I implemented the LSDD changepoint detection method decribed in [1] in Julia, to see if I could make it faster than the existing python implementation [2], which is based on a grid search that looks for the optimal parameters.
I obtain the desired results but despite my best efforts, my grid search version of it takes about the same time to compute as the python one, which is still way too long for real applications.
I also tried using the Optimize package which only makes things worse (2 or 3 times slower).
Here is the grid search that I implemented :
using Random
using LinearAlgebra
function squared_distance(X::Array{Float64,1},C::Array{Float64,1})
sqd = zeros(length(X),length(C))
for i in 1:length(X)
for j in 1:length(C)
sqd[i,j] = X[i]^2 + C[j]^2 - 2*X[i]*C[j]
end
end
return sqd
end
function lsdd(x::Array{Float64,1},y::Array{Float64,1}; folds = 5, sigma_list = nothing , lambda_list = nothing)
lx,ly = length(x), length(y)
b = min(lx+ly,300)
C = shuffle(vcat(x,y))[1:b]
CC_dist2 = squared_distance(C,C)
xC_dist2, yC_dist2 = squared_distance(x,C), squared_distance(y,C)
Tx,Ty = length(x) - div(lx,folds), length(y) - div(ly,folds)
#Define the training and testing data sets
cv_split1, cv_split2 = floor.(collect(1:lx)*folds/lx), floor.(collect(1:ly)*folds/ly)
cv_index1, cv_index2 = shuffle(cv_split1), shuffle(cv_split2)
tr_idx1,tr_idx2 = [findall(x->x!=i,cv_index1) for i in 1:folds], [findall(x->x!=i,cv_index2) for i in 1:folds]
te_idx1,te_idx2 = [findall(x->x==i,cv_index1) for i in 1:folds], [findall(x->x==i,cv_index2) for i in 1:folds]
xTr_dist, yTr_dist = [xC_dist2[i,:] for i in tr_idx1], [yC_dist2[i,:] for i in tr_idx2]
xTe_dist, yTe_dist = [xC_dist2[i,:] for i in te_idx1], [yC_dist2[i,:] for i in te_idx2]
if sigma_list == nothing
sigma_list = [0.25, 0.5, 0.75, 1, 1.2, 1.5, 2, 2.5, 2.2, 3, 5]
end
if lambda_list == nothing
lambda_list = [1.00000000e-03, 3.16227766e-03, 1.00000000e-02, 3.16227766e-02,
1.00000000e-01, 3.16227766e-01, 1.00000000e+00, 3.16227766e+00,
1.00000000e+01]
end
#memory prealocation
score_cv = zeros(length(sigma_list),length(lambda_list))
H = zeros(b,b)
hx_tr, hy_tr = [zeros(b,1) for i in 1:folds], [zeros(b,1) for i in 1:folds]
hx_te, hy_te = [zeros(1,b) for i in 1:folds], [zeros(1,b) for i in 1:folds]
#h_tr,h_te = zeros(b,1), zeros(1,b)
theta = zeros(b)
for (sigma_idx,sigma) in enumerate(sigma_list)
#the expression of H is different for higher dimension
#H = sqrt((sigma^2)*pi)*exp.(-CC_dist2/(4*sigma^2))
set_H(H,CC_dist2,sigma,b)
#check if the sum is performed along the right dimension
set_htr(hx_tr,xTr_dist,sigma,Tx), set_htr(hy_tr,yTr_dist,sigma,Ty)
set_hte(hx_te,xTe_dist,sigma,lx-Tx), set_hte(hy_te,yTe_dist,sigma,ly-Ty)
for i in 1:folds
h_tr = hx_tr[i] - hy_tr[i]
h_te = hx_te[i] - hy_te[i]
#set_h(h_tr,hx_tr[i],hy_tr[i],b)
#set_h(h_te,hx_te[i],hy_te[i],b)
for (lambda_idx,lambda) in enumerate(lambda_list)
set_theta(theta,H,lambda,h_tr,b)
score_cv[sigma_idx,lambda_idx] += dot(theta,H*theta) - 2*dot(theta,h_te)
end
end
end
#retrieve the value of the optimal parameters
sigma_chosen = sigma_list[findmin(score_cv)[2][2]]
lambda_chosen = lambda_list[findmin(score_cv)[2][2]]
#calculating the new "optimal" solution
H = sqrt((sigma_chosen^2)*pi)*exp.(-CC_dist2/(4*sigma_chosen^2))
H_lambda = H + lambda_chosen*Matrix{Float64}(I, b, b)
h = (1/lx)*sum(exp.(-xC_dist2/(2*sigma_chosen^2)),dims = 1) - (1/ly)*sum(exp.(-yC_dist2/(2*sigma_chosen^2)),dims = 1)
theta_final = H_lambda\transpose(h)
f = transpose(theta_final).*sum(exp.(-vcat(xC_dist2,yC_dist2)/(2*sigma_chosen^2)),dims = 1)
L2 = 2*dot(theta_final,h) - dot(theta_final,H*theta_final)
return L2
end
function set_H(H::Array{Float64,2},dist::Array{Float64,2},sigma::Float64,b::Int16)
for i in 1:b
for j in 1:b
H[i,j] = sqrt((sigma^2)*pi)*exp(-dist[i,j]/(4*sigma^2))
end
end
end
function set_theta(theta::Array{Float64,1},H::Array{Float64,2},lambda::Float64,h::Array{Float64,2},b::Int64)
Hl = (H + lambda*Matrix{Float64}(I, b, b))
LAPACK.posv!('L', Hl, h)
theta = h
end
function set_htr(h::Array{Float64,1},dists::Array{Float64,2},sigma::Float64,T::Int16)
