Python中随机大小的部分拆分数字

Question

我有x = 10 和y = 100。

我可以在x“元素持有者”之间随机分配y元素吗？

我想创建x 类别，每个类别都有随机数量的项目；但是，创建的项目数应该正好是y。

我猜是这样的

# number of categories and items
x = 10, y = 100

# keep track of how many items we have left to add
y_left = y

# create all categories
for i in range(x):
    # create category

    # find number of items in this category
    num_items_in_category = random.randint(1, y_left)

    # create items
    for j in range(num_items_in_category):
        # create item

    # set new number of items left to add
    y_left -= num_items_in_category

Answer 1

这实际上并不太难。让我们创建我们的容器：

import random

num_containers = 10
num_objects = 100

containers = [[] for _ in range(num_containers)]
objects = (some_object() for _ in range(num_objects))
# we don't need a list of these, just have to iterate over it, so this is a genexp

for object in objects:
    random.choice(containers).append(object)

Answer 2

使用这个answer 中的函数，您可以生成一个总和为y 的x 个随机数的列表。

遍历此列表中的项目，并从每个持有者的元素群体中做出许多随机选择（移除）。

或者举例：

# requires RandIntVec() & RandFloats() from linked answer

# whatever population you are choosing from, example letter
population = list('qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM1234567890')

# sizes (whatever you want)
num_of_holders = 7
total_elements = len(population)

# empty holder to put results in
element_holder = [[] for _ in xrange(num_of_holders)]

# distribute total_elements elements in randomly-sized portions among num_of_holders 'element holders' 
random_portions_in_holder = RandIntVec(num_of_holders, total_elements, Distribution=RandFloats(num_of_holders))

# assign each portion of elements to each 'holder'
for h, portion in enumerate(random_portions_in_holder):
    for p in range(portion):
        index = random.randrange( len(population) )
        element_holder[h].append(population.pop(index))

# display
print 'Randomly-portioned elements'
for h in element_holder:
    print h

# verify
print '\nMatch desired result?'
print 'total_elements      :', total_elements
print 'elements in holders :', sum([len(h) for h in element_holder])
print 'match               :', total_elements == sum([len(h) for h in element_holder])

输出

Randomly-portioned elements
['M', 'N', 'f', 'V', 'v', 'h', 'i', 'H', '6', '5', 'j', '7', 'r']
['u', 'Z', 'C', 'I', 's', 'm', 'g', 'p', 'q', 'a', 'O', 'T', 'L']
['K', 'E', 'P', 'U']
['Y', 'D', 'A', 'l', 'J', 'R', 'b', 'c', 'z', 'F']
['0', '1', 'o', 'X', 'G', '4', 'W', '3', '2']
['d', 'Q']
['e', 'y', 'B', '8', 'x', 'k', 'w', 't', 'S', 'n', '9']

Match desired result?
total_elements      : 62
elements in holders : 62
match               : True

附注复制链接函数时出现了一些缩进错误，您可能需要更正它们。 elif Distribution.lower() == 'normal':下的代码需要un-缩进1级。我提交了一个经过编辑的版本（结束批准），所以根据您复制的时间，您可能需要也可能不需要编辑。

Answer 3

我在工作中遇到了类似的问题，用字典解决了这个问题。

import random
# number of categories (number of keys in the dictionary) and items (sum of all the value)
x, y = 10, 100

snpDict = {}
i = 0
while i < y:
    a = random.randint(0, x)
    if a in snpDict:
        snpDict[a] += 1
    else:
        snpDict[a] = 1

    i += 1

我有大量的类别和项目，性能不好。