在 randint() 函数中赋予较低的值较高的权重

Question

使用randint()，我如何为较低的值赋予较高的权重（被选中的机会较高）？

我有以下代码：

def grab_int():
    current = randint(0,10) 
    # I want the weight of the numbers to go down the higher they get (so I will get more 0s than 1s, more 1s than 2, etc)

OBS！

我想以一种比其他答案更优雅的方式来做这件事（例如：Python Weighted Random）。有没有办法通过导入某种重量module来做到这一点？

规格： 我想要一个randint()，它返回0 的机会是30%，然后它会线性下降到10，有1% 的机会。

Answer 1

以下方法满足您的要求。它使用拒绝采样方法：随机均匀地生成一个整数，并以与其权重成正比的概率接受它。如果该号码不被接受，我们拒绝它并重试（另请参阅this answer of mine）。

import random

def weighted_random(mn, mx, mnweight, mxweight):
 while True:
    # Get the highest weight.
    highestweight=max(mnweight,mxweight)
    # Generate a uniform random integer in the interval [mn, mx].
    r=random.randint(mn,mx)
    # Calculate the weight for this integer.  This ensures the min's
    # weight is mnweight and the max's weight is mxweight
    weight=mnweight+(mxweight-mnweight)*((i-mn)/(mx-mn))
    # Generate a random value between 0 and the highest weight
    v=random.random()*highestweight
    # Is it less than this weight?
    if v<weight:
       # Yes, so return it
       return r
    # No, so try again

（诚然，由于浮点除法以及输出浮点数的random.random()，实现并不完全“优雅”，但下面的示例是，一旦我们编写了它。实现可以通过在fractions 模块中使用Fractions 来改进。）

这个方法也可以使用Python中已有的random.choices方法实现如下。首先我们计算random.choices 所需的权重，然后我们传递这些权重。但是，如果最小值和最大值之间的范围非常大，这种方法的效率并不高。

import random

# Calculate weights for `random.choices`
def make_weights(mn, mx, mnweight, mxweight):
 r=(mx-mn)
 return [mnweight+(mxweight-mnweight)*(i/r) for i in range(mn, mx+1)]

def weighted_random(mn, mx, mnweight, mxweight):
 weights=make_weights(mn, mx, mnweight, mxweight)
 return random.choices(range(mn, mx+1), weights=weights)[0]

使用 NumPy 库，这甚至可以按如下方式实现：

import random
import numpy

def weighted_random(mn, mx, mnweight, mxweight):
 return random.choices(range(mn, mx+1), \
      weights=numpy.linspace(mnweight,mxweight,(mx-mn)+1))[0]

weighted_random 函数的示例如下：

# Generate 100 random integers in the interval [0, 10],
# where 0 is assigned the weight 30 and 10 is assigned the
# weight 10 and numbers in between are assigned
# decreasing weights.
print([weighted_random(0,10,30,10) for i in range(100)])

Answer 2

我发现最简单的方法是手动分配权重：

def grab_int():
    global percent
    global percentLeft
    global upby

    # I want the weight of the numbers to go down the higher they get (so I will get more 0s than 1s, more 1s than 2, etc)
    current = randint(0,100)
    if current < 30:
        upby = randint(1,2)
        #0
    elif current < 40:
        upby = 1
        #1
    elif current < 45:
        upby = 2
        #2
    elif current < 50:
        upby = 3
        #3
        upby = 4
        #4
    elif current < 60:
        upby = 5
        #5
    elif current < 65:
        upby = 6
        #6
    elif current < 70:
        upby = 7
        #7
    elif current < 75:
        upby = 8
        #8
    elif current < 90:
        upby = 9
        #9
    elif current < 95:
        upby = 10
        #10
    else: # I'm dumb so I accidentally only added up to 95%, This just gives 0 a 5% higher chance without having to rewrite all the other values
        upby = 0
        #0