Display Random Choice (Python)

I have a list[] of items from which I'd like to display one randomly, but the displayed item must not repeat more than once in last x requests. list1 = item1, item2, item3, item4

Solution 1:

collections.deque is the only sequence type in python that naturally supports being bounded (and only in Python 2.6 and up.) If using python 2.6 or newer:

# Setup
from collections import deque
from random import choice
used = deque(maxlen=7)

# Now your sampling bit
item = random.choice([x for x in list1 if x not in used])
used.append(item)

If using python 2.5 or less, you can't use the maxlen argument, and will need to do one more operation to chop off the front of the deque:

while len(used) > 7:
    used.popleft()

This isn't exactly the most efficient method, but it works. If you need speed, and your objects are hashable (most immutable types), consider using a dictionary instead as your "used" list.

Also, if you only need to do this once, the random.shuffle method works too.

Solution 2:

Is this what you want?

list1 = range(10)
import random
random.shuffle(list1)
list2 = list1[:7]
for item in list2:
    print item
print list1[7]

In other words, look at random.shuffle(). If you want to keep the original list intact, you can copy it: list_copy = list1[:].

Solution 3:

You could try using a generator function and call .next() whenever you need a new item.

import random
def randomizer(l, x):
    penalty_box = []
    random.shuffle(l)
    while True:
        element = l.pop(0)
        # for show
        print penalty_box, l
        yield element
        penalty_box.append(element)
        if len(penalty_box) > x:
            # penalty time over for the first element in the box
            # reinsert randomly into the list
            element = penalty_box.pop(0)
            i = random.randint(0, len(l))
            l.insert(i, element)

Usage example:

>>> r = randomizer([1,2, 3, 4, 5, 6, 7, 8], 3)
>>> r.next()
[] [1, 5, 2, 6, 4, 8, 7]
3
>>> r.next()
[3] [5, 2, 6, 4, 8, 7]
1
>>> r.next()
[3, 1] [2, 6, 4, 8, 7]
5
>>> r.next()
[3, 1, 5] [6, 4, 8, 7]
2
>>> r.next()
[1, 5, 2] [4, 3, 8, 7]
6
>>> r.next()
[5, 2, 6] [4, 3, 8, 7]
1
>>> r.next()
[2, 6, 1] [5, 3, 8, 7]
4
>>> r.next()
[6, 1, 4] [3, 8, 2, 7]
5

Solution 4:

Something like:

# Setup
import random
list1 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
list2 = []

# Loop for as long as you want to display items
while loopCondition:
    index = random.randint(0, len(list1)-1)
    item = list1.pop(index)

    print item

    list2.append(item)
    if(len(list2) > 7):
        list1.append(list2.pop(0))

Solution 5:

I'd use set objects to get a list of items in list1 but not in list2:

import random

list1 = set(["item1", "item2", "item3", "item4", "item5",
             "item6", "item7", "item8", "item9", "item10"])
list2 = []
while True:  # Or something
    selection = random.choice(tuple(list1.difference(set(list2))))
    print(selection)
    list2.append(selection)
    if len(list2) > 7:
        list2 = list2[-7:]