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I was working with Scilab and I decide to work with Julia however there are some errors which I didn't arrive to solve. For instance, I would like to fill out a vector using values of a given function but I got this error. Here is the code that I used:
using LinearAlgebra
A = [5/12 -1/12; 3/4 1/4]; c=[1/3;1]; b=[3/4; 1/4];
N = 10; T = 4; ts = (0:N)*T/N;
dt = T/N; λ = 10^(-14/(2*N+1));
m=length(c) ;
em0=b'/A # b^t * inv(A)
em1 = 1 .-em0*ones(m,1)
γ(z) =#. z/(1.0 -z*em1)
u_hat=complex(zeros(1,N+1));
u_hat[1]=γ(im)
The over-arching issue you are facing is that, coming from Scilab, you are probably not used to distinguishing scalars, vectors and matrices. Like in Matlab, Scilab scalars are really 1x1 matrices, and vectors are really Nx1 or 1xN matrices.
This is very different in Julia. A scalar is not the same as a 1x1 matrix, and a vector is not the same as a Nx1 matrix. You should therefore take care to distinguish them. In particular, you should avoid creating a matrix, zeros(M, 1), when what you really need is a vector, zeros(M).
The direct reason for the error message is that γ(im) is a matrix, because em1 is a matrix:
julia> γ(im)
1×1 Matrix{ComplexF64}:
0.0 + 1.0im
u_hat is also a matrix of ComplexF64, and you are trying to assign a matrix as one of its elements, which naturally won't work, only scalar values can be elements of a Matrix{ComplexF64}.
I took the liberty of writing a cleaned up version of your code:
A = [5/12 -1/12; 3/4 1/4]
# use commas when defining vectors (this is just about style)
b = [3/4, 1/4]
N = 10
## None of the below variables are used. Try to make your example minimal
c = [1/3, 1]
T = 4
dt = T/N;
ts = (0:N) .* dt
λ = 10^(-14/(2*N+1))
m = length(c)
############### <- not used
# prefer vectors over 1xN or Nx1 matrices
em0 = A' \ b
# dot product of a vector and a vector of ones is just a sum, but super-wasterful and slow.
em1 = 1 - sum(em0)
# don't use global variables(!!!), and remove the `#.`
γ(z, a) = z / (1 - z * a)
# use vectors, not 1xN matrices, and directly create a complex matrix instead of converting a real one.
û = zeros(ComplexF64, N+1)
# Now this works
û[1] = γ(im, em1)
I renamed u_hat to û for fun.
Also: remember to put your code in a function, always.
Just in the case of locating the root of the problem:
The problem is where you declared the em1 as em1 = 1 .-em0*ones(m,1). Since the output of the em0*ones(m,1) is expected to be a scalar, you can grasp it using the only function (I don't argue with your approach, and that's out of the interest of this answer):
julia> using LinearAlgebra
# Note that with this modification, there isn't any need for `#.` anymore.
julia> γ(z) = z/(1.0 -z*em1)
γ (generic function with 1 method)
julia> A = [5/12 -1/12; 3/4 1/4]; c=[1/3;1]; b=[3/4; 1/4];
N = 10; T = 4; ts = (0:N)*T/N;
dt = T/N; λ = 10^(-14/(2*N+1));
m=length(c);
em0=b'/A;
#This is where the problem can be solved
em1 = 1 - only(em0*ones(m,1));
u_hat=complex(zeros(1,N+1));
u_hat[1]=γ(im)
0.0 + 1.0im
julia> u_hat
1×11 Matrix{ComplexF64}:
0.0+1.0im 0.0+0.0im 0.0+0.0im 0.0+0.0im 0.0+0.0im … 0.0+0.0im 0.0+0.0im 0.0+0.0im 0.0+0.0im 0.0+0.0im
I need to identify the rows (/columns) that have defined values in a large sparse Boolean Matrix. I want to use this to 1. slice (actually view) the Matrix by those rows/columns; and 2. slice (/view) vectors and matrices that have the same dimensions as the margins of a Matrix. I.e. the result should probably be a Vector of indices / Bools or (preferably) an iterator.
I've tried the obvious:
a = sprand(10000, 10000, 0.01)
cols = unique(a.colptr)
rows = unique(a.rowvals)
but each of these take like 20ms on my machine, probably because they allocate about 1MB (at least they allocate cols and rows). This is inside a performance-critical function, so I'd like the code to be optimized. The Base code seems to have an nzrange iterator for sparse matrices, but it is not easy for me to see how to apply that to my case.
Is there a suggested way of doing this?
Second question: I'd need to also perform this operation on views of my sparse Matrix - would that be something like x = view(a,:,:); cols = unique(x.parent.colptr[x.indices[:,2]]) or is there specialized functionality for this? Views of sparse matrices appear to be tricky (cf https://discourse.julialang.org/t/slow-arithmetic-on-views-of-sparse-matrices/3644 – not a cross-post)
Thanks a lot!
