[英]Creating Dictionaries from Lists inside of Dictionaries
I'm quite new to Python and I have been stumped by a seemingly simple task. 我对Python还是很陌生,但我被一个看似简单的任务所困扰。 In part of my program, I would like to create Secondary Dictionaries from the values inside of lists, of which they are values of a Primary Dictionary. 在我的程序的一部分中,我想从列表内部的值创建辅助字典,这些值是主字典的值。
I would also like to default those values to 0 我也想将这些值默认为0
For the sake of simplicity, the Primary Dictionary looks something like this: 为了简单起见,主要字典看起来像这样:
primaryDict = {'list_a':['apple', 'orange'], 'list_b':['car', 'bus']}
What I would like my result to be is something like: 我希望结果是这样的:
{'list_a':[{'apple':0}, {'orange':0}], 'list_b':[{'car':0}, {'bus':0}]}
I understand the process should be to iterate through each list in the primaryDict, then iterate through the items in the list and then assign them as Dictionaries. 我了解该过程应该是遍历primaryDict中的每个列表,然后遍历列表中的项目,然后将它们分配为Dictionary。
I've tried many variations of "for" loops all looking similar to: 我尝试过许多“ for”循环的变体,它们看起来都类似于:
for listKey in primaryDict:
for word in listKey:
{word:0 for word in listKey}
I've also tried some methods of combining Dictionary and List comprehension, but when I try to index and print the Dictionaries with, for example: 我还尝试了一些结合字典和列表理解的方法,但是当我尝试使用以下方法对索引和索引字典进行打印时:
print(primaryDict['list_a']['apple'])
I get the "TypeError: list indices must be integers or slices, not str", which I interpret that my 'apple' is not actually a Dictionary, but still a string in a list. 我收到“ TypeError:列表索引必须是整数或切片,而不是str”,我将其解释为“苹果”实际上不是字典,而是列表中的字符串。 I tested that by replacing 'apple' with 0 and it just returns 'apple', proving it true. 我通过将'apple'替换为0进行测试,它仅返回'apple',证明它是正确的。
I would like help with regards to: 我需要以下方面的帮助:
-Whether or not the values in my list are assigned as Dictionaries with value '0' -是否将我列表中的值分配为“ 0”值的“字典”
or 要么
-Whether the mistake is in my indexing (in the loop or the print function), and what I am mistaken with -错误是在我的索引中(在循环中还是在打印功能中),以及我误会了什么
or 要么
-Everything I've done won't get me the desired outcome and I should attempt a different approach -我所做的一切都无法获得理想的结果,我应该尝试其他方法
Thanks 谢谢
Here is a dict comprehension that works: 这是一个有效的字典理解:
{k: [{v: 0} for v in vs] for k, vs in primaryDict.items()}
There are two problems with your current code. 您当前的代码有两个问题。 First, you are trying to iterate over listKey
, which is a string. 首先,您尝试遍历listKey
,它是一个字符串。 This produces a sequence of characters. 这将产生一个字符序列。
Second, you should use something like 其次,您应该使用类似
[{word: 0} for word in words]
in place of 代替
{word:0 for word in listKey}
You are close. 你近了 The main issue is the way you iterate your dictionary, and the fact you do not append or assign your sub-dictionaries to any variable. 主要问题是迭代字典的方式,以及您没有将子字典追加或分配给任何变量的事实。
This is one solution using only for
loops and list.append
. 这是仅for
循环和list.append
解决方案。
d = {}
for k, v in primaryDict.items():
d[k] = []
for w in v:
d[k].append({w: 0})
{'list_a': [{'apple': 0}, {'orange': 0}],
'list_b': [{'car': 0}, {'bus': 0}]}
A more Pythonic solution is to use a single list comprehension. 一种更Python化的解决方案是使用单个列表理解。
d = {k: [{w: 0} for w in v] for k, v in primaryDict.items()}
If you are using your dictionary for counting, which seems to be the implication, an even more Pythonic solution is to use collections.Counter
: 如果您正在使用字典进行计数,这似乎是在暗示,那么一个更Python化的解决方案是使用collections.Counter
:
from collections import Counter
d = {k: Counter(dict.fromkeys(v, 0)) for k, v in primaryDict.items()}
{'list_a': Counter({'apple': 0, 'orange': 0}),
'list_b': Counter({'bus': 0, 'car': 0})}
There are specific benefits attached to collections.Counter
relative to normal dictionaries. collections.Counter
有一些特殊的好处 。
primaryDict = {'list_a':['apple', 'orange'], 'list_b':['car', 'bus']}
for listKey in primaryDict:
primaryDict[i] = [{word:0} for word in primaryDict[listKey]]
print(primaryDict)
Output: 输出:
{'list_a':[{'apple':0}, {'orange':0}], 'list_b':[{'car':0}, {'bus':0}]}
Hope this helps! 希望这可以帮助!
You can get the data structure that you desire via: 您可以通过以下方式获取所需的数据结构:
primaryDict = {'list_a':['apple', 'orange'], 'list_b':['car', 'bus']}
for k, v in primaryDict.items():
primaryDict[k] = [{e: 0} for e in v]
# primaryDict
{'list_b': [{'car': 0}, {'bus': 0}], 'list_a': [{'apple': 0}, {'orange': 0}]}
But the correct nested access would be: 但是正确的嵌套访问将是:
print(primaryDict['list_a'][0]['apple']) # note the 0
If you actually want primaryDict['list_a']['apple']
to work, do instead 如果您实际上希望primaryDict['list_a']['apple']
工作,请改用
for k, v in primaryDict.items():
primaryDict[k] = {e: 0 for e in v}
# primaryDict
{'list_b': {'car': 0, 'bus': 0}, 'list_a': {'orange': 0, 'apple': 0}}
@qqc1037 , I checked and updated your code to make it working. @ qqc1037 ,我检查并更新了您的代码以使其正常运行。 I have mentioned the problem with your code as comments. 我已将您的代码问题作为注释提及。 Finally, I have also added one more example using list comprehension , map() & lambda function . 最后,我还添加了一个使用list comprehension , map()和lambda函数的示例。
import json
secondaryDict = {}
for listKey in primaryDict:
new_list = [] # You did not define any temporary list
for word in primaryDict [listKey]: # You forgot to use key that refers the list
new_list.append( {word:0}) # Here you forgot to append to list
secondaryDict2.update({listKey: new_list}) # Finally, you forgot to update the secondary dictionary
# Pretty printing dictionary
print(json.dumps(secondaryDict, indent=4));
"""
{
"list_a": [
{
"apple": 0
},
{
"orange": 0
}
],
"list_b": [
{
"car": 0
},
{
"bus": 0
}
]
}
"""
Another example: Using list comprehension, map(), lambda function 另一个示例:使用列表推导,map(),lambda函数
# Using Python 3.5.2
import json
primaryDict = {'list_a':['apple', 'orange'], 'list_b':['car', 'bus']}
secondaryDict = dict(map(lambda key: (key, [{item:0} for item in primaryDict[key]]), list(primaryDict) ))
# Pretty printing secondary dictionary
print(json.dumps(secondaryDict, indent=4))
"""
{
"list_a": [
{
"apple": 0
},
{
"orange": 0
}
],
"list_b": [
{
"car": 0
},
{
"bus": 0
}
]
}
"""
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