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".我正在通过迭代移动列表中的对象并使用这些键(因此，如果我想在 x 轴上移动某些东西，我必须遍历所有列，这可能会导致我不是能够解决).
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), ...]
您可以使用 sort 函数的 key 参数对元组进行排序.key 参数的作用是得出一个必须用来比较两个对象的值.所以，在你的情况下，如果你希望 sort 只使用元组中的第一个元素，你可以这样做
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])
如果你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您作为参数传递给 key 的 lambda 函数，它将使用它返回的值来比较两个列表中的对象.因此，在您的情况下，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)]
如果你想按第二个元素排序，那么你会这样做
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 1 处的元素，即 3 并将在比较期间表示这个元组.同样的方法，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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