for (CVidx,dist) in enumerate(dists)
for (idx,value) in enumerate((1/T)*sum(exp.(-dist/(2*sigma^2)),dims = 1))
h[CVidx][idx] = value
end
end
end
function set_hte(h::Array{Float64,1},dists::Array{Float64,2},sigma::Array{Float64,1},T::Int16)
for (CVidx,dist) in enumerate(dists)
for (idx,value) in enumerate((1/T)*sum(exp.(-dist/(2*sigma^2)),dims = 1))
h[CVidx][idx] = value
end
end
end
function set_h(h,h1,h2,b)
for i in 1:b
h[i] = h1[i] - h2[i]
end
end
The set_H, set_h and set_theta functions are there because I read somewhere that modifying prealocated memory in place with a function was faster, but it did not make a great difference.
To test it, I use two random distribution as input data :
x,y = rand(500),1.5*rand(500)
lsdd(x,y) #returns a value around 0.3
Now here is the version of the code where I try to use Optimizer :
function Theta(sigma::Float64,lambda::Float64,x::Array{Float64,1},y::Array{Float64,1},folds::Int8)
lx,ly = length(x), length(y)
b = min(lx+ly,300)
C = shuffle(vcat(x,y))[1:b]
CC_dist2 = squared_distance(C,C)
xC_dist2, yC_dist2 = squared_distance(x,C), squared_distance(y,C)
#the subsets are not be mutually exclusive !
Tx,Ty = length(x) - div(lx,folds), length(y) - div(ly,folds)
shuffled_x, shuffled_y = [shuffle(1:lx) for i in 1:folds], [shuffle(1:ly) for i in 1:folds]
cv_index1, cv_index2 = floor.(collect(1:lx)*folds/lx)[shuffle(1:lx)], floor.(collect(1:ly)*folds/ly)[shuffle(1:ly)]
tr_idx1,tr_idx2 = [i[1:Tx] for i in shuffled_x], [i[1:Ty] for i in shuffled_y]
te_idx1,te_idx2 = [i[Tx:end] for i in shuffled_x], [i[Ty:end] for i in shuffled_y]
xTr_dist, yTr_dist = [xC_dist2[i,:] for i in tr_idx1], [yC_dist2[i,:] for i in tr_idx2]
xTe_dist, yTe_dist = [xC_dist2[i,:] for i in te_idx1], [yC_dist2[i,:] for i in te_idx2]
score_cv = 0
Id = Matrix{Float64}(I, b, b)
H = sqrt((sigma^2)*pi)*exp.(-CC_dist2/(4*sigma^2))
hx_tr, hy_tr = [transpose((1/Tx)*sum(exp.(-dist/(2*sigma^2)),dims = 1)) for dist in xTr_dist], [transpose((1/Ty)*sum(exp.(-dist/(2*sigma^2)),dims = 1)) for dist in yTr_dist]
hx_te, hy_te = [(lx-Tx)*sum(exp.(-dist/(2*sigma^2)),dims = 1) for dist in xTe_dist], [(ly-Ty)*sum(exp.(-dist/(2*sigma^2)),dims = 1) for dist in yTe_dist]
for i in 1:folds
h_tr, h_te = hx_tr[i] - hy_tr[i], hx_te[i] - hy_te[i]
#theta = (H + lambda * Id)\h_tr
theta = copy(h_tr)
Hl = (H + lambda*Matrix{Float64}(I, b, b))
LAPACK.posv!('L', Hl, theta)
score_cv += dot(theta,H*theta) - 2*dot(theta,h_te)
end
return score_cv,(CC_dist2,xC_dist2,yC_dist2)
end
function cost(params::Array{Float64,1},x::Array{Float64,1},y::Array{Float64,1},folds::Int8)
s,l = params[1],params[2]
return Theta(s,l,x,y,folds)[1]
end
"""
Performs the optinization
"""
function lsdd3(x::Array{Float64,1},y::Array{Float64,1}; folds = 4)
start = [1,0.1]
b = min(length(x)+length(y),300)
lx,ly = length(x),length(y)
#result = optimize(params -> cost(params,x,y,folds),fill(0.0,2),fill(50.0,2),start, Fminbox(LBFGS(linesearch=LineSearches.BackTracking())); autodiff = :forward)
result = optimize(params -> cost(params,x,y,folds),start, BFGS(),Optim.Options(f_calls_limit = 5, iterations = 5))
#bboptimize(rosenbrock2d; SearchRange = [(-5.0, 5.0), (-2.0, 2.0)])
#result = optimize(cost,[0,0],[Inf,Inf],start, Fminbox(AcceleratedGradientDescent()))
sigma_chosen,lambda_chosen = Optim.minimizer(result)
CC_dist2, xC_dist2, yC_dist2 = Theta(sigma_chosen,lambda_chosen,x,y,folds)[2]
H = sqrt((sigma_chosen^2)*pi)*exp.(-CC_dist2/(4*sigma_chosen^2))
h = (1/lx)*sum(exp.(-xC_dist2/(2*sigma_chosen^2)),dims = 1) - (1/ly)*sum(exp.(-yC_dist2/(2*sigma_chosen^2)),dims = 1)
theta_final = (H + lambda_chosen*Matrix{Float64}(I, b, b))\transpose(h)
f = transpose(theta_final).*sum(exp.(-vcat(xC_dist2,yC_dist2)/(2*sigma_chosen^2)),dims = 1)
L2 = 2*dot(theta_final,h) - dot(theta_final,H*theta_final)
return L2
end
No matter, which kind of option I use in the optimizer, I always end up with something too slow. Maybe the grid search is the best option, but I don't know how to make it faster... Does anyone have an idea how I could proceed further ?
