split list by certain repeated index value

Question

I have a list of integers, in which some are consecutive numbers.

What I have:

myIntList = [21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7] etc...

What I want:

MyNewIntList = [[21,22,23,24],[0,1,2,3],[0,1,2,3,4,5,6,7]]

I want to be able to split this list by the element 0, i.e when looping, if the element is 0, to split the list into separate lists. Then, after splitting myIntList whatever number of times (based on the recurrences of finding the element 0), I want to append each 'split' or group of consecutive integers into a list within a list.

Also would I be able to do the same sort of thing with a 'list of strings' instead of integers? (Split the main string list into smaller lists based on a reoccurring element)

EDIT:

How would I go about splitting the list by consecutive numbers? There's a part in my list where it jumps from 322 to 51, there is no 0 in between. I want to split:

[[...319,320,321,322,51,52,53...]]

into

[[...319,320,321,322],[51,52,53...]]

basically, how do I split elements in a list by consecutive numbers?

Posted here: Split list of lists (integers) by consecutive order into separate lists

Padraic Cunningham · Accepted Answer · 2015-12-14T16:43:51.517

it  = iter(myIntList)
out = [[next(it)]]
for ele in it:
    if ele != 0:
        out[-1].append(ele)
    else:
        out.append([ele])

print(out)

Or in a function:

def split_at(i, l):
    it = iter(l)
    out = [next(it)]
    for ele in it:
        if ele != i:
            out.append(ele)
        else:
            yield out
            out = [ele]
    yield out

It will catch if you have a 0 at the start:

In [89]: list(split_at(0, myIntList))
Out[89]: [[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

In [90]: myIntList = [0,21, 22, 23, 24, 0, 1, 2, 3, 0, 1, 2, 3, 4, 5, 6, 7]

In [91]: list(split_at(0, myIntList))
Out[91]: [[0, 21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

Thanks for helping out @PadraicCunningham. I find definitions helpful. — Mike Issa, Dec 14 '15 at 16:35

score 3 · Answer 2 · answered Dec 14 '15 at 16:31

3

(I vaguely suspect I've done this before but I can't find it now.)

from itertools import groupby, accumulate

def itergroup(seq, val):
    it = iter(seq)    
    grouped = groupby(accumulate(x==val for x in seq))
    return [[next(it) for c in g] for k,g in grouped]

gives

>>> itergroup([21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7], 0)
[[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]
>>> itergroup([0,1,2,0,3,4], 0)
[[0, 1, 2], [0, 3, 4]]
>>> itergroup([0,0], 0)
[[0], [0]]

(That said, in practice I use the yield version of the same loop/branch that everyone else does, but I'll post the above for variety.)

answered Dec 14 '15 at 16:31

DSM

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Thanks @DSM. Variety in coding is always useful. :) – Mike Issa Dec 14 '15 at 16:39
for some reason I get an ImportError: cannot import name accumulate...Thoughts? – Mike Issa Dec 14 '15 at 19:26
Nevermind, I changed Python 2 to 3, and it worked. Thanks – Mike Issa Dec 14 '15 at 19:28

Fernando Matsumoto · Answer 3 · 2015-12-14T16:13:30.297

You can use slicing:

myIntList = [21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]
myNewIntList = []
lastIndex = 0
for i in range(len(myIntList)):
    if myIntList[i] == 0:
        myNewIntList.append(myIntList[lastIndex:i])
        lastIndex = i

myNewIntList.append(myIntList[lastIndex:])
print(myNewIntList)
# [[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

You can split strings using the str.split function:

s = 'stackoverflow'
s.split('o') # ['stack', 'verfl', 'w'] (removes the 'o's)

import re
[part for part in re.split('(o[^o]*)', s) if part] # ['stack', 'overfl', 'ow'] (keeps the 'o's)

Ravichandra · Answer 4 · 2015-12-14T16:56:17.420

1

myIntList = [21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]

new = []
m,j=0,0
for i in range(myIntList.count(0)+1):
    try:
        j= j+myIntList[j:].index(0)
        if m==j:
           j= j+myIntList[j+1:].index(0)+1



        new.append(myIntList[m:j])
        m,j=j,m+j
    except:
        new.append(myIntList[m:])
        break
print new

output

 [[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

output2

myIntList = [0,21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]

[[0, 21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

edited Dec 14 '15 at 16:56

answered Dec 14 '15 at 16:39

Ravichandra

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1

try `[0,21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]` – Padraic Cunningham Dec 14 '15 at 16:41

Bhargav Rao · Answer 5 · 2015-12-14T16:22:06.333

0

You can loop through the whole list, appending to a temp list till 0 is found. Then you again reset the temp list and continue.

