产生最小和最大之间的相等计数

Question

I have a minimum value and maximum value, I'd like to generate a list of numbers between them such that all the numbers have equal counts. Is there a numpy function or any function out there?

Example: GenerateNums(start=1, stop=5, nums=10)

Expected output: [1,1,2,2,3,3,4,4,5,5] ie each number has an almost equal count

Answer 1

Takes "almost equal" to heart -- the difference between the most common and least common number is at most 1. No guarantee about which number is the mode.

def gen_nums(start, stop, nums):
    binsize = (1 + stop - start) * 1.0 / nums
    return map(lambda x: int(start + binsize * x), xrange(nums))

gen_nums(1, 5, 10)
[1, 1, 2, 2, 3, 3, 4, 4, 5, 5]

Answer 2

There is a numpy function:

In [3]: np.arange(1,6).repeat(2)
Out[3]: array([1, 1, 2, 2, 3, 3, 4, 4, 5, 5])

Answer 3

For almost equal counts, you can sample from a uniform distribution. numpy.random.randint does this:

>>> import numpy as np
>>> np.random.randint(low=1, high=6, size=10)
array([4, 5, 5, 4, 5, 5, 2, 1, 4, 2])

To get these values in sorted order:

>>> sorted(np.random.randint(low=1, high=6, size=10))
[1, 1, 1, 2, 3, 3, 3, 3, 5, 5]

This process is just like rolling dice :) As you sample more times, the counts of each value should become very similar:

>>> from collections import Counter
>>> Counter(np.random.randint(low=1, high=6, size=10000))
Counter({1: 1978, 2: 1996, 3: 2034, 4: 1982, 5: 2010})

For exactly equal counts:

>>> range(1,6) * 2
[1, 2, 3, 4, 5, 1, 2, 3, 4, 5]
>>> sorted(range(1,6) * 2)
[1, 1, 2, 2, 3, 3, 4, 4, 5, 5]

Answer 4

def GenerateNums(start=1, stop=5, nums=10):
    result = []
    rep = nums/(stop - start + 1 )   
    for i in xraneg(start,stop):
        for j in range(rep):
            result.append(i)

    return result

Answer 5

def GenerateNums(start=0,stop=0,nums=0,result=[]):
    assert (nums and stop > 0), "ZeroDivisionError"
    # get repeating value
    iter_val =  int(round(nums/stop))
    # go through strt/end and repeat the item on adding
    [[result.append(x) for __ in range(iter_val)] for x in range(start,stop)]   
    return result       
print (GenerateNums(start=0, stop=5, nums=30))

>>> [0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4]