julia - 如何在 Julia 1.0 的数组中用 NaN 替换 Inf 值？

Question

我在网上看到了几个地方的解决方案

 a = [1 2 3; 4 5 Inf]
 a[isinf(a)] = NaN

但这在 Julia 1.0.1 上给了我一个错误:

 ERROR: MethodError: no method matching isinf(::Array{Float64,2})
 Closest candidates are:
   isinf(::BigFloat) at mpfr.jl:851
   isinf(::Missing) at missing.jl:79
   isinf(::ForwardDiff.Dual) at <path on my local machine>

是什么赋予了？

Answer 1

作为补充评论。执行此操作的标准函数是 replace! .你可以这样使用它:

julia>  a = [1 2 3; 4 5 Inf]
2×3 Array{Float64,2}:
 1.0  2.0    3.0
 4.0  5.0  Inf

julia> replace!(a, Inf=>NaN)
2×3 Array{Float64,2}:
 1.0  2.0    3.0
 4.0  5.0  NaN

对于大型阵列，它的性能将优于广播。

如果你真的需要速度，你可以写一个像这样的简单函数:

function inf2nan(x)
    for i in eachindex(x)
        @inbounds x[i] = ifelse(isinf(x[i]), NaN, x[i])
    end
end

现在让我们简单比较一下三个选项的性能:

julia> function bench()
           x = fill(Inf, 10^8)
           @time x[isinf.(x)] .= NaN
           x = fill(Inf, 10^8)
           @time replace!(x, Inf=>NaN)
           x = fill(Inf, 10^8)
           @time inf2nan(x)
       end
bench (generic function with 1 method)

julia> bench()
  0.980434 seconds (9 allocations: 774.865 MiB, 0.16% gc time)
  0.183578 seconds
  0.109929 seconds

julia> bench()
  0.971408 seconds (9 allocations: 774.865 MiB, 0.03% gc time)
  0.184163 seconds
  0.102161 seconds