python-3.x - 根据Python中的另一个列表消除一个列表

Question

我有这样的二维列表

x_irp_group = [['x1_1_4', 'x1_2_4', 'x1_3_4', 'x1_4_4', 'x1_5_4', 'x1_6_4', 'x1_7_4', 'x1_8_4', 'x1_9_4', 'x1_10_4', 'x1_1_5', 'x1_2_5', 'x1_3_5', 'x1_4_5', 'x1_5_5', 'x1_6_5', 'x1_7_5', 'x1_8_5', 'x1_9_5', 'x1_10_5', 'x1_1_6', 'x1_2_6', 'x1_3_6', 'x1_4_6', 'x1_5_6', 'x1_6_6', 'x1_7_6', 'x1_8_6', 'x1_9_6', 'x1_10_6', 'x1_1_7', 'x1_2_7', 'x1_3_7', 'x1_4_7', 'x1_5_7', 'x1_6_7', 'x1_7_7', 'x1_8_7', 'x1_9_7', 'x1_10_7', 'x1_1_8', 'x1_2_8', 'x1_3_8', 'x1_4_8', 'x1_5_8', 'x1_6_8', 'x1_7_8', 'x1_8_8', 'x1_9_8', 'x1_10_8'], ['x1_1_8', 'x1_2_8', 'x1_3_8', 'x1_4_8', 'x1_5_8', 'x1_6_8', 'x1_7_8', 'x1_8_8', 'x1_9_8', 'x1_10_8', 'x1_1_9', 'x1_2_9', 'x1_3_9', 'x1_4_9', 'x1_5_9', 'x1_6_9', 'x1_7_9', 'x1_8_9', 'x1_9_9', 'x1_10_9', 'x1_1_10', 'x1_2_10', 'x1_3_10', 'x1_4_10', 'x1_5_10', 'x1_6_10', 'x1_7_10', 'x1_8_10', 'x1_9_10', 'x1_10_10', 'x1_1_11', 'x1_2_11', 'x1_3_11', 'x1_4_11', 'x1_5_11', 'x1_6_11', 'x1_7_11', 'x1_8_11', 'x1_9_11', 'x1_10_11', 'x1_1_12', 'x1_2_12', 'x1_3_12', 'x1_4_12', 'x1_5_12', 'x1_6_12', 'x1_7_12', 'x1_8_12', 'x1_9_12', 'x1_10_12']]

如果另一个一维列表中的元素像这样，我想消除这个二维列表

x_irp_eliminated_list =   ['x1_1_4', 'x1_1_8', 'x1_1_12', 'x1_1_16', 'x1_1_19', 'x1_1_22', 'x1_1_26', 'x1_1_30', 'x1_1_34', 'x1_1_37', 'x1_1_43', 'x1_1_49', 'x1_1_55', 'x1_1_61', 'x1_1_68', 'x1_1_75', 'x1_1_81', 'x1_1_87', 'x1_1_92', 'x1_1_96', 'x1_1_101', 'x1_1_107', 'x1_1_112', 'x1_1_116', 'x1_1_121', 'x1_1_126', 'x1_1_131', 'x1_1_134', 'x1_1_137', 'x1_1_141', 'x1_1_145', 'x1_1_149', 'x1_1_152', 'x1_1_155', 'x1_1_160', 'x1_1_164', 'x1_1_169', 'x1_1_173', 'x1_1_181', 'x1_1_189', 'x1_1_197', 'x1_1_205', 'x1_2_8', 'x1_2_10', 'x1_2_13', 'x1_2_17', 'x1_2_21', 'x1_2_25', 'x1_2_28', 'x1_2_30', 'x1_2_34', 'x1_2_40', 'x1_2_45', 'x1_2_51', 'x1_2_58', 'x1_2_66', 'x1_2_71', 'x1_2_77', 'x1_2_82', 'x1_2_86', 'x1_2_91', 'x1_2_97', 'x1_2_102', 'x1_2_106', 'x1_2_111', 'x1_2_117', 'x1_2_122', 'x1_2_125', 'x1_2_129', 'x1_2_132', 'x1_2_135', 'x1_2_139', 'x1_2_143', 'x1_2_147', 'x1_2_151', 'x1_2_154', 'x1_2_157', 'x1_2_161', 'x1_2_166', 'x1_2_172', 'x1_2_177', 'x1_2_181', 'x1_2_189', 'x1_2_197', 'x1_2_205', 'x1_2_214', 'x1_3_1', 'x1_3_4', 'x1_3_8', 'x1_3_11', 'x1_3_15', 'x1_3_18', 'x1_3_22', 'x1_3_25', 'x1_3_28', 'x1_3_32', 'x1_3_35', 'x1_3_39', 'x1_3_42', 'x1_3_46', 'x1_3_49', 'x1_3_52', 'x1_3_56', 'x1_3_59', 'x1_3_63', 'x1_3_66', 'x1_3_70', 'x1_3_73', 'x1_3_77', 'x1_3_81', 'x1_3_85', 'x1_3_88', 'x1_3_91', 'x1_3_94', 'x1_3_97', 'x1_3_101', 'x1_3_105', 'x1_3_109', 'x1_3_112', 'x1_3_115', 'x1_3_118', 'x1_3_122', 'x1_3_126', 'x1_3_130', 'x1_3_134', 'x1_3_137', 'x1_3_140', 'x1_3_143', 'x1_3_147', 'x1_3_151', 'x1_3_156', 'x1_3_159', 'x1_3_163']

我写了这样的代码，但效果不佳。

x_final = [i for i, j in zip(x_irp_group, x_irp_eliminated_list) if i == j]

我缩短了列表。通常它们的尺寸比那个大得多

标签： python-3.xlist

Answer 1

您拥有的列表理解不起作用，因为您将元素压缩在一起，这不是操作所代表的（它们不是并行数组）您想要的是以下内容：

x_final = [i for i in x_irp_group[0] if (i not in x_irp_eliminated_list)]

请注意，对于 2d 列表，您可能需要像这样嵌套：

# writing normal loops you'd write:
# for row in x_irp_group:
#     for i in row:
#         if (...):
# so I typically try to indent the loops similarly since nested array comprehension
# gets complicated, honestly I'd likely prefer using generator functions for this anyway
x_final = [[i   for i in row
                    if (i not in x_irp_eliminated_list)
           ]for row in x_irp_group   
          ]

虽然知道i not in x_irp_eliminated_list列表会很慢，但将其更改为集合会提高性能：

x_irp_eliminated_set = set(x_irp_eliminated_list)
x_final = [i for i in x_irp_group[0] if (i not in x_irp_eliminated_set)]

或者，如果列表是简单排序的，那么您可以将它们都转换为集合，进行减法然后再次排序：

x_final = [ sorted(set(x_irp_group[0]) - set(x_irp_eliminated_list)) ] 

虽然如果你有超级大名单，这可能不太理想。