Order the sub-lists in a nested list-CodePudding

I have a series of lists, and I want to combine them in a larger nested list. However, I want to order them in a certain way. I want the first sub-list to be the one whose first element is zero. Then i want the second sub-list to be the one whose first element is the same as the LAST element of the previous list.

For example, here's four sub-lists;

[0, 3], [7, 0], [3, 8], [8, 7]

I want to end up with this;

[[0, 3], [3, 8], [8, 7], [7,0]]

I can't for the life of me see the code logic in my head that would achieve this for me.

Can anyone help please?

CodePudding user response：

I think of your list as being a collection of links which are to be arranged into a chain. Here is an approach which uses @quanrama 's idea of a dictionary keyed by the first element of that link:

links = [[0, 3], [7, 0], [3, 8], [8, 7]]

d = {link[0]:link for link in links}
chain = []
i = min(d)
while d:
    link = d[i]
    chain.append(link)
    del d[i]
    i = link[1]

print(chain) #[[0, 3], [3, 8], [8, 7], [7, 0]]

CodePudding user response：

Another approach with a generator function:

links = [[0, 3], [7, 0], [3, 8], [8, 7]]

def get_path(links, *, start=0, end=0):
    linkmap = dict(links)
    key = start
    while True:
        link = linkmap[key]
        yield [key,link]
        key = link
        if link == end:
            break
        
print(list(get_path(links)))
print(list(get_path(links,start=3,end=3)))

# [[0, 3], [3, 8], [8, 7], [7, 0]]
# [[3, 8], [8, 7], [7, 0], [0, 3]]

CodePudding user response：

You can try something like this:

source = [[0, 3], [7, 0], [3, 8], [8, 7]]

# Start at 0
last_val = 0
# this will be the output
l = []
while len(l)==0 or last_val!=0:
   # Find the first value where the first element is last_val
   l.append(next(i for i in source if i[0]==last_val))
   # set last val to the second element of the list
   last_val = l[-1][1]

print(l)