Creating lists within a list Python

Question

Given a list of integers A=[1,2,3,4,5,6,7,9] and a list p=[4,5,9] , how can I separate the values in A so that if they do not appear in p, they will be separated into a sub-list dictated by the position of the element of p in A. For example, in this case the output should be A=[[1, 2, 3], 4, 5, [6, 7, 8], 9].

s=25
# make it a string
s = str(s)

output = []
last = None

for c in A:
    if last is None:
        output.append(c)
    elif (last in s) == (c in s):
        output[-1] = output[-1] + c
    else:
        output.append(c)
    last = c

output # ['1', '2', '34', '5', '67']

This is a similar version of the problem that involves a list of strings.

Reference : Joining elements in Python list

Hey there, have you written any code to attempt this, that we can look at? — dsvp9xyjsqmfvi8p
– dsvp9xyjsqmfvi8p, Commented Jan 21, 2020 at 17:27
Questions seeking debugging help ("why isn't this code working?") must include the desired behavior, a specific problem or error and the shortest code necessary to reproduce it in the question itself. Questions without a clear problem statement are not useful to other readers. See: How to create a Minimal, Complete, and Verifiable example. — Sean Pianka
– Sean Pianka, Commented Jan 21, 2020 at 17:32
I saw a similar question on: stackoverflow.com/questions/59813650/… and was interested in playing around with this idea — bgmn
– bgmn, Commented Jan 21, 2020 at 17:33
It would be really helpful if you could show what you have coded so far. Edit your question to include it before this is getting closed. — Austin
– Austin, Commented Jan 21, 2020 at 17:35

bgmn · Accepted Answer · 2020-01-31 14:24:17Z

2

You could do this by keeping track of all the elements found so far, and resetting the temporary list when you find an element in p:

A=[1,2,3,4,5,6,7,9]
p=[4,5,9]

def join_elements(A, p):
    curr = []
    for i in A: # Loop over possible values
        if i in p:
            if curr: # yield and reset current list
                yield curr 
                curr = []
            yield i # yield value from p 
        else: # Add to current list
            curr.append(i)
    if curr: # if we ended with an element that was not in p
        yield curr

print(list(join_elements(A, p)))

# [[1, 2, 3], 4, 5, [6, 7], 9]

edited Jan 31, 2020 at 14:24

bgmn

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answered Jan 21, 2020 at 19:48

rassar

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1 Comment

rassar Over a year ago

@Newbie123 Updated my answer.

Rakesh tripathi · Accepted Answer · 2020-01-21 20:08:59Z

0

Here you could find a general solution:

sub_list = [4, 5, 9]

temp = []
result = []
for i in range(1, 10):

    if i not in sub_list:
        temp.append(i)
        if temp not in result:
            result.append(temp)

    else:
        if len(temp) != 0:
            temp = []
        result.append(i)

print(result)



Output: [[1, 2, 3], 4, 5, [6, 7, 8], 9]

I believe this is what you are looking for.

edited Jan 21, 2020 at 20:08

answered Jan 21, 2020 at 19:43

Rakesh tripathi

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2 Comments

rassar Over a year ago

Shouldn't it be for i in A? (A is not guaranteed to be range(10)). Also, this fails if A ends with an element not in p.

Rakesh tripathi Over a year ago

Thanks @rassar - I have made some corrections and I believe it would work for all types of list. Please correct me if still there is any scope of improvement.

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