pseudo-randomize a list without repeats; while loop is not efficient

Question

I am creating an conditioning experiment with three conditions (0,1,2) and need to pseudo-randomize condition order. I need a randomize list with each condition occurring only 2 times in a row. Here how I tried to achieve it. The code is running but it takes an eternity...

Any ideas why this code is not working well and any different approaches to solve the problem?

#create a list with 36 values
types  = [0] * 4 + [1] * 18 + [2]*14 #0=CS+ ohne Verstärkung; 1 = CS-, 2=CS+     mit shock

#random.shuffle(types)

while '1,1,1' or '2,2,2' or '0,0,0' in types:
    random.shuffle(types)
else: print(types)

Thank you in advance! Martina

another similar question: stackoverflow.com/questions/3313590/… — DevLounge
– DevLounge, Commented Mar 24, 2017 at 19:19

tdelaney · Accepted Answer · 2017-03-24 19:28:57Z

1

Your loop has several problems. First while '1,1,1' or '2,2,2' or '0,0,0' in types: is the same as while ('1,1,1') or ('2,2,2') or ('0,0,0' in types):. Non-zero strings are always True so your condition is always true and the while never stops. Even if it did, types is a list of integers. '0,0,0' is a string and is not an element of the list.

itertools.groupby is a good tool to solve this problem. Its an iterator that is designed to group a sequence into subiterators. You can use it to see if any clusters of numbers are too long.

import random
import itertools

#create a list with 36 values
types  = [0] * 4 + [1] * 18 + [2]*14 #
print(types)

while True:
    random.shuffle(types)
    # try to find spans that are too long
    for key, subiter in itertools.groupby(types):
        if len(list(subiter)) >= 3:
            break # found one, staty in while 
    else:
        break # found none, leave while

print(types)

answered Mar 24, 2017 at 19:28

tdelaney

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Martina Takeaway Over a year ago

Really cool! It totally solved my problem! Never heard of the itertools module before... Learnt a lot! ;)

TemporalWolf · Accepted Answer · 2017-03-24 19:07:00Z

1

while '1,1,1' or '2,2,2' or '0,0,0' in types:
    random.shuffle(types)

evaluates as:

while True or True or '0,0,0' in types:
    random.shuffle(types)

and short-circuits at while True

Instead, use: any() which returns True if any of the inner terms are True

Additionally, your types is numbers and you're comparing it to strings:

>>> types
[0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2]

so you need to map those numbers to a string which can be compared:

>>> ','.join(map(str, types))
'0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2'

Try:

while any(run in ','.join(map(str, types)) for run in ['0,0,0', '1,1,1', '2,2,2']):
    random.shuffle(types)

>>> types
[1, 2, 1, 2, 1, 2, 1, 1, 0, 2, 0, 1, 2, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 0, 2, 1, 1, 0, 2, 1, 1, 2, 2, 1, 1]

edited Mar 24, 2017 at 19:07

answered Mar 24, 2017 at 19:00

TemporalWolf

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2 Comments

Martina Takeaway Over a year ago

Thanks! That is a great solution as well! As a total python beginner, I do not really understand what's the map-function is doing. Does it apply "str" to every element in the list types? So is it iterating over all items or does a map function works differently?

TemporalWolf Over a year ago

@MartinaTakeaway Yes, map() applies the given function to every item in the supplied iterator.

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