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Octree quantization #7
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| struct OctreeQuantization <: AbstractColorQuantizer | ||
| numcolors::Int | ||
| function OctreeQuantization( | ||
| colorspace::Type{<:Colorant}, | ||
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| numcolors::Int = 256; | ||
| kwargs..., | ||
| ) | ||
| colorspace == RGB{N0f8} || error("Octree Algorithm only supports RGB colorspace") | ||
| return new(numcolors) | ||
| end | ||
| end | ||
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| const OCTREE_DEFAULT_COLORSPACE = RGB{N0f8} | ||
| # Color to RGB | ||
| # constructor for struct | ||
| function OctreeQuantization(numcolors::Int = 256; kwargs...) | ||
| return OctreeQuantization(OCTREE_DEFAULT_COLORSPACE, numcolors; kwargs...) | ||
| end | ||
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| function (alg::OctreeQuantization)(img::AbstractArray) | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. I've tried running using ColorQuantization, TestImages
img = testimage("fabio")
img = RGB{N0f8}.(img)
alg = OctreeQuantization(64)
alg(img)however, the algorithm doesn't terminate after waiting several minutes. There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. The problem is fabio having high number of colors combined with how I prune the tree, lines using allleaves(root) is where trouble is There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. It's actually horrifyingly slow ;-; |
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| return octreequantisation!(img; numcolors = alg.numcolors) | ||
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| end | ||
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| function octreequantisation!(img; numcolors = 256, precheck::Bool = false) | ||
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| # ensure the img is in RGB colorspace | ||
| if (eltype(img) != RGB{N0f8}) | ||
| error("Octree Algorithm requires img to be in RGB colorspace") | ||
| end | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Maybe There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. We could surely but then we are quantising a different image entirely in a different colorspace, we def could avoid deepcopying if we are not gonna accept it if it not right type |
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| # checks if image has more colors than in numcolors | ||
| if precheck == true | ||
| unumcolors = length(unique(img)) | ||
| # @show unumcolors | ||
| if unumcolors <= numcolors | ||
| @debug "Image has $unumcolors unique colors" | ||
| return unique(img) | ||
| end | ||
| end | ||
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| # step 1: creating the octree | ||
| root = Cell( | ||
| SVector(0.0, 0.0, 0.0), | ||
| SVector(1.0, 1.0, 1.0), | ||
| ["root", 0, [], RGB{N0f8}.(0.0, 0.0, 0.0), 0], | ||
| ) | ||
| cases = map( | ||
| p -> [bitstring(UInt8(p))[6:end], 0, Vector{Int}([]), RGB{N0f8}.(0.0, 0.0, 0.0), 1], | ||
| 1:8, | ||
| ) | ||
| split!(root, cases) | ||
| inds = collect(1:length(img)) | ||
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| function putin(root, in) | ||
| r, g, b = map(p -> bitstring(UInt8(p * 255)), channelview([img[in]])) | ||
| rgb = r[1] * g[1] * b[1] | ||
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| # finding the entry to the tree | ||
| ind = 0 | ||
| for i = 1:8 | ||
| if (root.children[i].data[1] == rgb) | ||
| root.children[i].data[2] += 1 | ||
| ind = i | ||
| break | ||
| end | ||
| end | ||
| curr = root.children[ind] | ||
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| for i = 2:8 | ||
| cases = map( | ||
| p -> [ | ||
| bitstring(UInt8(p))[6:end], | ||
| 0, | ||
| Vector{Int}([]), | ||
| RGB{N0f8}.(0.0, 0.0, 0.0), | ||
| i, | ||
| ], | ||
| 1:8, | ||
| ) | ||
| rgb = r[i] * g[i] * b[i] | ||
| if (isleaf(curr) == true && i <= 8) | ||
| split!(curr, cases) | ||
| end | ||
| if (i == 8) | ||
| for j = 1:8 | ||
| if (curr.children[j].data[1] == rgb) | ||
| curr = curr.children[j] | ||
| curr.data[2] += 1 | ||
| push!(curr.data[3], in) | ||
| curr.data[4] = img[in] | ||
| return | ||
| end | ||
| end | ||
| end | ||
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| # handle 1:7 cases for rgb to handle first seven bits | ||
| for j = 1:8 | ||
| if (curr.children[j].data[1] == rgb) | ||
| curr.children[j].data[2] += 1 | ||
| curr = curr.children[j] | ||
| break | ||
| end | ||
| end | ||
| end | ||
| end | ||
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| # build the tree | ||
| for i in inds | ||
| root.data[2] += 1 | ||
| putin(root, i) | ||
| end | ||
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| # step 2: reducing tree to a certain number of colors | ||
| # there is scope for improvements in allleaves as it's found again n again | ||
| leafs = [p for p in allleaves(root)] | ||
| filter!(p -> !iszero(p.data[2]), leafs) | ||
| tobe_reduced = leafs[1] | ||
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| while (length(leafs) > numcolors) | ||
| parents = unique([parent(p) for p in leafs]) | ||
| parents = sort(parents; by = c -> c.data[2]) | ||
| tobe_reduced = parents[1] | ||
| # @show tobe_reduced.data | ||
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| for i = 1:8 | ||
| append!(tobe_reduced.data[3], tobe_reduced.children[i].data[3]) | ||
| tobe_reduced.data[4] += | ||
| tobe_reduced.children[i].data[4] * tobe_reduced.children[i].data[2] | ||
| end | ||
| tobe_reduced.data[4] /= tobe_reduced.data[2] | ||
| tobe_reduced.children = nothing | ||
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| # we don't want to do this again n again | ||
| leafs = [p for p in allleaves(root)] | ||
| filter!(p -> !iszero(p.data[2]), leafs) | ||
| end | ||
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| # step 3: palette formation and quantisation now | ||
| da = [p.data for p in leafs] | ||
| for i in da | ||
| for j in i[3] | ||
| img[j] = i[4] | ||
| end | ||
| end | ||
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| colors = [p[4] for p in da] | ||
| return colors | ||
| end | ||
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| function octreequantisation(img; kwargs...) | ||
| img_copy = deepcopy(img) | ||
| palette = octreequantisation!(img_copy; kwargs...) | ||
| return img_copy, palette | ||
| end | ||
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