如何按元组的第一个元素对元组列表进行排序？

Question

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:

[(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:

[(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”。我正在通过迭代和使用这些键移动列表中的对象（因此，如果我想在x轴上移动某个对象，我必须遍历所有列，这可能会导致一个可怕的问题，我无法解决）。

我怎样才能用这种方法对元组排序？以下内容：

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

Answer 1

这就可以了

import operator
self.gridKeys.sort(key=operator.itemgetter(1))

Answer 2

而福特西的解决方案是正确的严格意义上，它正是你所要求的标题。这可能不是你想要的。最好是通过相反的元组来进行排序。

self.gridKeys.sort(key=lambda x:tuple(reversed(x)))

这就迫使你有如下命令：

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

而不是让第一个元素无序如下：

[(4, 0), (9, 0), (6, 0), (1, 0), (3, 0), ...]

这就是我在使用时得到的：

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

默认情况下，Python从左到右进行词典排序。有效地反转元组使Python从右到左进行词典排序。

Answer 3

可以使用sort函数的key参数对元组进行排序。参数key的作用是产生一个值，该值必须用于比较两个对象。所以，在您的例子中，如果您希望sort只使用元组中的第一个元素，那么您可以这样做

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

如果只想使用元组中的第二个元素，那么

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

sort函数将把列表中的每个元素传递给作为参数传递给key的lambda函数，并使用它返回的值来比较列表中的两个对象。所以，在你的例子中，假设你在列表中有两项

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

如果你想按第二个元素排序，你可以

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

首先，它将(1, 3)传递给lambda函数，lambda函数返回索引1处的元素，即3，该函数将在比较期间表示此元组。同样，2将用于第二个元组。