pick random item from a list with constantly reapeating items, but not repeat one twice immediately after

Please see the sample code below.

 import random import sys def special_select(d, num_to_show=100): selected = None for i in range(num_to_show): selected = random.choice(list(d.difference([selected]))) print selected 

You may try the function like follows.

 >> d = set(range(5)) >> special_select(d) 

I'd go for:

from itertools import groupby, islice from random import choice choices = range(5) items = (k for k, g in groupby(iter(lambda: choice(choices), None))) for item in islice(items, 100): print item 

This lets itertools.groupby handle consecutive duplicate values and islice means you can just skim some off, so if you wanted 50 of them, just use:

random50 = list(islice(items, 50)) 

Remember the previous item and keep picking a new choice until it differs:

import random def pick_random(options): last = next = None while True: while next == last: next = random.choice(options) yield next last = next 

Use this as a generator:

>>> options = (1, 2, 3, 4, 5) >>> picks = pick_random(options) >>> for i in range(100): ... print(next(picks)) ... 1 5 4 3 2 3 (...) 3 4 5 

Using itertools.islice() makes it easier to pick just 100:

from itertools import islice for pick in islice(pick_random(options), 100): print(pick) 

Not the most efficient one in Python, but what is done here is: when append the next item, the next item is sampled from the pool excluding the previous item.

def rand_chain(n, chos_from): rd_ls=[] rd_ls.append(random.choice(chos_from)) for i in range(n-1): rd_ls.append(random.choice([v for v in chos_from if v!=rd_ls[i]])) return rd_ls >>> rand_chain(10, range(10)) [8, 2, 9, 2, 9, 5, 9, 8, 2, 4] 

If you use numpy, replace [v for v in chos_from if v!=rd_ls[i]] with more efficient stuff.

Yes. Plenty.

• Determine number of times to iterate.
• Start a random number generator.
• Set variable previous to None or similar.
• Start iteration.
• Generate next random number and compare it with previous. If it is the same, skip it.