[1] : http://www.mcduplessis.com/wp-content/uploads/2016/05/Journal-IEICE-2014-CLSDD-1.pdf
[2] : http://www.ms.k.u-tokyo.ac.jp/software.html
What is the fastest way to iterate over all elements in a 3D NumPy array? If array.shape = (r,c,z), there must be something faster than this:
x = np.asarray(range(12)).reshape((1,4,3))
#function that sums nearest neighbor values
x = np.asarray(range(12)).reshape((1, 4,3))
#e is my element location, d is the distance
def nn(arr, e, d=1):
d = e[0]
r = e[1]
c = e[2]
return sum(arr[d,r-1,c-1:c+2]) + sum(arr[d,r+1, c-1:c+2]) + sum(arr[d,r,c-1]) + sum(arr[d,r,c+1])
Instead of creating a nested for loop like the one below to create my values of e to run the function nn for each pixel :
for dim in range(z):
for row in range(r):
for col in range(c):
e = (dim, row, col)
I'd like to vectorize my nn function in a way that extracts location information for each element (e = (0,1,1) for example) and iterates over ALL elements in my matrix without having to manually input each locational value of e OR creating a messy nested for loop. I'm not sure how to apply np.vectorize to this problem. Thanks!
It is easy to vectorize over the d dimension:
def nn(arr, e):
r,c = e # (e[0],e[1])
return np.sum(arr[:,r-1,c-1:c+2],axis=2) + np.sum(arr[:,r+1,c-1:c+2],axis=2) +
np.sum(arr[:,r,c-1],axis=?) + np.sum(arr[:,r,c+1],axis=?)
now just iterate over the row and col dimensions, returning a vector, that is assigned to the appropriate slot in x.
for row in <correct range>:
for col in <correct range>:
x[:,row,col] = nn(data, (row,col))
The next step is to make
rows = [:,None]
cols =
arr[:,rows-1,cols+2] + arr[:,rows,cols+2] etc.
This kind of problem has come up many times, with various descriptions - convolution, smoothing, filtering etc.
We could do some searches to find the best, or it you prefer, we could guide you through the steps.
Converting a nested loop calculation to Numpy for speedup
is a question similar to yours. There's only 2 levels of looping, and sum expression is different, but I think it has the same issues:
for h in xrange(1, height-1):
for w in xrange(1, width-1):
new_gr[h][w] = gr[h][w] + gr[h][w-1] + gr[h-1][w] +
t * gr[h+1][w-1]-2 * (gr[h][w-1] + t * gr[h-1][w])
Here's what I ended up doing. Since I'm returning the xv vector and slipping it in to the larger 3D array lag, this should speed up the process, right? data is my input dataset.
def nn3d(arr, e):
r,c = e
n = np.copy(arr[:,r-1:r+2,c-1:c+2])
n[:,1,1] = 0
n3d = np.ma.masked_where(n == nodata, n)
xv = np.zeros(arr.shape[0])
for d in range(arr.shape[0]):
if np.ma.count(n3d[d,:,:]) < 2:
element = nodata
else:
element = np.sum(n3d[d,:,:])/(np.ma.count(n3d[d,:,:])-1)
xv[d] = element
return xv
lag = np.zeros(shape = data.shape)
for r in range(1,data.shape[1]-1): #boundary effects
for c in range(1,data.shape[2]-1):
lag[:,r,c] = nn3d(data,(r,c))
What you are looking for is probably array.nditer:
a = np.arange(6).reshape(2,3)
for x in np.nditer(a):
print(x, end=' ')
which prints
0 1 2 3 4 5