英文:
How to create a nested dictionary in a for loop (not using defaultdict)?
问题
以下是您的翻译好的代码部分：
my input looks like this:test_file = [['ref1', 'test1', 2],['ref2', 'test1', 3],['ref3', 'test2', 4],['ref1', 'test2', 4],['ref2', 'test2', 1],['ref1', 'test1', 4],['ref1', 'test1', 5]]And I'm trying to get a nested dictionary like this:desired_output = {'ref1':{'test1':[2,5,4]},'ref1':{'test2':[4]},'ref2':{'test1':[3]},'ref2':{'test2':[1]},'ref3':{'test2':[4]}}I tried to use defaultdict by appending the values to the second key, however, I got this error:> AttributeError: 'collections.defaultdict' object has no attribute> 'append'So, I tried this:for entry in test_file:nest1 = {}try:nest1[entry[1]].append(entry[2])except KeyError:nest1[entry[1]] = [entry[2]]try:mynestdict[entry[0]].append(nest1)except KeyError:mynestdict[entry[0]] = [nest1]print(dict(mynestdict))But I'm getting this:{'ref1': [{'test1': [2]}, {'test2': [4]}, {'test1': [4]}, {'test1': [5]}, 'ref2': [{'test1': [3]}, {'test2': [1]}, 'ref3': [{'test2': [4]}]}I'm not familiar with nested dictionaries and I really would like to understand them, any suggestions?
请注意，您的期望输出是不太合理的，因为在Python中，字典中的键必须是唯一的。因此，无法使用相同的键多次。如果您希望将多个值与相同的键相关联，您可以使用列表或其他数据结构。如果您有关于嵌套字典的问题或其他需求，请随时提问。
英文:
my input looks like this:
test_file = [['ref1', 'test1', 2],['ref2', 'test1', 3],['ref3', 'test2', 4],['ref1', 'test2', 4],['ref2', 'test2', 1],['ref1', 'test1', 4],['ref1', 'test1', 5]]
And I'm trying to get a nested dictionary like this:
desired_output = {'ref1':{'test1':[2,5,4]},'ref1':{'test2':[4]},'ref2':{'test1':[3]},'ref2':{'test2':[1]},'ref3':{'test2':[4]}}
I tried to use defaultdict by appending the values to the second key, however, I got this error:
> AttributeError: 'collections.defaultdict' object has no attribute
> 'append'
So, I tried this:
for entry in test_file:nest1 = {}try:nest1[entry[1]].append(entry[2])except KeyError:nest1[entry[1]] = [entry[2]]try:mynestdict[entry[0]].append(nest1)except KeyError:mynestdict[entry[0]] = [nest1]print(dict(mynestdict))
But I'm getting this:
{'ref1': [{'test1': [2]}, {'test2': [4]}, {'test1': [4]}, {'test1': [5]}],'ref2': [{'test1': [3]}, {'test2': [1]}],'ref3': [{'test2': [4]}]}
I'm not familiar with nested dictionaries and I really would like to understand them, any suggestions?
答案1
得分: 1
Since you're creating nested dictionaries, you need nested defaultdict().
result = defaultdict(lambda: defaultdict(list))for key1, key2, value in test_file:result[key1][key2].append(value)pprint(default_to_regular(result))# Output:{'ref1': {'test1': [2, 4, 5], 'test2': [4]},'ref2': {'test1': [3], 'test2': [1]},'ref3': {'test2': [4]}}
获取default_to_regular()函数，请访问 https://stackoverflow.com/questions/26496831/how-to-convert-defaultdict-of-defaultdicts-of-defaultdicts-to-dict-of-dicts-o
英文:
Since you're creating nested dictionaries, you need nested defaultdict().
result = defaultdict(lambda: defaultdict(list))for key1, key2, value in test_file:result[key1][key2].append(value)pprint(default_to_regular(result))# Output:{'ref1': {'test1': [2, 4, 5], 'test2': [4]},'ref2': {'test1': [3], 'test2': [1]},'ref3': {'test2': [4]}}
Get default_to_regular() from https://stackoverflow.com/questions/26496831/how-to-convert-defaultdict-of-defaultdicts-of-defaultdicts-to-dict-of-dicts-o
答案2
得分: 1
我认为你想要的输出是这样的，合并重复的键：
{'ref1': {'test1': [2, 4, 5], 'test2': [4]}, 'ref2': {'test1': [3], 'test2': [1]}, 'ref3': {'test2': [4]}}
要实现这个效果，只需检查嵌套键是否存在，如果不存在，则创建它们。你要求不使用 `collections.defaultdict`，以下是不使用它的代码：
test_file = [['ref1', 'test1', 2],
['ref2', 'test1', 3],
['ref3', 'test2', 4],
['ref1', 'test2', 4],
['ref2', 'test2', 1],
['ref1', 'test1', 4],
['ref1', 'test1', 5]]
d = {}
for el in test_file:
if el[0] not in d:
d[el[0]] = {}
if el[1] not in d[el[0]]:
d[el[0]][el[1]] = []
d[el[0]][el[1]].append(el[2])
print(d)
<details><summary>英文:</summary>I think the output you're looking for is this, combining the duplicate keys:
{'ref1': {'test1': [2, 4, 5], 'test2': [4]}, 'ref2': {'test1': [3], 'test2': [1]}, 'ref3': {'test2': [4]}}
To get that, just check if the nested keys exist, and if they don't, then create them. You asked for code that doesn't use `collections.defaultdict` and it doesn't make it too much more:
test_file = [['ref1', 'test1', 2],
['ref2', 'test1', 3],
['ref3', 'test2', 4],
['ref1', 'test2', 4],
['ref2', 'test2', 1],
['ref1', 'test1', 4],
['ref1', 'test1', 5]]
d = {}
for el in test_file:
if el[0] not in d:
d[el[0]] = {}
if el[1] not in d[el[0]]:
d[el[0]][el[1]] = []
d[el[0]][el[1]].append(el[2])
print(d)
</details># 答案3**得分**: 1I understand your request. Here's the translated code portion:```pythontest_file = [['ref1', 'test1', 2],['ref2', 'test1', 3],['ref3', 'test2', 4],['ref1', 'test2', 4],['ref2', 'test2', 1],['ref1', 'test1', 4],['ref1', 'test1', 5]]df = pd.DataFrame(test_file)df.groupby([0, 1])[2].apply(np.array).unstack().agg(lambda x: x.dropna().to_dict(), axis=1).to_json()
Please note that this code translates the given data into a JSON format.
英文:
test_file = [['ref1', 'test1', 2],['ref2', 'test1', 3],['ref3', 'test2', 4],['ref1', 'test2', 4],['ref2', 'test2', 1],['ref1', 'test1', 4],['ref1', 'test1', 5]]df = pd.DataFrame(test_file)df.groupby([0, 1])[2].apply(np.array).unstack().agg(lambda x: x.dropna().to_dict(), axis=1).to_json()'{"ref1":{"test1":[2,4,5],"test2":[4]},"ref2":{"test1":[3],"test2":[1]},"ref3":{"test2":[4]}}'
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