我有一个字符串列表，像这样:
X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
Y = [ 0, 1, 1, 0, 1, 2, 2, 0, 1 ]
使用Y中的值对X进行排序以得到以下输出的最短方法是什么?
["a", "d", "h", "b", "c", "e", "i", "f", "g"]
具有相同“键”的元素的顺序并不重要。我可以使用for结构，但我很好奇是否有更短的方法。有什么建议吗?
当前回答
More_itertools有一个并行排序可迭代对象的工具:
鉴于
from more_itertools import sort_together
X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
Y = [ 0, 1, 1, 0, 1, 2, 2, 0, 1]
Demo
sort_together([Y, X])[1]
# ('a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g')
其他回答
More_itertools有一个并行排序可迭代对象的工具:
鉴于
from more_itertools import sort_together
X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
Y = [ 0, 1, 1, 0, 1, 2, 2, 0, 1]
Demo
sort_together([Y, X])[1]
# ('a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g')
把两个列表压缩在一起，排序，然后取你想要的部分:
>>> yx = zip(Y, X)
>>> yx
[(0, 'a'), (1, 'b'), (1, 'c'), (0, 'd'), (1, 'e'), (2, 'f'), (2, 'g'), (0, 'h'), (1, 'i')]
>>> yx.sort()
>>> yx
[(0, 'a'), (0, 'd'), (0, 'h'), (1, 'b'), (1, 'c'), (1, 'e'), (1, 'i'), (2, 'f'), (2, 'g')]
>>> x_sorted = [x for y, x in yx]
>>> x_sorted
['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g']
把这些结合起来得到:
[x for y, x in sorted(zip(Y, X))]
X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
Y = [ 0, 1, 1, 0, 1, 2, 2, 0, 1 ]
你可以用一行写出来:
X, Y = zip(*sorted(zip(Y, X)))
我创建了一个更通用的函数，它根据另一个列表对两个以上的列表进行排序，灵感来自@Whatang的答案。
def parallel_sort(*lists):
"""
Sorts the given lists, based on the first one.
:param lists: lists to be sorted
:return: a tuple containing the sorted lists
"""
# Create the initially empty lists to later store the sorted items
sorted_lists = tuple([] for _ in range(len(lists)))
# Unpack the lists, sort them, zip them and iterate over them
for t in sorted(zip(*lists)):
# list items are now sorted based on the first list
for i, item in enumerate(t): # for each item...
sorted_lists[i].append(item) # ...store it in the appropriate list
return sorted_lists
上面的大多数解决方案都很复杂，我认为如果列表的长度不同或不包含完全相同的项目，它们就不会起作用。下面的解决方案很简单，不需要任何导入。
list1 = ['B', 'A', 'C'] # Required sort order
list2 = ['C', 'B'] # Items to be sorted according to list1
result = list1
for item in list1:
if item not in list2: result.remove(item)
print(result)
输出:
['B', 'C']
注意:任何不在list1中的项都将被忽略，因为算法不知道要使用的排序顺序。
