如何比循环更快地生成嵌套字典？

Question

我想知道如何比使用简单循环更快地创建嵌套字典（这对于大量数据来说可能很麻烦）。我曾经做过以下事情：

list_1 = {1, 2, 3, 4, 5, 6, ....}
list_2 = {"a", "b", "c", "d", "e", "f", ....}
list_3 = {10, 20, 30, 40, 50, 60 ....}

for element_1 in list_1:
    dico[element_1] = {}
    for element_2 in list_2:
        dico[element_1][element_2]={}
        for element_3 in list_3:
            dico[element_1][element_2][element_3] = {}

问题是，我认为如果有多个步骤和大量数据，它可能真的很慢...

谢谢

Answer 1

试试这个嵌套字典理解：

print({i: {x: {y: {} for y in list_3} for x in list_2} for i in list_1})

输出：

{1: {'d': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'c': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'a': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'e': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'b': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'f': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}}, 2: {'d': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'c': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'a': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'e': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'b': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'f': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}}, 3: {'d': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'c': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'a': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'e': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'b': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'f': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}}, 4: {'d': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'c': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'a': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'e': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'b': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'f': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}}, 5: {'d': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'c': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'a': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'e': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'b': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'f': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}}, 6: {'d': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'c': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'a': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'e': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'b': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}, 'f': {40: {}, 10: {}, 50: {}, 20: {}, 60: {}, 30: {}}}}

Answer 2

仍然有循环但更快更短：

dico = {}
for element_1 in list_1:
    dico1 = dico[element_1] = {}
    for element_2 in list_2:
        dico2 = dico1[element_2]={}
        for element_3 in list_3:
            dico2[element_3] = {}

基准测试结果：

Round 1  Round 2  Round 3 
  59 ms    58 ms    55 ms  original
  40 ms    41 ms    39 ms  optimized
  46 ms    44 ms    43 ms  U11

基准代码（Try it online!）：

from timeit import timeit

list_1 = {1, 2, 3, 4, 5, 6, ...}
list_2 = {"a", "b", "c", "d", "e", "f", ...}
list_3 = {10, 20, 30, 40, 50, 60, ...}

def original():
    dico = {}
    for element_1 in list_1:
        dico[element_1] = {}
        for element_2 in list_2:
            dico[element_1][element_2]={}
            for element_3 in list_3:
                dico[element_1][element_2][element_3] = {}
    return dico

def optimized():
    dico = {}
    for element_1 in list_1:
        dico1 = dico[element_1] = {}
        for element_2 in list_2:
            dico2 = dico1[element_2]={}
            for element_3 in list_3:
                dico2[element_3] = {}
    return dico

def U11():
    return {i: {x: {y: {} for y in list_3} for x in list_2} for i in list_1}

# config
funcs = original, optimized, U11
number = 1000

# correctness
expect = original()
for func in funcs:
    result = func()
    print(result == expect, func.__name__)
print()

# speed
tss = [[] for _ in funcs]
for r in range(1, 4):
    print(*(f'Round {i} ' for i in range(1, r+1)))
    for func, ts in zip(funcs, tss):
        t = timeit(func, number=number) / number
        ts.append(t)
        print(*('%4d ms ' % (t * 1e6) for t in ts), func.__name__)
    print()

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