按连续顺序对元组列表进行排序

Question

我想按连续顺序对元组列表进行排序，因此每个元组的第一个元素等于前一个元素的最后一个元素。

例如：

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

我开发了这样的搜索：

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)    

有没有pythonic方式来实现这样的秩序？ 并且有一种方法可以“就地”＃34; （只有一个清单）？

在我的问题中，输入可以使用所有元组以循环方式重新排序，因此选择的第一个元素并不重要。

Answer 1

假设list中的元组为圆形，您可以使用dict在 O（n）的复杂度内实现它：

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持有的最终价值为：

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

Answer 2

如果你不害怕浪费一些内存，你可以创建一个字典start_dict，其中包含起始整数作为键，元组作为值，并执行以下操作：

tpl = [(10, 7), (4, 9), (13, 4), (7, 13), (9, 10)]
start_dict = {item[0]: item for item in tpl}

start = tpl[0][0]
res = []
while start_dict:
    item = start_dict[start]
    del start_dict[start]
    res.append(item)
    start = item[-1]

print(res)

如果两个元组以相同的数字开头，你将丢失其中一个...如果不是所有的起始编号都被使用，则循环将不会终止。

但也许这是要建立的东西。

Answer 3

实际上，关于您打算将什么作为输出有很多问题，如果输入列表具有无效结构以满足您的需要，那该怎么办。

假设您有一对输入，其中每个数字仅包含两次。因此，我们可以将此类输入视为图形，其中数字是节点，每对都是边缘。据我了解你的问题，你认为这个图是循环的，看起来像这样：

10 - 7 - 13 - 4 - 9 - 10 (same 10 as at the beginning)

这表明您可以缩小列表以将图表存储到[10, 7, 13, 4, 9]。以下是对输入列表进行排序的脚本：

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

# sorting and archiving
first = input[0][0]
last = input[0][1]
output_in_place = [first, last]

while last != first:
    for item in input:
        if item[0] == last:
            last = item[1]
            if last != first:
                output_in_place.append(last)

print(output_in_place)

# output
output = []
for i in range(len(output_in_place) - 1):
    output.append((output_in_place[i], output_in_place[i+1]))
output.append((output_in_place[-1], output_in_place[0]))

print(output)

Answer 4

我首先要创建一个表格

的字典

{first_value: [list of tuples with that first value], ...}

然后从那里开始工作：

from collections import defaultdict

chosen_tuples = input[:1]  # Start from the first

first_values = defaultdict()
for tup in input[1:]:
    first_values[tup[0]].append(tup)

while first_values:  # Loop will end when all lists are removed
    value = chosen_tuples[-1][1]  # Second item of last tuple
    tuples_with_that_value = first_values[value]
    chosen_tuples.append(tuples_with_that_value.pop())
    if not chosen_with_that_value:
        del first_values[value]  # List empty, remove it

Answer 5

你可以试试这个：

mapData = function() {
        "combinationData": {
            "combinationName": "2_2",
            "dataGroups": {
                "tableType": "2",

                "twoAxisData": {
                    for (var i = 0; i < 5; i++) { //Need to add a loop like this to iterate till its value
                        "vAxis": $scope.vAxis,
                        "dataValue": {
                            "hAxis": "2",
                            "dataValue": $scope.hrows
                        },
                    }

                }
            }

<强>输出：

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

output = [input[0]]  # output contains the first element of input
temp = input[1:]  # temp contains the rest of elements in input

while temp:
    item = [i for i in temp if i[0] == output[-1][1]].pop()  # We compare each element with output[-1]
    output.append(item)  # We add the right item to output
    temp.remove(item)  # We remove each handled element from temp

Answer 6

这是一个（效率低于字典版本）变体，其中列表就地更改：

tpl = [(10, 7), (4, 9), (13, 4), (7, 13), (9, 10)]

for i in range(1, len(tpl)-1):   # iterate over the indices of the list
    item = tpl[i]
    for j, next_item in enumerate(tpl[i+1:]):  # find the next item 
                                               # in the remaining list
        if next_item[0] == item[1]:
            next_index = i + j
            break
    tpl[i], tpl[next_index] = tpl[next_index], tpl[i]  # now swap the items

