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
另外，如果你不介意使用numpy数组(或者实际上已经在处理numpy数组…)，这里有另一个很好的解决方案:
people = ['Jim', 'Pam', 'Micheal', 'Dwight']
ages = [27, 25, 4, 9]
import numpy
people = numpy.array(people)
ages = numpy.array(ages)
inds = ages.argsort()
sortedPeople = people[inds]
我在这里找到的: http://scienceoss.com/sort-one-list-by-another-list/
其他回答
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))]
下面是Whatangs的答案，如果你想获得两个排序的列表(python3)。
X = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]
Y = [ 0, 1, 1, 0, 1, 2, 2, 0, 1]
Zx, Zy = zip(*[(x, y) for x, y in sorted(zip(Y, X))])
print(list(Zx)) # [0, 0, 0, 1, 1, 1, 1, 2, 2]
print(list(Zy)) # ['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g']
记住Zx和Zy是元组。 我也在想是否有更好的方法来做到这一点。
警告:如果你用空列表运行它，它会崩溃。
list1 = ['a','b','c','d','e','f','g','h','i']
list2 = [0,1,1,0,1,2,2,0,1]
output=[]
cur_loclist = []
获取list2中的唯一值
list_set = set(list2)
查找list2中索引的loc
list_str = ''.join(str(s) for s in list2)
索引在list2中的位置使用cur_loclist跟踪
[0, 3, 7, 1, 2, 4, 8, 5, 6]
for i in list_set:
cur_loc = list_str.find(str(i))
while cur_loc >= 0:
cur_loclist.append(cur_loc)
cur_loc = list_str.find(str(i),cur_loc+1)
print(cur_loclist)
for i in range(0,len(cur_loclist)):
output.append(list1[cur_loclist[i]])
print(output)
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
另外，如果你不介意使用numpy数组(或者实际上已经在处理numpy数组…)，这里有另一个很好的解决方案:
people = ['Jim', 'Pam', 'Micheal', 'Dwight']
ages = [27, 25, 4, 9]
import numpy
people = numpy.array(people)
ages = numpy.array(ages)
inds = ages.argsort()
sortedPeople = people[inds]
我在这里找到的: http://scienceoss.com/sort-one-list-by-another-list/
