key6': 'item7'}

我需要从字典中删除重复的值，并用空值（）替换它们。在这里找到了一对夫妇的解决方案，但是他们正在按照预期的方式工作。

I need to remove duplicate values from the dictionary and replace them with an empty value (""). Found a couple solution on here but they are working as intended

for key, value in myDict.items():
    if values not in key newDict.values():
        myDict[key] = value
    else:
        myDict[key] = ""
print newDict

这是删除所有的值，并输出

This is removing all the values and is outputting

# newDict:{key1: '', key2: '', key3: '', key4: '', key5: '', key6: '')

我正在寻找输出为

# newDict = {'key1': '', 'key2':['item4', 'item5', 'item6'], 'key3': '', 'key4':  
'item8', key5: ['item1', 'item2', 'item3'], 'key6': 'item7'}

推荐答案

你有正确的总体想法，但你的代码有三个问题：

You have the right overall idea, but there are three problems with your code:

1. 您正在将值存储回 myDict 而不是 newDict 。


2. 在第2行，你正在检查值而不是值。


3. 同样在第2行，键不应该在那里，并抛出一个 SyntaxError 。

1. You're storing values back into myDict instead of into newDict.
2. On line 2, you're checking values instead of value.
3. Also on line 2, key shouldn't be there, and throws a SyntaxError.

这是正确的代码：

newDict = {}
for key, value in myDict.iteritems():
    if value not in newDict.values():
        newDict[key] = value
    else:
        newDict[key] = ""
print newDict

如果您不在反三位运营商营地，您还可以将其缩短为：

If you're not in the anti-ternary operator camp, you could also shorten it to this:

newDict = {}
for key, value in myDict.iteritems():
    newDict[key] = value if value not in newDict.values() else ""
print newDict

或者，如果您宁愿从原始 dict （ myDict ）中删除值，而不是构建一个ne w $（$ code> newDict ），你可以这样做：

Or, if you would rather just remove the values from the original dict (myDict) instead of building a new one (newDict), you could do this:

foundValues = []
for key, value in myDict.iteritems():
    if value not in foundValues:
        foundValues.append(myDict[key])
    else:
        myDict[key] = ""
print myDict

如果您需要在具体订单，请查看 OrderedDict s 。

If you need duplicate values removed in a specific order, check out OrderedDicts.

更新：

更新的要求 - 从最初的 dict 中删除​​该值，如果您能够简单地初始化 myDict 与 OrderedDict 而不是 dict ，所有您需要做的是替换为： p>

In light of the updated requirements -- that values be removed from the original dict, starting from the beginning -- if you're able to simply initialize myDict with an OrderedDict instead of a dict, all you need to do is replace this:

myDict = {'key1': ['item1', 'item2', 'item3'], 'key2': ['item4', 'item5', 'item6'], 'key3': 'item7', 'key4': 'item8', 'key5': ['item1', 'item2', 'item3'], 'key6': 'item7'}

与此相关：

from collections import OrderedDict

…

myDict = OrderedDict([('key1', ['item1', 'item2', 'item3']), ('key2', ['item4', 'item5', 'item6']), ('key3', 'item7'), ('key4', 'item8'), ('key5', ['item1', 'item2', 'item3']), ('key6', 'item7')])

然后使用与上述相同的代码。

and then use the same code provided above.

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