Julia中最多可包含多个图像或数组 [英] Maximum of multiple images or arrays in Julia
问题描述
我想找到最多几个图像:将它们加载到数组中，然后沿第一维查找最大图像.
例如Python代码:
import cv2
import sys
import numpy as np
imgs_paths = sys.argv[1:]
imgs = list(map(cv2.imread, imgs_paths))
imgs_arr = np.array(imgs, dtype=np.float32)
imgs_max = np.max(imgs_arr, 0)
我所做的是以下事情:
using Colors, Images
function im_to_array(im)
img_array = permutedims(channelview(im), (2,3,1))
img_array = Float32.(img_array)
return img_array
end
imgs = map(Images.load, imgs_paths)
imgs_arr = map(im_to_array, imgs)
a = imgs_arr
b = reshape(cat(a..., dims=1), tuple(length(a), size(a[1])...))
imgs_max = maximum(b, dims=1)
但这很丑.
我发现了一种获取最大值的更简单方法(下面的代码)，但是它的性能太差了.可能不是我所期望的.
function im_to_array(im)
img_array = permutedims(channelview(im), (2,3,1))
img_array = Float32.(img_array)
return img_array
end
imgs = map(Images.load, imgs_paths)
imgs_arr = map(im_to_array, imgs)
imgs_max = max.(imgs_arr...)
在笔记本电脑上，第一种方法对120张FHD图像的运行时间约为5秒.而且我无法弄清第二种方法的运行时间，因为我等待了大约30分钟，而且它没有停止.我正在Julia 1.4.1上进行测试
是否有更好的方法来查找多个图像?
UPD :这是我想要的简单情况:
a = [zeros(Int8, 8, 8, 3), zeros(Int8, 8, 8, 3), zeros(Int8, 8, 8, 3)] # 3 black images with shape 8x8
max.(a) #doesn't work
max.(a...) #works with this simple input but when I test it on 120 FHD images it's extremely slow
UPD2 :我在较少数量的图像上测试了这两种方法.
function max1(imgs_arr)
a = imgs_arr
b = reshape(cat(a..., dims=1), tuple(length(a), size(a[1])...))
imgs_max = maximum(b, dims=1)
return imgs_max
end
function max2(imgs_arr)
return max.(imgs_arr...)
end
imgs_arr = my_imgs_arrays[1:5]
@time max1(imgs_arr)
@time max2(imgs_arr)
0.247060 seconds (5.29 k allocations: 142.657 MiB)
0.154158 seconds (44.85 k allocations: 26.388 MiB)
imgs_arr = my_imgs_arrays[1:15]
@time max1(imgs_arr)
@time max2(imgs_arr)
0.600093 seconds (72.38 k allocations: 382.923 MiB)
0.769446 seconds (1.24 M allocations: 71.374 MiB)
imgs_arr = my_imgs_arrays[1:25]
@time max1(imgs_arr)
@time max2(imgs_arr)
1.057548 seconds (23.08 k allocations: 618.309 MiB)
5.270050 seconds (151.52 M allocations: 2.329 GiB, 4.77% gc time)
因此，我使用的图像更多-运作起来更慢.
似乎您要对多个图像进行成对最大减少.首先，这是一个生成随机图像"的函数:
rand_images(k, dims...) = [rand(UInt8, dims...) for _ = 1:k]
我将生成一个包含三个随机10x12图像的矢量:
julia> images = rand_images(3, 10, 12)
3-element Array{Array{UInt8,2},1}:
[0x51 0xdc … 0xf7 0x1e; 0xe1 0x10 … 0xd8 0x98; … ; 0x54 0x45 … 0x7a 0xaf; 0x7b 0xfc … 0x0a 0x81]
[0xc8 0xa5 … 0xa8 0x81; 0x92 0x89 … 0x9f 0xbe; … ; 0x6a 0x03 … 0xb1 0xfd; 0x34 0xa9 … 0xa3 0x50]
[0x26 0x9b … 0x2a 0x7c; 0x5c 0x7d … 0x8d 0x2b; … ; 0x32 0x1b … 0x57 0xdf; 0x96 0xa1 … 0x2a 0xc9]
一种简单的方法是进行成对最大减少:
julia> using BenchmarkTools
julia> @btime reduce(images) do a, b
max.(a, b)
end
400.485 ns (2 allocations: 416 bytes)
10×12 Array{UInt8,2}:
0xc8 0xdc 0x82 0xa7 0xa6 0xce 0xcd 0xb2 0x6e 0xba 0xf7 0x81
0xe1 0x89 0x9f 0xeb 0x89 0xdf 0xd2 0xd2 0xab 0xea 0xd8 0xbe
0xeb 0xdd 0x9e 0xe2 0xf5 0x4b 0xd2 0xe8 0xe4 0xf8 0xb9 0xf8
0x63 0xa3 0xd7 0xea 0xf0 0x93 0xed 0xf7 0xfb 0xfb 0x9f 0xbb
0xf2 0x51 0xf0 0xd4 0xfc 0xcf 0xf4 0xdd 0xeb 0xc3 0xe9 0xf9
0xf8 0x72 0xfa 0x92 0x72 0xaa 0xa2 0xed 0xa1 0xdf 0xf1 0xd0
0xef 0xe6 0x64 0xb3 0xd0 0x6a 0xce 0x9e 0x96 0xba 0xed 0xf9
0xdb 0xc5 0x52 0xb3 0xf7 0xd1 0xdd 0xba 0xac 0xbc 0xd3 0xa1
0x6a 0x45 0x88 0xda 0xf5 0xc6 0xcf 0x64 0xbc 0xf9 0xb1 0xfd
