假设我有两个列表:
list1 = [3, 2, 4, 1, 1]
list2 = ['three', 'two', 'four', 'one', 'one2']
如果我运行list1.sort()，它会把它排序到[1,1,2,3,4]，但是否有一种方法让list2同步(所以我可以说项目4属于' 3 ')?因此，期望输出为:
list1 = [1, 1, 2, 3, 4]
list2 = ['one', 'one2', 'two', 'three', 'four']
我的问题是，我有一个相当复杂的程序，它可以很好地处理列表，但我需要开始引用一些数据。我知道这对字典来说是一个完美的情况，但我试图在我的处理中避免字典，因为我确实需要对键值进行排序(如果我必须使用字典，我知道如何使用它们)。
Basically the nature of this program is, the data comes in a random order (like above), I need to sort it, process it and then send out the results (order doesn't matter but users need to know which result belongs to which key). I thought about putting it in a dictionary first, then sorting list one but I would have no way of differentiating of items in the with the same value if order is not maintained (it may have an impact when communicating the results to users). So ideally, once I get the lists I would rather figure out a way to sort both lists together. Is this possible?
Schwartzian变换。内置的Python排序是稳定的，所以两个1不会造成问题。
>>> l1 = [3, 2, 4, 1, 1]
>>> l2 = ['three', 'two', 'four', 'one', 'second one']
>>> zip(*sorted(zip(l1, l2)))
[(1, 1, 2, 3, 4), ('one', 'second one', 'two', 'three', 'four')]
是什么:
list1 = [3,2,4,1, 1]
list2 = ['three', 'two', 'four', 'one', 'one2']
sortedRes = sorted(zip(list1, list2), key=lambda x: x[0]) # use 0 or 1 depending on what you want to sort
>>> [(1, 'one'), (1, 'one2'), (2, 'two'), (3, 'three'), (4, 'four')]
我使用senderle给出的答案已经很长时间了，直到我发现了np.argsort。
下面是它的工作原理。
# idx works on np.array and not lists.
list1 = np.array([3,2,4,1])
list2 = np.array(["three","two","four","one"])
idx = np.argsort(list1)
list1 = np.array(list1)[idx]
list2 = np.array(list2)[idx]
我觉得这个解决方案更直观，而且效果很好。本:
def sorting(l1, l2):
# l1 and l2 has to be numpy arrays
idx = np.argsort(l1)
return l1[idx], l2[idx]
# list1 and list2 are np.arrays here...
%timeit sorting(list1, list2)
100000 loops, best of 3: 3.53 us per loop
# This works best when the lists are NOT np.array
%timeit zip(*sorted(zip(list1, list2)))
100000 loops, best of 3: 2.41 us per loop
# 0.01us better for np.array (I think this is negligible)
%timeit tups = zip(list1, list2); tups.sort(); zip(*tups)
100000 loops, best for 3 loops: 1.96 us per loop
即使np。argsort不是最快的，我发现它更容易使用。
我想扩展open jfs的答案，这对我的问题很有效:按第三个装饰列表排序两个列表:
我们可以以任何方式创建装饰列表，但在本例中，我们将从两个原始列表之一的元素创建它，我们想要排序:
# say we have the following list and we want to sort both by the algorithms name
# (if we were to sort by the string_list, it would sort by the numerical
# value in the strings)
string_list = ["0.123 Algo. XYZ", "0.345 Algo. BCD", "0.987 Algo. ABC"]
dict_list = [{"dict_xyz": "XYZ"}, {"dict_bcd": "BCD"}, {"dict_abc": "ABC"}]
# thus we need to create the decorator list, which we can now use to sort
decorated = [text[6:] for text in string_list]
# decorated list to sort
>>> decorated
['Algo. XYZ', 'Algo. BCD', 'Algo. ABC']
现在，我们可以应用jfs的解决方案对两个列表按第三个排序
# create and sort the list of indices
sorted_indices = list(range(len(string_list)))
sorted_indices.sort(key=decorated.__getitem__)
# map sorted indices to the two, original lists
sorted_stringList = list(map(string_list.__getitem__, sorted_indices))
sorted_dictList = list(map(dict_list.__getitem__, sorted_indices))
# output
>>> sorted_stringList
['0.987 Algo. ABC', '0.345 Algo. BCD', '0.123 Algo. XYZ']
>>> sorted_dictList
[{'dict_abc': 'ABC'}, {'dict_bcd': 'BCD'}, {'dict_xyz': 'XYZ'}]
算法解决方案:
list1 = [3,2,4,1, 1]
list2 = ['three', 'two', 'four', 'one', 'one2']
lis = [(list1[i], list2[i]) for i in range(len(list1))]
list1.sort()
list2 = [x[1] for i in range(len(list1)) for x in lis if x[0] == i]
输出:—>输出速度:0.2s
>>>list1
>>>[1, 1, 2, 3, 4]
>>>list2
>>>['one', 'one2', 'two', 'three', 'four']
