How to check if there are 3 or 4 or 5 or 6 consecutive numbers in a list with 6 items?
I’m using python.
I tried to generate combinations of numbers 1-42. I have 5million+ combinations. I am trying to reduce the number of combinations by removing the ones that have 3-6 consecutive numbers.
Given lists:
(5,8,12,28,29,30) has 3 consecutive
(1,8,9,10,11,23) has 4 consecutive
(2,12,13,14,15,16) has 5 consecutive
(3,4,5,6,7,8) has 6 consecutive
(3,9,11,14,15,21) has 2 consecutive
(2,5,7,12,21,34)
(3,5,8,10,12,34)
By removing the lists that have 3-6 consecutive numbers the output should be:
(3,9,11,14,15,21)
(2,5,7,12,21,34)
(3,5,8,10,12,34)
Here's what you're asking for:
import itertools
def maxRunSize(t):
return 1+max(len(tuple(g)) if k else 0 for k,g in itertools.groupby(zip(t, t[1:]), key=lambda t: t[1]-t[0]==1))
Output:
In [87]: maxRunSize((5,8,12,28,29,30))
Out[87]: 3
In [88]: maxRunSize((1,8,9,10,11,23))
Out[88]: 4
In [89]: maxRunSize((2,12,13,14,14,16))
Out[89]: 3
In [90]: maxRunSize((2,12,13,14,15,16))
Out[90]: 5
In [91]: maxRunSize((3,4,5,6,7,8))
Out[91]: 6
But if you'd rather just generate combinations without long runs (so that you won't have to filter them out later), check this out:
import random
def generate(n, k, window):
"""
Generate n-choose-k combinations in [1,n] such that there is no window-length of consecutive numbers
"""
answer = []
while True:
if len(answer) >= window:
if answer[-window:] == list(range(answer[-window], window+1)):
answer = []
elif len(answer) == k:
yield answer
answer = []
elif len(answer) == window-1:
if answer[1-window:] == list(range(answer[1-window], window)):
answer.append(random.choice([i for i in range(1,n+1) if i!=answer[-1]+1]))
else:
answer.append(random.choice(range(1,n+1)))