How to check if list contains consecutive numbers Python

Question

I have below list:

l = [1, 2, 3, 4, 10, 11, 12]

By looking at the above list, we can say it’s not consecutive. In order to find that using python, we can use below line of code:

print(sorted(l) == list(range(min(l), max(l)+1)))
# Output: False

This gives output False because 5, 6, 7, 8, 9 are missing. I want to further extend this functionality to check how many integers are missing. Also to note, no duplicates are allowed in the list. For ex:

l = [1, 2, 3, 4, 10, 11, 12, 14]

output of above list should be [5, 1] because 5 integers are missing between 4 and 10 and 1 is missing between 12 and 14

Asked By: S Andrew

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Source

Answer 1

This answers the question from the comments of how to find out how many are missing at multiple points in the list. Here we assume the list arr is sorted and has no duplicates:

it1, it2 = iter(arr), iter(arr)
next(it2, None) # advance past the first element
counts_of_missing = [j - i - 1 for i, j in zip(it1, it2) if j - i > 1]
total_missing = sum(counts_of_missing)

The iterators allow us to avoid making an extra copy of arr. If we can be wasteful of memory, omit the first two lines and change zip(it1, it2) to zip(arr, arr[1:]):

counts_of_missing = [j - i - 1 for i, j in zip(arr, arr[1:]) if j - i > 1]

Answered By: Steven Rumbalski

Answer 2

I think this will help you

L = [1, 2, 3, 4, 10, 11, 12, 14]
C = []
D = True
for _ in range(1,len(L)):
    if L[_]-1!=L[_-1]:
        C.append(L[_]-L[_-1]-1)
        D = False
print(D)
print(C)

Here I have checked that a number at ith index minus 1 is equal to its previous index. if not then D = false and add it to list

Answered By: Bhaskar Gupta

Answer 3

here is my attempt:

from itertools import groupby

l = [1, 2, 3, 4, 10, 11, 12, 14]

not_in = [i not in l for i in range(min(l),max(l)+1)]
missed = [sum(g) for i,g in groupby(not_in) if i]

>>> missed
'''
[5, 1]

Answered By: SergFSM

How to check if list contains consecutive numbers Python

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