显示随机选择 (Python)
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Display random choice (Python)
提问by eozzy
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.
我有一个项目列表 []，我想从中随机显示一个项目，但显示的项目在最近的 x 次请求中不能重复多次。
- list1 = item1, item2, item3, item4, item5, item6, item7, item8, item9, item 10
- Display a random selection from the list above
- list2 = store the last displayed item in list2 which should only store 7 items, not more
- Display a random selection from the list but make sure it doesn't exist in the list2
- list1 = item1, item2, item3, item4, item5, item6, item7, item8, item9, item 10
- 显示上面列表中的随机选择
- list2 = 将最后显示的项目存储在 list2 中，它应该只存储 7 个项目，而不是更多
- 显示列表中的随机选择，但确保它不存在于列表 2 中
Is that the right way to do it? Either way, I'd like to know how to limit a list to store only 7 items?
这是正确的做法吗？无论哪种方式，我都想知道如何将列表限制为仅存储 7 个项目？
Thanks
谢谢
回答by Crast
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:
collections.deque 是 python 中唯一自然支持有界的序列类型（并且仅在 Python 2.6 及更高版本中。）如果使用 python 2.6 或更新版本：
# 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:
如果使用 python 2.5 或更低版本，则不能使用 maxlen 参数，并且需要再执行一项操作来切掉 deque 的前面：
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.
这并不是最有效的方法，但它确实有效。如果您需要速度，并且您的对象是可散列的（大多数不可变类型），请考虑使用字典作为“已使用”列表。
Also, if you only need to do this once, the random.shuffle method works too.
此外，如果您只需要执行一次，random.shuffle 方法也可以。
回答by Alok Singhal
Is this what you want?
这是你想要的吗？
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[:].
换句话说，看看random.shuffle()。如果你想保持原始列表完整，你可以复制它：list_copy = list1[:]。
回答by tkerwin
You could try using a generator function and call .next()whenever you need a new item.
您可以尝试使用生成器函数并.next()在需要新项目时调用。
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:
用法示例：
>>> 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
回答by Teddy
I'd use set objects to get a list of items in list1 but not in list2:
我会使用 set 对象来获取 list1 中但不在 list2 中的项目列表：
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:]
回答by Sapph
Something like:
就像是：
# 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))
