根据另一个列表

Question

假设我有一个嵌套列表，如：

my_list = [[1, 1, "c1"], [1, 2, "c1"], [5, 1, "c2"], [5, 2, "c2"], [6, 1, "c3"], [6, 2, "c3"], [2, 1, "c4"], [2, 2, "c4"], [3, 1, "c5"], [3, 2, "c5"], [4, 1, "c6"], [4, 2, "c6"]]

我还有另一个列表，根据需要对my_list列表的每个元素执行哪些排序来保持位置。假设它被定义为：

ordering = [2, 1]

现在我想通过多个参数对列表进行排序。首先，我想按列表ordering进行排序，它应该在索引[1]的ma_list中排序项目，之后我想按索引[0]中列表中的项目排序。总结一下，我最终想要的是：

list = [[1, 2, "c1"], [2, 2, "c4"], [3, 2, "c5"], [4, 2, "c6"], [5, 2, "c2"], [6, 2, "c3"], [[1, 1, "c1"], [2, 1, "c4"], [3, 1, "c5"], [4, 1, "c6"], [5, 1, "c2"], [6, 1, "c3"]

有没有（更好的Pythonic）方法呢？建议欢迎！

Answer 1

使用Lambda表达式

您可以将sorted()函数与下面的lambda表达式一起用作id age iq 1 85 100 来实现它：

key

此lambda表达式将根据# v 'i-1' since your `ordering` list is holding # v position instead of `index` lambda x: [x[i-1] for i in ordering] 列表中的索引返回my_list列表中每个元素的值列表。根据返回的列表，将执行排序。

示例运行：

ordering

使用operator.itemgetter

使用operator.itemgetter()作为：

更好

>>> my_list = [[1, 1, "c1"], [1, 2, "c1"], [5, 1, "c2"], [5, 2, "c2"], [6, 1, "c3"], [6, 2, "c3"], [2, 1, "c4"], [2, 2, "c4"], [3, 1, "c5"], [3, 2, "c5"], [4, 1, "c6"], [4, 2, "c6"]]
>>> ordering = [2, 1]

>>> sorted(my_list, key=lambda x: [x[i-1] for i in ordering])
[[1, 1, 'c1'], [2, 1, 'c4'], [3, 1, 'c5'], [4, 1, 'c6'], [5, 1, 'c2'], [6, 1, 'c3'], [1, 2, 'c1'], [2, 2, 'c4'], [3, 2, 'c5'], [4, 2, 'c6'], [5, 2, 'c2'], [6, 2, 'c3']]

Answer 2

要根据ordering进行排序，您可以尝试：

l = [[1, 1, "c1"], [1, 2, "c1"], [5, 1, "c2"], [5, 2, "c2"], [6, 1, "c3"], [6, 2, "c3"], [2, 1, "c4"], [2, 2, "c4"], [3, 1, "c5"], [3, 2, "c5"], [4, 1, "c6"], [4, 2, "c6"]]
ordering = [2, 1]
new_l = sorted(l, key=lambda x:[x[ordering[0]-1], x[ordering[1]-1]])

输出：

[[1, 1, 'c1'], [2, 1, 'c4'], [3, 1, 'c5'], [4, 1, 'c6'], [5, 1, 'c2'], [6, 1, 'c3'], [1, 2, 'c1'], [2, 2, 'c4'], [3, 2, 'c5'], [4, 2, 'c6'], [5, 2, 'c2'], [6, 2, 'c3']]

Answer 3

您可以使用运算符模块中的itemgetter来构造已排序的键。

from operator import itemgetter
l =  [[1, 1, "c1"], [1, 2, "c1"], [5, 1, "c2"], [5, 2, "c2"], [6, 1, "c3"], [6, 2, "c3"], [2, 1, "c4"], [2, 2, "c4"], [3, 1, "c5"], [3, 2, "c5"], [4, 1, "c6"], [4, 2, "c6"]]
ordering = [1, 0]
new_l = sorted(l, key=itemgetter(*order))