Python：对列表进行排序并相应地更改另一个列表

Question

我有两个列表：一个包含一组x点，另一个包含y点。 Python以某种方式管理混合x点，或用户可以。我需要按照从最低到最高的顺序对它们进行排序，并且移动y点以跟随它们的x个对应点。他们在两个单独的名单..我该怎么做？

Answer 1

您可以压缩列表并对结果进行排序。排序元组时，默认情况下，应该排序第一个成员。

>>> xs = [3,2,1] 
>>> ys = [1,2,3] 
>>> points = zip(xs,ys) 
>>> points 
[(3, 1), (2, 2), (1, 3)] 
>>> sorted(points) 
[(1, 3), (2, 2), (3, 1)] 

然后再解压它们：

>>> sorted_points = sorted(points) 
>>> new_xs = [point[0] for point in sorted_points] 
>>> new_ys = [point[1] for point in sorted_points] 
>>> new_xs 
[1, 2, 3] 
>>> new_ys 
[3, 2, 1] 

Answer 2

如果x和y的意思是一个单元（如一个点），它将使更多的意义将它们存储为元组而不是两个单独的列表。

无论如何，这里是你应该做的：

x = [4, 2, 5, 4, 5,…] 
y = [4, 5, 2, 3, 1,…] 

zipped_list = zip(x,y) 
sorted_list = sorted(zipped_list) 

Answer 3

>>> xs = [5, 2, 1, 4, 6, 3] 
>>> ys = [1, 2, 3, 4, 5, 6] 
>>> xs, ys = zip(*sorted(zip(xs, ys))) 
>>> xs 
(1, 2, 3, 4, 5, 6) 
>>> ys 
(3, 2, 6, 4, 1, 5) 

Answer 4

>>> import numpy 

>>> sorted_index = numpy.argsort(xs) 
>>> xs = [xs[i] for i in sorted_index] 
>>> ys = [ys[i] for i in sorted_index] 

如果你能numpy.array工作

>>> xs = numpy.array([3,2,1]) 
>>> xs = numpy.array([1,2,3]) 
>>> sorted_index = numpy.argsort(xs) 
>>> xs = xs[sorted_index] 
>>> ys = ys[sorted_index]