在1行代码中计算向量中每个唯一整数的实例? [英] Count instances of each unique integer in a vector in 1 line of code?

问题描述

是否有一种巧妙的方法可以重写此Julia函数(也许仅使用一行代码)而又不会使其变慢呢? (我刚开始使用Julia.这太棒了！)K是一个正整数，而zd是一个不大于K的正整数向量.谢谢！

Is there a slick way to rewrite this Julia function, perhaps using just 1 line of code, without making it much slower? (I just started using Julia. It's great!) K is a positive integer and zd is a vector of positive integers no greater than K. Thanks!

function tally(zd)
    ret = zeros(Int64, K)
    for k in zd
        ret[k] += 1
    end
    return ret 
end

示例:

julia> K = 5
julia> zd = [1,2,2,2,2,3];
julia> tally(zd)
5-element Array{Float64,1}:
 1
 4
 1
 0
 0

推荐答案

我尚未测试性能，但是使用hist函数应该可以:

I haven't tested the performance, but using the hist function should work:

hist(zd,0.5:K+0.5)[2]

给予:

5元素数组{Int64,1}: 1个 4 1个 0 0

5-element Array{Int64,1}: 1 4 1 0 0

或者，如果零不重要，则只需使用

or, if the zeros are unimportant, just use

hist(zd)[2]
3-element Array{Int64,1}:
 1
 4
 1

这篇关于在1行代码中计算向量中每个唯一整数的实例?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！