Display random selection (Python)

Question

I have a list [] of elements from which I would like to display one randomly, but the displayed element should not be repeated more than once in the last x queries.

• list1 = item1, item2, item3, item4, item5, item6, item7, item8, item9, item 10
• Display random selection from the list above.
• list2 = save the last displayed item in list2, which should only store 7 items, no more
• Show random selection from the list, but sure that it does not exist in the song2

Is this the right way to do this? In any case, I would like to know how to limit the list to store only 7 items?

thank

Answer 1

collections.deque - python, ( Python 2.6 ). python 2.6 new:

# 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)

python 2.5 , maxlen, , deque:

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

, . , ( ), , "" .

, , random.shuffle.

Answer 2

, ?

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

, random.shuffle(). , : list_copy = list1[:].

Answer 3

You can try to use the generator function and call .next()whenever you need a new element.

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

Answer 4

Sort of:

# 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))

Answer 5

I would use the given 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:]