如何从包含许多已排序列表的列表中获取排序列表？

Question

For example, I have a list with many many elements

x = [[1, 2, 3], [3, 4, 5],[2, 3, 4], [4, 7, 8], ...]

each element has already sorted.

How can I get a list named y that is from x 's elements and sorted it?

Answer 1

If I understand your question correctly, you should look into heapq.merge :

>>> import heapq
>>> lists = x = [[1, 2, 3], [3, 4, 5], [2, 3, 4], [4, 7, 8]]
>>> list(heapq.merge(*lists))
[1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 7, 8]

Answer 2

Here is a simple solution without any library:

x = [[1,2,3], [3,4,5], [2,3,4], [4,7,8]]

print(sorted([item for sublist in x for item in sublist]))

Returns: [1, 2, 2, 3, 3, 3, 4, 4, 4, 5, 7, 8]

Answer 3

If you just want to sort use sorted ,try this

x = [[1, 2, 3], [3, 4, 5],[2, 3, 4], [4, 7, 8]]
y = sorted(x)
print y

If you want to get all elements in sorted way,try this

x = [[1, 2, 3], [3, 4, 5],[2, 3, 4], [4, 7, 8]]
y = sorted([item for sublist in x for item in sublist])
print y

Answer 4

You need a "merge" operation: just start with two lists and keep taking the element that is smaller; when one list is completed copy the eventually remaining elements from the other one...

def merge(A, B):
    result = []
    ia = ib = 0
    while ia < len(A) and ib < len(B):
        if a[ia] < b[ib]:
            result.append(A[ia])
            ia += 1
        else:
            result.append(B[ib])
            ib += 1
    # copy "tail"
    while ia < len(A):
        result.append(A[ia])
        ia += 1
    while ib < len(B):
        result.append(B[ib])
        ib += 1
    return result

If you have a collection of N lists just merge them in pairs and you will end up with half of them. Repeat until you get just one list.

Answer 5

I am going to give you some pointers.

First, define a function merge(x, y) that gets two sorted lists and returns a combined sorted list.

Then, run merge on couples

[merge(lst[i], lst[i+1] ) for i in range(0,len(lst),2) ] 

This process should be repeated as long as the list has more than two elements, and if the number of elements is odd, append [] to it

Answer 6

Method 1:

import heapq
x = [[1, 2, 3], [3, 4, 5],[2, 3, 4], [4, 7, 8]]
y = list(heapq.merge(*x))

Method 2:

import itertools
y = sorted(itertools.chain(*x))