python之有没有办法以特定方式订购此列表

所以我有这个列表:

[7, 31, 31, 61, 61, 79, 29, 79, 29, 103, 37, 103, 37, 47, 47, 53, 53, 89, 5, 89, 5, 13, 13, 83, 83, 101, 53, 101, 17, 53, 11, 17, 11, 59, 17, 59, 17, 41, 41, 79, 79, 97, 3, 97, 3, 41, 19, 41, 19, 53, 53, 67, 29, 67, 29, 73, 23, 73, 23, 43, 43, 71] 

我想知道是否有一种方法可以对它们进行排序，例如一对中的最后一个元素是第二对中的第一个，就像这样:

[7, 31, 31, 61, 61, 79, 79, 29, 29, 103, 103, 37 ....] 
[pair1] [pair2] [pair3] [pair4] [pair5]  [pair6] ... 

有办法吗？

我试过 for 循环，但它的时间复杂度非常高。

这是我的代码:

 
lis=[7, 31, 31, 61, 61, 79, 29, 79, 29, 103, 37, 103, 37, 47, 47, 53, 53, 89, 5, 89, 5, 13, 13, 83, 83, 101, 53, 101, 17, 53, 11, 17, 11, 59, 17, 59, 17, 41, 41, 79, 79, 97, 3, 97, 3, 41, 19, 41, 19, 53, 53, 67, 29, 67, 29, 73, 23, 73, 23, 43, 43, 71] 
for i in range(0, len(lis)-4,4): 
    if lis[i]==lis[i+2]: 
        p=lis[i] 
        lis[i]=lis[i+1] 
        lis[i+1]=p 
    elif lis[i]==lis[i+3]: 
        p = lis[i] 
        lis[i] = lis[i + 1] 
        lis[i + 1] = p 
        p=lis[i+2] 
        lis[i+2]=lis[i+3] 
        lis[i+3]=p 
    elif lis[i+1]==lis[i+3]: 
        p=lis[i+2] 
        lis[i+2]=lis[i+3] 
        lis[i+3]=p 
print(lis) 

请您参考如下方法:

您可以这样做，例如根据原始 list 中的 index 进行排序，

>>> x 
[7, 31, 31, 61, 61, 79, 29, 79, 29, 103, 37, 103, 37, 47, 47, 53, 53, 89, 5, 89, 5, 13, 13, 83, 83, 101, 53, 101, 17, 53, 11, 17, 11, 59, 17, 59, 17, 41, 41, 79, 79, 97, 3, 97, 3, 41, 19, 41, 19, 53, 53, 67, 29, 67, 29, 73, 23, 73, 23, 43, 43, 71] 
 
# the `sorted` function takes a `key` function, that one can use to manipulate how the sort should be based on 
# In this case, you could use `index` of the `element` in the `list` `x`, 
# So, while the sort goes through each element, it check against the index in the original list, which is `x, thus it aligns each item based on the `index`. 
>>> sorted(x, key=x.index) # note that you have more than a pair, so :) 
[7, 31, 31, 61, 61, 79, 79, 79, 79, 29, 29, 29, 29, 103, 103, 37, 37, 47, 47, 53, 53, 53, 53, 53, 53, 89, 89, 5, 5, 13, 13, 83, 83, 101, 101, 17, 17, 17, 17, 11, 11, 59, 59, 41, 41, 41, 41, 97, 97, 3, 3, 19, 19, 67, 67, 73, 73, 23, 23, 43, 43, 71] 

如果您真的喜欢对，那么，

>>> data = [] 
>>> for k in sorted(x, key=x.index): # `using direct x.index instead of useless lambda(i previously used) in this case, as @Austin suggested 
...   if data.count(k) < 2: # add the element until it reaches count 2, which is `pair`, which is what you asked for 
...     data.append(k) 
...  
>>> data 
[7, 31, 31, 61, 61, 79, 79, 29, 29, 103, 103, 37, 37, 47, 47, 53, 53, 89, 89, 5, 5, 13, 13, 83, 83, 101, 101, 17, 17, 11, 11, 59, 59, 41, 41, 97, 97, 3, 3, 19, 19, 67, 67, 73, 73, 23, 23, 43, 43, 71] 

推理:

list.index will return the first matching index. Say you have a list, x = [1,2,1,3] and x.index(1) will always return 0. Thus it will be sorted differently if one use list.index as the keyfunc for sorting