我如何将每个 n 元素设置为 dict 理解中的 dict 值答案

【问题标题】：How i can setup every n elements as dict values in dict comprehension我如何将每个 n 元素设置为 dict 理解中的 dict 值
【发布时间】：2020-03-31 20:54:13
【问题描述】：

一个例子：

data_1 = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36]
data_2 = ['key1','key2','key3','key4','key5','key6','key7','key8','key9','key10','key11','key12']
data_3 = {'123':{'a':1},'222':{'b':2}, '333':{'b':3}}

我需要根据下一个条件设置这个列表来 dict 键，data_1 中的每 12 个元素应该是 data_2 中每个键的值，这个 dict 应该是数据 3 中的键值。

预期输出：

{'123':{'key1':1,'key2':2,'key3':3,'key4':4,'key5':5,'key6':6,'key7':7,'key8':8,'key9':9,'key10':10,'key11':11,'key12':12},
{'222':{'key1':13,'key2':14,'key3':15,'key4':16,'key5':17,'key6':18,'key7':19,'key8':20,'key9':21,'key10':22,'key11':23,'key12':24},
{'333':{'key1':25,'key2':26,'key3':27,'key4':28,'key5':29,'key6':30,'key7':31,'key8':32,'key9':33,'key10':34,'key11':35,'key12':36}

我的尝试：

    result = {
        key: dict(zip(data_2,data_1) for key, value in data_3.items()
    }

我的结果：

{'123': {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5, 'key6': 6, 'key7': 7, 'key8': 8, 'key9': 9, 'key10': 10, 'key11': 11, 'key12': 12}, '222': {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5, 'key6': 6, 'key7': 7, 'key8': 8, 'key9': 9, 'key10': 10, 'key11': 11, 'key12': 12}, '333': {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5, 'key6': 6, 'key7': 7, 'key8': 8, 'key9': 9, 'key10': 10, 'key11': 11, 'key12': 12}}

我明白这部分陈述中有什么错误：

键：dict(zip(data_2,data_1)

我怎样才能达到我期望的结果？

【问题讨论】：

标签： python dictionary dictionary-comprehension

【解决方案1】：

一种方法是使用zip这一事实，只会迭代直到较小的可迭代对象被消耗，因此您可以执行以下操作：

iterable_data1 = iter(data_1)

result = {key: dict(zip(data_2, iterable_data1)) for key in data_3}

for key, value in result.items():
    print(key, value)

这非常高效，因为它不会创建任何额外的列表。另外，你可以这样做：

result = {key: dict(zip(data_2, data_1[12 * i:])) for i, key in enumerate(data_3)}

for key, value in result.items():
    print(key, value)

输出

123 {'key1': 1, 'key2': 2, 'key3': 3, 'key4': 4, 'key5': 5, 'key6': 6, 'key7': 7, 'key8': 8, 'key9': 9, 'key10': 10, 'key11': 11, 'key12': 12}
222 {'key1': 13, 'key2': 14, 'key3': 15, 'key4': 16, 'key5': 17, 'key6': 18, 'key7': 19, 'key8': 20, 'key9': 21, 'key10': 22, 'key11': 23, 'key12': 24}
333 {'key1': 25, 'key2': 26, 'key3': 27, 'key4': 28, 'key5': 29, 'key6': 30, 'key7': 31, 'key8': 32, 'key9': 33, 'key10': 34, 'key11': 35, 'key12': 36}

第二种方法的想法是，对于data_3 中的每个key，将指针移动到data_1 12 个位置。

【讨论】：