如何在Ruby中使用分布宝石创建指数分布？

Question

I need the mean to be 2, with starting at 1 up to 10,000 random numbers.

require 'distribution'

mean = 2

# generate a rng with exponential distribution
rng = Distribution::Exponential.rng(mean)

Answer 1

Well, I'm not much of a statistician, so I'm a little lost in the code here.

Can you tell me, then, is the first parameter to rng (it is named l ) supposed to stand for "limit"?

If so, I don't really know, I'm getting the same sort of results that you must be getting, but again, I'm not a statistician or even a mathematician, I just like to code, so I am probably not of too much help.

EDIT: So, again, I dont really know what is going on or is supposed to be going on here. Here is what gives me a relatively close mean of 2 (I got this just by messing around with the l value):

require 'distribution'
rng=Distribution::Exponential.rng(0.5)
sum=0
1000.times { sum += rng.call }
sum /= 1000
puts sum

Seems to give a value generally between 1.9 and 2.1. Hope this helps :)

PRINT EACH VALUE:

require 'distribution'
rng = Distribution::Exponential.rng(0.5)
values = [] # we'll store each value in an array

# calculate the results
how_many = 1000 # let's store 1000 in a variable since we'll reuse the value
how_many.times { values << rng.call } # rng is now a Proc, so we use call to get our value

# print out the results
values.each_with_index { |v,i| puts "#{i+1}: #{v}" } # print each value one by one
puts
puts "Total sum: #{values.inject(&:+)}"
puts "Mean: #{values.inject(&:+)/how_many}"

Again, not sure if this is exactly "right" or what you're looking for, but it definitely seems to approach 2. Try a bigger number than 1000, like 100000.