在Julia中更改数据集中的值

本文介绍了在Julia中更改数据集中的值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在将R中的函数转换为Julia，但是我不知道如何转换以下R代码:

I am converting a function in R to Julia, but I do not know how to convert the following R code:

x[x==0]=4

基本上，x包含数字行，但是只要有0，我就需要将其更改为4.数据集x来自二项式分布.有人可以帮我在Julia中定义上面的代码吗?

Basically, x contains rows of numbers, but whenever there is a 0, I need to change it to a 4. The data set x comes from a binomial distribution. Can someone help me define the above code in Julia?

推荐答案

两个需要发表评论的小问题是:

Two small issues two long for a comment are:

在Julia 0.7中，您应该写x[x .== 0] .= 4(也要在作业中使用第二个点)

In Julia 0.7 you should write x[x .== 0] .= 4 (using a second dot in assignment also)

通常，使用速度更快，例如foreach或循环，而不是分配带有x .== 0的向量，例如:

In general it is faster to use e.g. foreach or a loop than to allocate a vector with x .== 0, e.g.:

julia> using BenchmarkTools julia> x = rand(1:4, 10^8); julia> function f1(x) x[x .== 4] .= 0 end f1 (generic function with 1 method) julia> function f2(x) foreach(i -> x[i] == 0 && (x[i] = 4), eachindex(x)) end f2 (generic function with 1 method) julia> @benchmark f1($x) BenchmarkTools.Trial: memory estimate: 11.93 MiB allocs estimate: 10 -------------- minimum time: 137.889 ms (0.00% GC) median time: 142.335 ms (0.00% GC) mean time: 143.145 ms (1.08% GC) maximum time: 160.591 ms (0.00% GC) -------------- samples: 35 evals/sample: 1 julia> @benchmark f2($x) BenchmarkTools.Trial: memory estimate: 0 bytes allocs estimate: 0 -------------- minimum time: 86.904 ms (0.00% GC) median time: 87.916 ms (0.00% GC) mean time: 88.504 ms (0.00% GC) maximum time: 91.289 ms (0.00% GC) -------------- samples: 57 evals/sample: 1