按连续顺序对元组列表进行排序 [英] Sort a list of tuples in consecutive order

问题描述

我想按连续的顺序对元组列表进行排序，因此每个元组的第一个元素都等于前一个的最后一个元素.

I want to sort a list of tuples in a consecutive order, so the first element of each tuple is equal to the last element of the previous one.

例如:

input = [(10, 7), (4, 9), (13, 4), (7, 13), (9, 10)]
output = [(10, 7), (7, 13), (13, 4), (4, 9), (9, 10)]

我已经进行了这样的搜索:

I have developed a search like this:

output=[]
given = [(10, 7), (4, 9), (13, 4), (7, 13), (9, 10)]
t = given[0][0]
for i in range(len(given)):
      # search tuples starting with element t
      output += [e for e in given if e[0] == t]
      t = output[-1][-1] # Get the next element to search

print(output)    

是否有实现这种命令的Python方式? 以及一种就地"(仅包含列表)的方法?

Is there a pythonic way to achieve such order? And a way to do it "in-place" (with only a list)?

在我的问题中，可以使用所有元组以循环方式对输入进行重新排序，因此选择第一个元素并不重要.

In my problem, the input can be reordered in a circular way using all the tuples, so it is not important the first element chosen.

推荐答案

假定list中的元组是圆形的，则可以使用dict在 O(n)复杂度内实现它em>为:

Assuming your tuples in the list will be circular, you may use dict to achieve it within complexity of O(n) as:

input = [(10, 7), (4, 9), (13, 4), (7, 13), (9, 10)]
input_dict = dict(input)  # Convert list of `tuples` to dict

elem = input[0][0]  # start point in the new list

new_list = []  # List of tuples for holding the values in required order

for _ in range(len(input)):
    new_list.append((elem, input_dict[elem]))
    elem = input_dict[elem]
    if elem not in input_dict:
        # Raise exception in case list of tuples is not circular
        raise Exception('key {} not found in dict'.format(elem))

new_list保留的最终值将是:

>>> new_list
[(10, 7), (7, 13), (13, 4), (4, 9), (9, 10)]

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