CODEWITHSUNDEEP
pythonarraysalgorithm
hcsaral
hcsaral

Reputation: 63

Count longest streak of repeating numbers in a python list of integers

I have a lists of integers like mylist = [40,46,46,50,50,54,54,67,75,75,79,79,87...

The lists are strictly increasing, where if a number repeats it only repeats one time (like 46 in ..40,46,46,50..) and differences between continues repeats are always the same ( like for mylist its 4 ..40,46,46,50,50,54,54.. because 50-46 = 4 , 54-50 = 4 and so on).

I would like to find out longest streak of this doubling pattern. For example for this part of mylist there are two streaks (46,46,50,50,54,54) where the repeat streak is 3 and (75,75,79,79) where the repeat streak is 2 .

I have different lists where the differences can change (can be 4 or 5 or 6) but this doubling and increasing pattern don't change.

How can I find longest repeating streak (which is 3 for mylist example) ? I couldn't find a concise and simple solution without using a lot of for loops and if conditions.

Upvotes: 1

Views: 656

Answers (2)

DDD1
DDD1

Reputation: 450

You can do it as follows:

from itertools import groupby

aux = [sum(1 for _ in b) for a,b in groupby(mylist)]
aux2 = [sum(1 for _ in b if a != 1) for a,b in groupby(aux)]
consecutive = max(aux2)
print(consecutive)

Output

3

Where consecutive is the value you are looking for. By repeating the same operation twice, it first transforms your list into a list of integers, where each repeated value is now represented as a 2:

print(aux)
[1,2,2,2,1,1,1,...]

It then does the exact same thing if the value is not a 1, so the second list is now:

print(aux2)
[1,3,...]

And then you return the max value of that list.

Upvotes: 0

Dani Mesejo
Dani Mesejo

Reputation: 61910

You could use groupby twice and then max with len as the key:

from itertools import groupby
from operator import itemgetter

mylist = [40, 46, 46, 50, 50, 54, 54, 67, 75, 75, 79, 79, 87]

groups = ((k, sum(1 for _ in v)) for k, v in groupby(mylist))

streaks = [list(v) for k, v in groupby(groups, key=itemgetter(1)) if k == 2]

result = max(streaks, key=len)
print(result)

Output

[(46, 2), (50, 2), (54, 2)]

The first iteration, groups the numbers in consecutive equals values, the second one does it by streak size, finally max just retrieve the largest of those groups.

If you wish to unpack the result, just do:

print([n for n, count in result for _ in range(count)])

Output

[46, 46, 50, 50, 54, 54]

Upvotes: 1

Related Questions