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?
你可以使用值作为键对索引进行排序:
indexes = range(len(list1))
indexes.sort(key=list1.__getitem__)
要获得给定已排序索引的已排序列表:
sorted_list1 = map(list1.__getitem__, indexes)
sorted_list2 = map(list2.__getitem__, indexes)
在你的例子中，你不应该有list1, list2，而是一个单独的对列表:
data = [(3, 'three'), (2, 'two'), (4, 'four'), (1, 'one'), (1, 'one2')]
它很容易创造;在Python中很容易排序:
data.sort() # sort using a pair as a key
仅按第一个值排序:
data.sort(key=lambda pair: pair[0])
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')]
你可以使用zip()和sort()函数来实现:
Python 2.6.5 (r265:79063, Jun 12 2010, 17:07:01)
[GCC 4.3.4 20090804 (release) 1] on cygwin
>>> list1 = [3,2,4,1,1]
>>> list2 = ['three', 'two', 'four', 'one', 'one2']
>>> zipped = zip(list1, list2)
>>> zipped.sort()
>>> slist1 = [i for (i, s) in zipped]
>>> slist1
[1, 1, 2, 3, 4]
>>> slist2 = [s for (i, s) in zipped]
>>> slist2
['one', 'one2', 'two', 'three', 'four']
希望这能有所帮助
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')]
你可以使用值作为键对索引进行排序:
indexes = range(len(list1))
indexes.sort(key=list1.__getitem__)
要获得给定已排序索引的已排序列表:
sorted_list1 = map(list1.__getitem__, indexes)
sorted_list2 = map(list2.__getitem__, indexes)
在你的例子中，你不应该有list1, list2，而是一个单独的对列表:
data = [(3, 'three'), (2, 'two'), (4, 'four'), (1, 'one'), (1, 'one2')]
它很容易创造;在Python中很容易排序:
data.sort() # sort using a pair as a key
仅按第一个值排序:
data.sort(key=lambda pair: pair[0])
