Python - How do I split a nested list into further list?

Question

I have a nested list that is similar to this and sorted on time ascending:

[[TIME, 'b', 0],
[TIME, 'b', 1],
[TIME, 'b', 1],
[TIME, 'b', 1],
[TIME, 'b', 10],
[TIME, 'b', 0],
[TIME, 'b', 1],
[TIME, 'b', 1],
[TIME, 'b', 10],
[TIME, 'b', 0],
[TIME, 'b', 1],
[TIME, 'b', 1],
[TIME, 'b', 1],
[TIME, 'b', 1],
[TIME, 'b', 10]]

The third item in each list is either a 0, 1 or 10. 0 represents the start, 1 represents moving and 10 represents the finish. How can I split this into a further nested list so I would end up with a list of journeys similar to this:

[[[TIME, 'b', 0],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 10]],
[[TIME, 'b', 0],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 10]],
[[TIME, 'b', 0],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 10]]]

Ive been trying to get my head around these but have not been successful: http://stackoverflow.com/questions/5936771/python-how-to-split-a-list-into-an-unknown-number-of-smaller-lists-based-on-a-d http://stackoverflow.com/questions/949098/python-split-a-list-based-on-a-condition — 7wonders, Aug 26 '12 at 23:47
That first link is a good one, and has several viable solutions. Maybe the simplest nontrivial one to start with is the list comprehension one: make a list of the indices which have elements with the last term equal to zero, and then make a new list slicing from each index to the next. — DSM, Aug 26 '12 at 23:54
You can just iterate over the list and add the elements to the sublist until you encounter an item with `10` (like shown in the first of your links). — Felix Kling, Aug 26 '12 at 23:55
I would suggest that you pay attention to the answer by [Cedric Julien](http://stackoverflow.com/a/5936831/577088). The other answers have merits, but they aren't as simple and clear. — senderle, Aug 26 '12 at 23:57

Levon · Accepted Answer · 2012-08-27T04:28:10.373

4

Probably there's a more pythonic way to do this, but this is functional, straight forward and should be easy to follow. biglist holds the initial data.

newlist = []
sublist = []
for i in biglist:
    sublist.append(i)
    if i[2] == 10:
        newlist.append(sublist)
        sublist = []

gives:

[[[TIME, 'b', 0],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 10]],
[[TIME, 'b', 0],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 10]],
[[TIME, 'b', 0],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 1],[TIME, 'b', 10]]]

edited Aug 27 '12 at 04:28

answered Aug 26 '12 at 23:57

Levon

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1

Well, this was one of the first answers so it was the first I tried and it works perfectly. – 7wonders Aug 27 '12 at 00:06
It does seem to add an extra set of opening and closing [] though around the main list – 7wonders Aug 27 '12 at 00:15
1

Upvoted because there's a lot to be said for straightforwardness. Even a non-Python programmer can look at that and figure out what's happening. More importantly, you'll be able to figure it out yourself when you come back next year to maintain it. – Kirk Strauser Aug 27 '12 at 00:15
for some reason I seem to get an extra [ at the beginning and ] at the end so instead of just three its four – 7wonders Aug 27 '12 at 00:19
@Levon: I can reproduce the extra nesting from this code, FWIW. – DSM Aug 27 '12 at 00:22
1

But I fully agree with Kirk, I can actually read this, understand whats happening and learn from it. thikonom's answer below works perfectly too and if I could share my accept with him I would. – 7wonders Aug 27 '12 at 00:22
@DSM Never mind ... my problem was related to *how* I was printing out the result for verification .. "facepalm" – Levon Aug 27 '12 at 00:30

score 1 · Answer 2 · answered Aug 26 '12 at 23:59

1

#x is your initial list
k, new_list = 0, []
for i,j in enumerate(x):
    if j[2] == 10:
        new_list.append(x[k:i+1])
        k = i+1

answered Aug 26 '12 at 23:59

thikonom

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score 1 · Answer 3 · answered Aug 27 '12 at 00:04

1

Contrary to the other solution, here we test on whether the journey begins (d[2] == 0) instead of when it ends:

output = []
for d in data:
    if d[2] == 0:
        # Starting a new journey: we add a new list
        output.append([d])
    else:
        # Continuing a journey: we extend the last list
        output[-1].extend([d])

Of course, this gonna fail if the first journey never starts (you'll get an IndexError from the else statement)...

answered Aug 27 '12 at 00:04

Pierre GM

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It seems to miss the closing of the list on the 10's – 7wonders Aug 27 '12 at 00:11
As stated, you don't close the list on a 10, you start a new one (therefore, closing the previous one) on a 0. – Pierre GM Aug 27 '12 at 00:14
I like this one, although I'd use `.append(d)` instead of `.extend([d])`. – georg Aug 27 '12 at 08:16

score 0 · Answer 4 · answered Aug 26 '12 at 23:56

0

The most obvious solution, if I have understood the question properly:

def iter_journeys(lst):
    for e in lst:
        act = e[2]

        if act == 0:
            journey = []

        journey.append(e)

        if act == 10:
            yield journey

answered Aug 26 '12 at 23:56

Eldar Abusalimov

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score 0 · Answer 5 · answered Aug 26 '12 at 23:58

0

I'm sure there's a more efficient way of doing this, but here's a basic solution:

def create_journeys(paths):
  journeys = [[]]

  for path in paths:
    journeys[-1].append(path)

    if path[-1] == 10:
      journeys.append([])

  return journeys

answered Aug 26 '12 at 23:58

Blender

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Python - How do I split a nested list into further list?

5 Answers5