如何按元组的第一个元素对列表进行排序? [英] How to sort a list of tuples by their first element?

问题描述

我有一个元组列表:

self.gridKeys = self.gridMap.keys() # The keys of the instance of the GridMap (It returns the product of every possible combination of positions in the specified grid, in tuples.)
print self.gridKeys

self.gridKeys:

self.gridKeys:

[(7, 3), (6, 9), (0, 7), (1, 6), (3, 7), (2, 5), (8, 5), (5, 8), (4, 0), (9, 0), (6, 7), (5, 5), (7, 6), (0, 4), (1, 1), (3, 2), (2, 6), (8, 2), (4, 5), (9, 3), (6, 0), (7, 5), (0, 1), (3, 1), (9, 9), (7, 8), (2, 1), (8, 9), (9, 4), (5, 1), (7, 2), (1, 5), (3, 6), (2, 2), (8, 6), (4, 1), (9, 7), (6, 4), (5, 4), (7, 1), (0, 5), (1, 0), (0, 8), (3, 5), (2, 7), (8, 3), (4, 6), (9, 2), (6, 1), (5, 7), (7, 4), (0, 2), (1, 3), (4, 8), (3, 0), (2, 8), (9, 8), (8, 0), (6, 2), (5, 0), (1, 4), (3, 9), (2, 3), (1, 9), (8, 7), (4, 2), (9, 6), (6, 5), (5, 3), (7, 0), (6, 8), (0, 6), (1, 7), (0, 9), (3, 4), (2, 4), (8, 4), (5, 9), (4, 7), (9, 1), (6, 6), (5, 6), (7, 7), (0, 3), (1, 2), (4, 9), (3, 3), (2, 9), (8, 1), (4, 4), (6, 3), (0, 0), (7, 9), (3, 8), (2, 0), (1, 8), (8, 8), (4, 3), (9, 5), (5, 2)]

排序后:

self.gridKeys = self.gridMap.keys() # The keys of the instance of the GridMap (It returns the product of every possible combination of positions in the specified grid, in tuples.)
self.gridKeys.sort() # They're dicts, so they need to be properly ordered for further XML-analysis.
print self.gridKeys

self.gridKeys:

self.gridKeys:

[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (5, 0), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (5, 7), (5, 8), (5, 9), (6, 0), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (6, 7), (6, 8), (6, 9), (7, 0), (7, 1), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 7), (7, 8), (7, 9), (8, 0), (8, 1), (8, 2), (8, 3), (8, 4), (8, 5), (8, 6), (8, 7), (8, 8), (8, 9), (9, 0), (9, 1), (9, 2), (9, 3), (9, 4), (9, 5), (9, 6), (9, 7), (9, 8), (9, 9)]

每个元组的第一个元素是"x"，第二个元素是"y".我通过迭代并使用这些键在列表中移动对象(因此，如果我想在x轴上移动某物，则必须遍历所有列，这可能会导致一个可怕的问题，我不是能够解决).

The first element of each tuple is the "x", and the second the "y". I'm moving objects in a list through iteration and using these keys (So, if I want to move something in the x axis, I have to go through all the column, and that might be causing a horrid problem that I'm not being able to solve).

如何以这种方式对元组进行排序?:

How can I sort the tuples in this way?:

[(1, 0), (2, 0), (3, 0), (4, 0), (5, 0), ...]

推荐答案

您可以使用sort函数的key参数对元组进行排序. key参数的功能是提供一个必须用于比较两个对象的值.因此，在您的情况下，如果您希望sort仅使用元组中的第一个元素，则可以执行类似的操作

You can use the key parameter of the sort function, to sort the tuples. The function of key parameter, is to come up with a value which has to be used to compare two objects. So, in your case, if you want the sort to use only the first element in the tuple, you can do something like this

self.gridKeys.sort(key=lambda x: x[0])

如果您只想使用元组中的第二个元素，那么

If you want to use only the second element in the tuple, then

self.gridKeys.sort(key=lambda x: x[1])

sort函数会将列表中的每个元素传递给作为参数传递给key的lambda函数，它将使用它返回的值来比较列表中的两个对象.因此，就您而言，假设您在列表中有两个这样的项目

sort function will pass each and every element in the list to the lambda function you pass as parameter to key and it will use the value it returns, to compare two objects in the list. So, in your case, lets say you have two items in the list like this

data = [(1, 3), (1, 2)]

如果您要按第二个元素排序，则可以

and if you want to sort by the second element, then you would do

data.sort(key=lambda x: x[1])

首先，它将(1, 3)传递给lambda函数，该函数返回索引为3的元素，该元素为3，它将在比较期间表示该元组.同样，2将用于第二个元组.

First it passes (1, 3) to the lambda function which returns the element at index 1, which is 3 and that will represent this tuple during the comparison. The same way, 2 will be used for the second tuple.

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