Adding same values to an array

Question

I have this array:

            countOverlaps = [numA, numB, numC, numD, numE, numF, numG, numH, numI, numJ, numK, numL]

and then I condense this array by getting rid of all 0 values:

            countOverlaps = [x for x in countOverlaps if x != 0]

When I do this, I get an output like this: [2, 1, 3, 2, 3, 1, 1]

Which is what it should, so that makes sense. Now I want to add values to the array so that each number adds itself to the array the number of times it appears.

Like this:

Original: [2, 1, 3, 2, 3, 1, 1]

What I want: [2,2,1,3,3,3,2,2,3,3,3,1,1]

Is something like this possible in python?

Thanks

It looks like you want us to write some code for you. While many users are willing to produce code for a coder in distress, they usually only help when the poster has already tried to solve the problem individually. A good way to show this effort is to include a Minimal, complete, verifiable example. Check the intro tour you finished before posting, especially How to Ask. — Prune
– Prune, Commented Oct 19, 2018 at 1:15
It sounds like you are trying to count the number of times a thing occurs. If so, do you really want to append the number of occurrences to a list? Seems a dictionary would be more appropriate. If so, your output would be this {1: 3, 2: 2, 3: 2} — user9074332
– user9074332, Commented Oct 19, 2018 at 1:19

shaik moeed · Accepted Answer · 2018-10-19 01:39:40Z

1

Updated

Please check below:

>>> a = [2, 1, 3, 2, 3, 1, 1]
>>> [b for b in a for _ in range(b)]
[2, 2, 1, 3, 3, 3, 2, 2, 3, 3, 3, 1, 1]

edited Oct 19, 2018 at 1:39

answered Oct 19, 2018 at 1:28

shaik moeed

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2 Comments

A. Bhatnagar Over a year ago

Works perfectly! Thank you so much

Chiheb Nexus Over a year ago

This could be rewritten like [k for k in a for _ in range(k)]

PL200 · Accepted Answer · 2018-10-19 01:37:04Z

0

This can be done using list comprehension. So far you had:

countOverlaps = [10,25,11,0,10,6,9,0,12,6,0,6,6,11,18]
countOverlaps = [x for x in countOverlaps if x != 0]

This gives us all non=0 numbers. Then we can do what you want with the following code:

mylist = [number for number in list(set(countOverlaps)) for i in range(0, countOverlaps.count(number)) ]

This turns 'mylist' into the following output, which is what you're after:

[6, 6, 6, 6, 9, 10, 10, 11, 11, 12, 18, 25]

answered Oct 19, 2018 at 1:37

PL200

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