生成具有固定和的随机 integer numpy 向量

Question

Based on the codes, i want to randomize the integer where there are specific range specify but in the last three array I need to have 100 in total. The sum of the last three array cannot be exceed than 100 . I don't know how to solve this problem.

self.position = np.array([rd.randint(0,2), rd.randint(0,2), rd.randint(0,100), rd.randint(0,50), rd.randint(0,1)])

Answer 1

Although the solution Rmano suggest is valid, it lack total randomness because once you draw the first random integer the other are pseudo-random or not random at all because of the dependent.

A more random solution would be, in my opinion:

import numpy as np

def generate_fix_sum_random_vec(limit, num_elem, tries=10):
    v = np.random.randint(0, limit, num_elem)
    s = sum(v)
    if (np.sum(np.round(v/s*limit)) == limit):
        return np.round(v / s * limit)
    elif (np.sum(np.floor(v/s*limit)) == limit):
        return np.floor(v / s * limit)
    elif (np.sum(np.ceil(v/s*limit)) == limit):
        return np.ceil(v / s * limit)
    else:
        return generate_fix_sum_random_vec(limit, num_elem, tries-1)



for i in range(25):
    test_vec = generate_fix_sum_random_vec(100, 3)
    test_vec = test_vec.astype(int)
    print("vec: ", test_vec, "sum of vector: ", np.sum(test_vec))

This solution is more robust, and can be valid to any limit on the sum of the vector and it easily applicable to vector in any length by changing the num_elem . Moreover it doesn't assume anything and just trying until it find a match.

The output of the loop:

vec:  [53 32 15] sum of vector:  100
vec:  [40 38 22] sum of vector:  100
vec:  [56 38  6] sum of vector:  100
vec:  [ 5 17 78] sum of vector:  100
vec:  [12 29 59] sum of vector:  100
vec:  [ 1 34 65] sum of vector:  100
vec:  [ 3 56 41] sum of vector:  100
vec:  [35 65  0] sum of vector:  100
vec:  [54  9 37] sum of vector:  100
vec:  [45  8 47] sum of vector:  100
vec:  [30 56 14] sum of vector:  100
vec:  [34 63  3] sum of vector:  100
vec:  [17 40 43] sum of vector:  100
vec:  [56 36  8] sum of vector:  100
vec:  [52 45  3] sum of vector:  100
vec:  [35 34 31] sum of vector:  100
vec:  [25 41 34] sum of vector:  100
vec:  [ 1 78 21] sum of vector:  100
vec:  [ 1 49 50] sum of vector:  100
vec:  [51 31 18] sum of vector:  100
vec:  [50 10 40] sum of vector:  100
vec:  [36 63  1] sum of vector:  100
vec:  [30 30 40] sum of vector:  100
vec:  [27 30 43] sum of vector:  100
vec:  [21 27 52] sum of vector:  100

There is a need to check that indeed the resulted vector sum to the desired value, because this isn't guaranteed

Answer 2

Answer to the comment below from the OP:

I am not sure how much randomness you get like this, but:

def three_to_100_b():
    a3 = random.randint(0,1)
    a2 = random.randint(0,50)
    a1 = 100-a3-a2
    return a1, a2, a3

will give you three pseudo -random numbers, the first in the range 0-100, the second in the range 0-50, the last one 0 or 1, and summing 100.

But notice that the randomness between a1 and a2 is really small --- they are on a straight line +0 or -1 by definition...

Notice, new answer --- the old one has flaws in randomness.

One method to have three numbers that sum up to 100 is to generate three random numbers, and then scale them so that their sums is 100. This is overly verbose, but the idea is:

#! /usr/bin/env python3

import random

def three_to_100():
    r1 = random.randint(0,100)
    r2 = random.randint(0,100)
    r3 = random.randint(0,100)
    a1 = r1*100/(r1+r2+r3)
    a2 = r2*100/(r1+r2+r3)
    a3 = 100 - a1 - a2
    return a1, a2,a3

#main
print("x\t y\t z")
for i in range(1000):
    # print(three_to_100())
    a=three_to_100()
    print("{}\t  {}\t {}".format(*a))

I plotted the (x,y) pairs for the first 1000 numbers (the third number is represented by the color of the dot) and I have:

old answer

You can generate a number from 0 to 100, then another one from 0 to (100 - the first number), and so on. The last one won't be random:

#! /usr/bin/env python3

import random

def three_to_100():
    r1 = random.randint(0,100)
    r2 = random.randint(0,100-r1)
    r3 = 100 - r1 - r2
    return r1, r2,r3

#main
for i in range(10):
    print(three_to_100())

The output will be:

[romano:~/tmp] % ./randsum.py 
(26, 7, 67)
(85, 13, 2)
(43, 41, 16)
(75, 22, 3)
(66, 19, 15)
(30, 16, 54)
(43, 30, 27)
(36, 2, 62)
(31, 2, 67)
(91, 0, 9)

This is a plot of the first two numbers — they seem pretty random, but maybe it's true that they seem a bit crowded on the xy axis.