合并字典而不覆盖以前的值，其中 value 是一个列表 [英] Merge dictionaries without overwriting previous value where value is a list

问题描述

我知道合并字典而不覆盖值，合并几个"python字典，如何合并多个具有相同键的字典? 和 如何在一个表达式中合并两个 Python 字典?.

不过，我的问题略有不同.

假设我有这三本词典

dict_a = {'a': [3.212], 'b': [0.0]}dict_b = {'a': [923.22, 3.212], 'c': [123.32]}dict_c = {'b': [0.0]}

合并后的结果应该是

result_dict = {'a': [3.212, 3.212, 923.22], 'b': [0.0, 0.0], 'c': [123.32]}

这里的代码有效，但会在列表中嵌套列表

result_dict = {}dicts = [dict_a, dict_b, dict_c]对于字典中的 d:对于 d.iteritems() 中的 k, v:result_dict.setdefault(k, []).append(v)

使用 extend 而不是 append 会阻止嵌套列表，但如果键尚不存在则不起作用.所以基本上它应该在没有覆盖的情况下执行 update，就像在其他线程中一样.

抱歉，这是我的错误.昨天太晚了，我没有注意到抛出错误的行不是我认为的那样，因此假设我的字典已经具有上述结构.事实上，mgilson 是正确的，假设它与 TypeError 相关.准确地说，是一个可统一的浮动".

解决方案

我很确定 .extend 在这里工作......

<预><代码>>>>dict_a = {'a': [3.212], 'b': [0.0]}>>>dict_b = {'a': [923.22, 3.212], 'c': [123.32]}>>>dict_c = {'b': [0.0]}>>>result_dict = {}>>>dicts = [dict_a, dict_b, dict_c]>>>>>>对于字典中的 d:...对于 k, v 在 d.iteritems():... result_dict.setdefault(k, []).extend(v)...>>>结果字典{'a':[3.212, 923.22, 3.212]，'c':[123.32]，'b':[0.0, 0.0]}

神奇之处在于 dict.setdefault 方法.如果该键不存在，setdefault 添加一个具有您提供的默认值的新键.然后它返回那个可以修改的默认值.

<小时>

请注意，如果项目 v 不可迭代，则此解决方案会有问题.也许这就是你的意思?例如如果 dict_a = {'a': [3.212], 'b': 0.0}.

在这种情况下，我认为您需要在 try-except 子句中捕获 TypeError: type object is not iterable :

for d in dicts:对于 d.iteritems() 中的 k, v:尝试:result_dict.setdefault(k, []).extend(v)除了类型错误:result_dict[k].append(v)

I am aware of Merge dictionaries without overwriting values, merging "several" python dictionaries, How to merge multiple dicts with same key?, and How to merge two Python dictionaries in a single expression?.

My problem is slightly different, however.

Lets say I have these three dictionaries

dict_a = {'a': [3.212], 'b': [0.0]}
dict_b = {'a': [923.22, 3.212], 'c': [123.32]}
dict_c = {'b': [0.0]}

The merged result should be

result_dict = {'a': [3.212, 3.212, 923.22], 'b': [0.0, 0.0], 'c': [123.32]}

This code here works, but would nest lists within the lists

result_dict = {}
dicts = [dict_a, dict_b, dict_c]

for d in dicts:
    for k, v in d.iteritems():
        result_dict.setdefault(k, []).append(v)

Using extend instead of append would prevent the nested lists, but doesn't work if a key doesn't exist yet. So basically it should do a update without the overwriting, as in the other threads.

Sorry, it was a mistake on my side. It was too late yesterday and I didn't notice the line that threw the error wasn't the one I thought it did, therefore assumed my dictionaries would already have the above structure. In fact, mgilson was correct assuming that it was related to a TypeError. To be exact, an 'uniterable float'.

解决方案

I'm pretty sure that .extend works here ...

>>> dict_a = {'a': [3.212], 'b': [0.0]}
>>> dict_b = {'a': [923.22, 3.212], 'c': [123.32]}
>>> dict_c = {'b': [0.0]}
>>> result_dict = {}
>>> dicts = [dict_a, dict_b, dict_c]
>>> 
>>> for d in dicts:
...     for k, v in d.iteritems():
...         result_dict.setdefault(k, []).extend(v)
... 
>>> result_dict
{'a': [3.212, 923.22, 3.212], 'c': [123.32], 'b': [0.0, 0.0]}

The magic is in the dict.setdefault method. If the key doesn't exist, setdefault adds a new key with the default value you provide. It then returns that default value which can then be modified.

---

Note that this solution will have a problem if the item v is not iterable. Perhaps that's what you meant? e.g. if dict_a = {'a': [3.212], 'b': 0.0}.

In this case, I think you'll need to resort to catching the TypeError: type object is not iterable in a try-except clause:

for d in dicts:
    for k, v in d.iteritems():
        try:
            result_dict.setdefault(k, []).extend(v)
        except TypeError:
            result_dict[k].append(v)

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