这是同一个想法的更高效版本：

tpl = [(10, 7), (4, 9), (13, 4), (7, 13), (9, 10)]
start_index = {item[0]: i for i, item in enumerate(tpl)}

item = tpl[0]
next_index = start_index[item[-1]]
for i in range(1, len(tpl)-1):
    tpl[i], tpl[next_index] = tpl[next_index], tpl[i]
    # need to update the start indices:
    start_index[tpl[next_index][0]] = next_index
    start_index[tpl[i][0]] = i
    next_index = start_index[tpl[i][-1]]
print(tpl)

列表就地更改;字典只包含元组的起始值及其在列表中的索引。

Answer 7

以下是使用sorted函数和自定义键函数的强大解决方案：

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

def consec_sort(lst):
    def key(x):
        nonlocal index
        if index <= lower_index:
            index += 1
            return -1
        return abs(x[0] - lst[index - 1][1])
    for lower_index in range(len(lst) - 2):
        index = 0
        lst = sorted(lst, key=key)
    return lst

output = consec_sort(input)
print(output)

未修改原始列表。请注意，对于sorted长度为5的列表，input被调用3次。在每次调用中，正确放置一个额外的元组。第一个元组保持其原始位置。

我使用了nonlocal关键字，这意味着此代码仅适用于Python 3（可以使用global代替使其成为合法的Python 2代码。）

Answer 8

我的两分钱：

def match_tuples(input):
    # making a copy to not mess up with the original one
    tuples = input[:]          # [(10,7), (4,9), (13, 4), (7, 13), (9, 10)]
    last_elem = tuples.pop(0)  # (10,7)

    # { "first tuple's element": "index in list"}
    indexes = {tup[0]: i for i, tup in enumerate(tuples)} # {9: 3, 4: 0, 13: 1, 7: 2}

    yield last_elem  # yields de firts element

    for i in range(len(tuples)):
        # get where in the list is the tuple which first element match the last element in the last tuple
        list_index = indexes.get(last_elem[1])
        last_elem = tuples[list_index] # just get that tuple
        yield last_elem

<强>输出：

input = [(10,7), (4,9), (13, 4), (7, 13), (9, 10)]
print(list(match_tuples(input)))
# output: [(10, 7), (7, 13), (13, 4), (4, 9), (9, 10)]

Answer 9

要获得O(n)算法，需要确保一个算法不对数组进行双循环。实现此目的的一种方法是将已处理的值保留在某种查找表中（dict将是一个不错的选择。）

例如类似的东西（我希望内联评论很好地解释了这个功能）。这会就地修改列表，并应避免在列表上进行不必要的（甚至是隐式的）循环：

inp = [(10, 7), (4, 9), (13, 4), (7, 13), (9, 10)]

# A dictionary containing processed elements, first element is
# the key and the value represents the tuple. This is used to
# avoid the double loop
seen = {}

# The second value of the first tuple. This must match the first
# item of the next tuple
current = inp[0][1]

# Iteration to insert the next element
for insert_idx in range(1, len(inp)):
    # print('insert', insert_idx, seen)
    # If the next value was already found no need to search, just
    # pop it from the seen dictionary and continue with the next loop
    if current in seen:
        item = seen.pop(current)
        inp[insert_idx] = item
        current = item[1]
        continue

    # Search the list until the next value is found saving all
    # other items in the dictionary so we avoid to do unnecessary iterations
    # over the list.
    for search_idx in range(insert_idx, len(inp)):
        # print('search', search_idx, inp[search_idx])
        item = inp[search_idx]
        first, second = item
        if first == current:
            # Found the next tuple, break out of the inner loop!
            inp[insert_idx] = item
            current = second
            break
        else:
            seen[first] = item