0x96 0xfc 0xb1 0xc0 0xc4 0xcf 0x89 0xb4 0xe8 0xad 0xa3 0xc9
那非常快:400ns.我会将其放在与您正在做的图像大小相当的图像上，但是您没有提到我可以看到的图像大小(代码与数据无关，因此图像中的数据无关紧要)./p>
约简计算出最大切片，一次对一张图像进行缩小，这可能不是最快的方法.似乎一次计算所有图像中的每个最大像素"可能更快，这有点复杂，但也可以做到:
function max_images(images::Vector{<:Array})
M = copy(images[1])
for i = 1:length(M)
for j = 2:length(images)
M[i] = max(M[i], images[j][i])
end
end
return M
end
这可以工作，但是要花费421纳秒，这比数组精简版本要慢！哎呀.原因之一是，不能保证图像的大小都是相同的，因此必须在对每个图像进行索引的内部循环中进行边界检查.我们可以通过在@inbounds M[i] = max(M[i], images[j][i])上添加入站注释来跳过此过程，后果自负.这使时间降至282 ns.通过告诉编译器可以通过将@simd宏放在每个for循环上来安全地对两个循环重新排序以利用指令级并行性，从而获得更高的速度.这将时间缩短至240 ns.该代码的最终版本是:
function max_images(images::Vector{<:Array})
M = copy(images[1])
@simd for i = 1:length(M)
@simd for j = 2:length(images)
@inbounds M[i] = max(M[i], images[j][i])
end
end
return M
end
I want to find a maximum of several images: load them into an array and find a maximum along first dimension.
Python code for example:
import cv2
import sys
import numpy as np
imgs_paths = sys.argv[1:]
imgs = list(map(cv2.imread, imgs_paths))
imgs_arr = np.array(imgs, dtype=np.float32)
imgs_max = np.max(imgs_arr, 0)
What I did is the following:
using Colors, Images
function im_to_array(im)
img_array = permutedims(channelview(im), (2,3,1))
img_array = Float32.(img_array)
return img_array
end
imgs = map(Images.load, imgs_paths)
imgs_arr = map(im_to_array, imgs)
a = imgs_arr
b = reshape(cat(a..., dims=1), tuple(length(a), size(a[1])...))
imgs_max = maximum(b, dims=1)
But it's ugly.
I found more simple way to get a maximum (code is below) but it's performance is awful. May be it does not what I'm expecting.
function im_to_array(im)
img_array = permutedims(channelview(im), (2,3,1))
img_array = Float32.(img_array)
return img_array
end
imgs = map(Images.load, imgs_paths)
imgs_arr = map(im_to_array, imgs)
imgs_max = max.(imgs_arr...)
Run time of the first method on 120 FHD images is ~5 seconds on my laptop. And I can't figure out run time of the second method because I was waiting for ~30 minutes and it didn't stop. I'm testing it on Julia 1.4.1
Is there a better way to find a maximum of multiple images?
UPD: here is simple case of what I want:
a = [zeros(Int8, 8, 8, 3), zeros(Int8, 8, 8, 3), zeros(Int8, 8, 8, 3)] # 3 black images with shape 8x8
max.(a) #doesn't work
max.(a...) #works with this simple input but when I test it on 120 FHD images it's extremely slow
UPD2: I tested both methods on a smaller number of images.
function max1(imgs_arr)
a = imgs_arr
b = reshape(cat(a..., dims=1), tuple(length(a), size(a[1])...))
imgs_max = maximum(b, dims=1)
return imgs_max
end
function max2(imgs_arr)
return max.(imgs_arr...)
end
imgs_arr = my_imgs_arrays[1:5]
@time max1(imgs_arr)
@time max2(imgs_arr)
0.247060 seconds (5.29 k allocations: 142.657 MiB)
0.154158 seconds (44.85 k allocations: 26.388 MiB)
imgs_arr = my_imgs_arrays[1:15]
@time max1(imgs_arr)
@time max2(imgs_arr)
0.600093 seconds (72.38 k allocations: 382.923 MiB)
0.769446 seconds (1.24 M allocations: 71.374 MiB)
imgs_arr = my_imgs_arrays[1:25]
@time max1(imgs_arr)
@time max2(imgs_arr)
1.057548 seconds (23.08 k allocations: 618.309 MiB)
5.270050 seconds (151.52 M allocations: 2.329 GiB, 4.77% gc time)
So, more images I use - more slowly it works.
