Python 3.x 使用嵌套索引列表查找列表中的值？

我正在设法找到这个问题的解决办法。 给出一个数字列表：

numbers = [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501, 440, 771, 601, 582, 658, 954, 1154, 1175, 1459, 1786, 1667, 2186, 2558, 2774, 3388, 3448, 4070, 4785, 6060, 4268, 5642, 5236, 4774, 5849, 5966, 6361, 6198, 6411, 5841, 7099, 7933, 9623, 10633, 10581, 10102, 10559, 11231, 10699, 10817, 11012, 11656, 10899, 10028, 9974, 10598, 9200, 9709, 8926, 9263, 8764, 8849, 8894, 9434, 8599, 8946, 8915, 8338, 8371, 8572, 8952, 9268, 8485, 8858, 8529, 8823, 8718, 8846, 8971, 8970, 8587, 8393, 8777, 8961, 8697, 8809, 8217, 8241, 7824, 7772, 7971, 7870, 7717, 7586, 7413, 7165, 7105, 6788, 6843, 6784, 6683, 6683, 6550, 6752, 6710, 6623, 6719, 6569, 6363, 6534, 6491, 6623, 6586, 6587, 6511, 6240, 6410, 6415, 6389, 6214, 6096, 5901, 5828, 5850, 5830, 5779, 5833, 5741, 5607, 5380, 5449, 5484, 5468, 5429, 5387, 5364, 5121, 5186, 5239, 5191, 5185, 5155, 5081, 4892, 5054, 5017, 5016, 5030, 4911, 4839, 4718, 4790, 4767, 4838, 4860, 4797, 4688, 4639, 4642, 4675, 4758, 4843, 4897, 4932, 4670, 4893, 4958, 5064, 5144, 5097, 5106, 5020, 5172, 5310, 5421, 5406, 5361, 5414, 5435, 5612, 5667, 5803, 5960, 6043, 6090, 6109, 6330, 6489, 7112, 7421, 7764, 8026, 8129, 8371, 8835, 9294, 9735, 10376, 10757, 11481, 10981, 11345, 11969, 12673, 13442, 13406, 13690, 14041, 13556, 14937, 14703, 14804, 15843, 16108, 15444, 15704, 17077, 16260, 16392, 17148, 16342, 15886, 17418, 17987, 17834, 18381, 18017, 18431, 19483, 19116, 20368, 20109, 20248, 21577, 20765, 19583, 21333, 21717, 22441, 22313, 22562, 22201, 20717, 23337, 24059, 24538, 24295, 24891, 24087, 23393, 25195, 27267, 26809, 26390, 26046, 26126, 25043, 27829, 27078, 28450, 28701, 27798, 25752, 25838, 27562, 28206, 27729, 27651, 26902, 26265, 26074, 27787, 28116, 27772, 28510, 28917, 28340, 26814, 29499, 28595, 28833, 27849, 27363, 26588, 26095, 27329, 26613, 25938, 23845, 23015, 23955, 23902, 23218, 23330, 23012, 22540]

以及索引的嵌套列表：

nested_list = [[0, 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, 37, 38, 39, 40, 41, 42, 43, 44, 45], [46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76], [77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106], [107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137], [138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168], [169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198], [199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229], [230, 231, 232, 233, 234, 235, 236, 237, 238, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 256, 257, 258, 259], [260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290], [291, 292, 293, 294, 295, 296, 297, 298, 299, 300]]

使用嵌套索引列表，我需要返回列表中项目的值。 示例输出为：

[[27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501],[440, 771, 601, 582, 658, 954, 1154, 1175, 1459, 1786, 1667, 2186, 2558, 2774, 3388, 3448, 4070, 4785, 6060, 4268, 5642, 5236, 4774, 5849, 5966, 6361, 6198, 6411, 5841, 7099], etc....

我现在的代码是

values_from_list = []
        
    for i in nested_list:
        values_from_list.append([numbers[i] for i in nested_list[0]])
    print(values_from_list)

它只返回嵌套的_列表中i需要重复的一组值 我当前获得的输出：

[[27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501], [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501], [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501], [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501], [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501], [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501], [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501], [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501], [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501], [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501], [27, 24, 33, 52, 54, 53, 61, 71, 57, 163, 182, 196, 228, 270, 302, 501]]

我希望我的问题能被理解，

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非常感谢

在使用嵌套列表[0]时，应使用

i

；然后你应该使用另一个变量，在理解中使用

i

。许多错误隐藏在不清楚的变量命名中；我们应该努力确保变量名始终正确标识其用途。我将改写如下：

values_from_list = []
    
for sublist in nested_list:
    values_from_list.append([numbers[index] for index in sublist])
print(values_from_list)

或者，更简洁地说

values_from_list = [
    [numbers[index] for index in sublist]
    for sublist in nested_list
]

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这是伟大的阿玛丹，谢谢你的清晰和简洁的代码和解释！我知道我现在哪里出错了！