從字典中過濾某些鍵的最快方法，這些字典存儲在字典列表中，而字典列表又存儲在字典中

Question

我正在尋找從“外部”字典內的列表中的“內部”字典中過濾某些鍵的最快方法，這是一種棘手的數據結構。

數據結構（輸入）如下所示：所有內部字典具有相同的鍵，但列表可以具有不同數量的字典。

d = {1:[{1:33 ,2:33, 3:33, 4:33}, {1:33 ,2:33, 3:33, 4:33}, {1:33 ,2:33, 3:33, 4:33}],
 2:[{1:33 ,2:33, 3:33, 4:33}],
 3:[{1:33 ,2:33, 3:33, 4:33}, {1:33 ,2:33, 3:33, 4:33}]}

我想要的 output 是相同的結構，但只有內部字典鍵 1 和 2，我不再需要 3 和 4：

d_new = {1: [{1: 33, 2: 33}, {1: 33, 2: 33}, {1: 33, 2: 33}],
 2: [{1: 33, 2: 33}],
 3: [{1: 33, 2: 33}, {1: 33, 2: 33}, {1: 33, 2: 33}]}

到目前為止，我有兩種解決問題的方法，但速度很慢：

1. 列表和字典理解的組合：

d_new = {n: [{1: d[1], 2: d[2]} for d in d[n]] for n in d.keys()}

2. For循環

d_new = {}

keys = d.keys()

for key in keys:
    storage = []
    for x in d[key]:
        storage.append({1: x[1], 2: x[2]})
    d_new[key] = storage

它們都解決了這個問題，但是在應用於大型詞典時非常慢，所以我正在尋找一種更快更有效的方法，也許也可以在這里應用某種並行化。 很高興有任何幫助！

Answer 1

編碼

from copy import deepcopy

d = {1:[{1:33 ,2:33, 3:33, 4:33}, {1:33 ,2:33, 3:33, 4:33}, {1:33 ,2:33, 3:33, 4:33}],
 2:[{1:33 ,2:33, 3:33, 4:33}],
 3:[{1:33 ,2:33, 3:33, 4:33}, {1:33 ,2:33, 3:33, 4:33}]}

d_new = deepcopy(d) #Done so that we don't run into issues with list items

d_items = list(d_new.items())
for i in d_items:
    for j in i[1]:
        pop_l = [] #Created a list of entries to pop since popping them while in the for loop changes the loop length and gives a runtime error
        for k in j:
            if k > 2:
                pop_l.append(k)
        for x in pop_l: #Popping all unwanted keys
            j.pop(x)

print(d_new)

Output

{1: [{1: 33, 2: 33}, {1: 33, 2: 33}, {1: 33, 2: 33}], 2: [{1: 33, 2: 33}], 3: [{1: 33, 2: 33}, {1: 33, 2: 33}]}