It seems like you're looking to do a pairwise max reduction across a number of images. First, here's a function to generate random "images":
rand_images(k, dims...) = [rand(UInt8, dims...) for _ = 1:k]
I'll generate a vector of three random 10x12 images:
julia> images = rand_images(3, 10, 12)
3-element Array{Array{UInt8,2},1}:
[0x51 0xdc … 0xf7 0x1e; 0xe1 0x10 … 0xd8 0x98; … ; 0x54 0x45 … 0x7a 0xaf; 0x7b 0xfc … 0x0a 0x81]
[0xc8 0xa5 … 0xa8 0x81; 0x92 0x89 … 0x9f 0xbe; … ; 0x6a 0x03 … 0xb1 0xfd; 0x34 0xa9 … 0xa3 0x50]
[0x26 0x9b … 0x2a 0x7c; 0x5c 0x7d … 0x8d 0x2b; … ; 0x32 0x1b … 0x57 0xdf; 0x96 0xa1 … 0x2a 0xc9]
One straightforward way to do this is to do a pairwise max reduction:
julia> using BenchmarkTools
julia> @btime reduce(images) do a, b
max.(a, b)
end
400.485 ns (2 allocations: 416 bytes)
10×12 Array{UInt8,2}:
0xc8 0xdc 0x82 0xa7 0xa6 0xce 0xcd 0xb2 0x6e 0xba 0xf7 0x81
0xe1 0x89 0x9f 0xeb 0x89 0xdf 0xd2 0xd2 0xab 0xea 0xd8 0xbe
0xeb 0xdd 0x9e 0xe2 0xf5 0x4b 0xd2 0xe8 0xe4 0xf8 0xb9 0xf8
0x63 0xa3 0xd7 0xea 0xf0 0x93 0xed 0xf7 0xfb 0xfb 0x9f 0xbb
0xf2 0x51 0xf0 0xd4 0xfc 0xcf 0xf4 0xdd 0xeb 0xc3 0xe9 0xf9
0xf8 0x72 0xfa 0x92 0x72 0xaa 0xa2 0xed 0xa1 0xdf 0xf1 0xd0
0xef 0xe6 0x64 0xb3 0xd0 0x6a 0xce 0x9e 0x96 0xba 0xed 0xf9
0xdb 0xc5 0x52 0xb3 0xf7 0xd1 0xdd 0xba 0xac 0xbc 0xd3 0xa1
0x6a 0x45 0x88 0xda 0xf5 0xc6 0xcf 0x64 0xbc 0xf9 0xb1 0xfd
0x96 0xfc 0xb1 0xc0 0xc4 0xcf 0x89 0xb4 0xe8 0xad 0xa3 0xc9
That's pretty fast: 400ns. I would time it on images of size comparable to what you're doing, but you didn't mention images sizes that I can see (the code isn't data dependent, so the data in the images shouldn't matter).
The reduction computes a maximal slice, reducing that with an image at a time, which may not be the fastest way to do this. It seems like it may be faster to compute each maximal "pixel" one at a time across all the images, which is a bit more complicated but can also be done:
function max_images(images::Vector{<:Array})
M = copy(images[1])
for i = 1:length(M)
for j = 2:length(images)
M[i] = max(M[i], images[j][i])
end
end
return M
end
This works but it takes 421 nanoseconds which is slower than the array reduce version! Oops. One of the reasons is that there's no guarantee that the images are all the same size so there's bounds checking in the inner loop indexing into each image. We can skip that at our own risk by putting an inbounds annotation on @inbounds M[i] = max(M[i], images[j][i]). That brings the time down to 282 ns. There's a bit more speed that can be gained by telling the compiler that it can safely reorder both of the loops to take advantage of instruction-level parallelism by putting the @simd macro on each for loop. That brings the time down to 240 ns. The final version of the code is:
function max_images(images::Vector{<:Array})
M = copy(images[1])
@simd for i = 1:length(M)
@simd for j = 2:length(images)
@inbounds M[i] = max(M[i], images[j][i])
end
end
return M
end
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