>>> myIntList = [21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]
>>> newlist = [] 
>>> templist = []
>>> for i in myIntList:
...      if i==0:
...          newlist.append(templist)
...          templist = []
...      templist.append(i)
... 
>>> newlist.append(templist)
>>> newlist
[[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

and for strings you can use the same approach by using the list call

>>> s = "winterbash"
>>> list(s)
['w', 'i', 'n', 't', 'e', 'r', 'b', 'a', 's', 'h']

Also using itertools

>>> import itertools
>>> myIntList = [21,22,23,24,0,1,2,3,0,1,2,3,4,5,6,7]
>>> temp=[list(g) for k,g in itertools.groupby(myIntList,lambda x:x== 0) if not k]
>>> if myIntList[0]!=0:
...     newlist = [temp[0]] + [[0]+i for i in temp[1:]]
... else:
...     newlist = [[0]+i for i in temp]
... 
>>> newlist
[[21, 22, 23, 24], [0, 1, 2, 3], [0, 1, 2, 3, 4, 5, 6, 7]]

edited Dec 14 '15 at 16:22

answered Dec 14 '15 at 15:54

Bhargav Rao

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Thank you kindly. Very interesting technique. I was following something along the lines of this recipe here: http://stackoverflow.com/questions/14529523/python-split-for-lists the first response helped me, but I just wanted to see at least one other approach. – Mike Issa Dec 14 '15 at 16:01
1

@MikeIssa Added. Is it what you want? – Bhargav Rao Dec 14 '15 at 16:05
1

@BhargavRao: what if 0 is the first element? (On your itertools version.) – DSM Dec 14 '15 at 16:13
1

@DSM Yep. Did not see that edge case. I will change it soon. Thanks – Bhargav Rao Dec 14 '15 at 16:14
@DSM, Any thoughts how to make it work using the itertools version? – Mike Issa Dec 14 '15 at 16:20
1

@DSM I did it trivially. Please mention if there is any other better way. – Bhargav Rao Dec 14 '15 at 16:22
1

@BhargavRao, you should use a generator expression for the first, you can call `next(gen)` to get the first element – Padraic Cunningham Dec 14 '15 at 16:23
1

@Padraic Guess that'll make it similar to your answer. I'll edit it anyway. – Bhargav Rao Dec 14 '15 at 16:25
1

@BhargavRao, I mean the groupby solution, then you wont be storing multiple copies of the data – Padraic Cunningham Dec 14 '15 at 16:26
1

@PadraicCunningham Thanks, will edit in a few seconds. – Bhargav Rao Dec 14 '15 at 16:31
@ BhargavRao Thank you for going the extra mile and showing the itertools method. I was stuck because I couldn't get the 0 in my list initially, but you cleared things up exceptionally well. Kudos! – Mike Issa Dec 14 '15 at 16:32
1

@MikeIssa DSM has added a better approach. Do follow his answer. Thanks – Bhargav Rao Dec 14 '15 at 16:33
New post, if anyone can help. http://stackoverflow.com/questions/34277467/split-list-of-integers-by-consecutive-order-into-separate-lists – Mike Issa Dec 14 '15 at 22:06

score -3 · Answer 6 · answered Dec 14 '15 at 16:10

You could try:

i = 0
j = 0
loop = True
newList = []

while loop:
    try:
        i = myIntList.index(0, j)
        newList.append(myIntList[j:i])
        j = i + 1
    except ValueError as e:
        newList.append(myIntList[j:])
        loop = False

print newList
[[21, 22, 23, 24], [1, 2, 3], [1, 2, 3, 4, 5, 6, 7]]

split list by certain repeated index value

6 Answers6