Regarding getting the non-zero rows and columns of a sparse matrix, the following functions should be pretty efficient:
nzcols(a::SparseMatrixCSC) = collect(i
for i in 1:a.n if a.colptr[i]<a.colptr[i+1])
function nzrows(a::SparseMatrixCSC)
active = falses(a.m)
for r in a.rowval
active[r] = true
end
return find(active)
end
For a 10_000x10_000 matrix with 0.1 density it takes 0.2ms and 2.9ms for cols and rows, respectively. It should also be quicker than method in question (apart from the correctness issue as well).
Regarding views of sparse matrices, a quick solution would be to turn view into a sparse matrix (e.g. using b = sparse(view(a,100:199,100:199))) and use functions above. In code:
nzcols(b::SubArray{T,2,P}) where {T,P<:AbstractSparseArray} = nzcols(sparse(b))
nzrows(b::SubArray{T,2,P}) where {T,P<:AbstractSparseArray} = nzrows(sparse(b))
A better solution would be to customize the functions according to view. For example, when the view uses UnitRanges for both rows and columns:
# utility predicate returning true if element of sorted v in range r
inrange(v,r) = searchsortedlast(v,last(r))>=searchsortedfirst(v,first(r))
function nzcols(b::SubArray{T,2,P,Tuple{UnitRange{Int64},UnitRange{Int64}}}
) where {T,P<:SparseMatrixCSC}
return collect(i+1-start(b.indexes[2])
for i in b.indexes[2]
if b.parent.colptr[i]<b.parent.colptr[i+1] &&
inrange(b.parent.rowval[nzrange(b.parent,i)],b.indexes[1]))
end
function nzrows(b::SubArray{T,2,P,Tuple{UnitRange{Int64},UnitRange{Int64}}}
) where {T,P<:SparseMatrixCSC}
active = falses(length(b.indexes[1]))
for c in b.indexes[2]
for r in nzrange(b.parent,c)
if b.parent.rowval[r] in b.indexes[1]
active[b.parent.rowval[r]+1-start(b.indexes[1])] = true
end
end
end
return find(active)
end
which work faster than the versions for the full matrices (for 100x100 submatrix of above 10,000x10,000 matrix cols and rows take 16μs and 12μs, respectively on my machine, but these are unstable results).
A proper benchmark would use fixed matrices (or at least fix the random seed). I'll edit this line with such a benchmark if I do it.
In case the indices are not ranges, the fallback to converting to a sparse matrix works, but here are versions for indices which are Vectors. If the indices are mixed, yet another set of versions needs to be made. Quite repetitive, but this is the strength of Julia, when the versions are done, the code will choose optimized methods correctly using the types in the caller without too much effort.
function sortedintersecting(v1, v2)
i,j = start(v1), start(v2)
while i <= length(v1) && j <= length(v2)
if v1[i] == v2[j] return true
elseif v1[i] > v2[j] j += 1
else i += 1
end
end
return false
end
function nzcols(b::SubArray{T,2,P,Tuple{Vector{Int64},Vector{Int64}}}
) where {T,P<:SparseMatrixCSC}
brows = sort(unique(b.indexes[1]))
return [k
for (k,i) in enumerate(b.indexes[2])
if b.parent.colptr[i]<b.parent.colptr[i+1] &&
sortedintersecting(brows,b.parent.rowval[nzrange(b.parent,i)])]
end
function nzrows(b::SubArray{T,2,P,Tuple{Vector{Int64},Vector{Int64}}}
) where {T,P<:SparseMatrixCSC}
active = falses(length(b.indexes[1]))
for c in b.indexes[2]
active[findin(b.indexes[1],b.parent.rowval[nzrange(b.parent,c)])] = true
end
return find(active)
end
-- ADDENDUM --
Since it was noted nzrows for Vector{Int} indices is a bit slow, this is an attempt to improve its speed by replacing findin with a version exploiting sortedness:
function findin2(inds,v,w)
i,j = start(v),start(w)
res = Vector{Int}()
while i<=length(v) && j<=length(w)
if v[i]==w[j]
push!(res,inds[i])
i += 1
elseif (v[i]<w[j]) i += 1
else j += 1
end
end
return res
end
function nzrows(b::SubArray{T,2,P,Tuple{Vector{Int64},Vector{Int64}}}
) where {T,P<:SparseMatrixCSC}
active = falses(length(b.indexes[1]))
inds = sortperm(b.indexes[1])
brows = (b.indexes[1])[inds]
for c in b.indexes[2]
active[findin2(inds,brows,b.parent.rowval[nzrange(b.parent,c)])] = true
end
return find(active)
end
Given an n*2 data matrix X I'd like to calculate the bivariate empirical cdf for each observation, i.e. for each i in 1:n, return the percentage of observations with 1st element not greater than X[i,1] and 2nd element not greater than X[i,2].
Because of the nested search involved it gets terribly slow for n ~ 100k, even after porting it to Fortran. Does anyone know if there's a better way of handling sample sizes like this?
Edit: I believe this problem is similar (in terms of complexity) to finding Kendall's tau, which is of order O(n^2). In that case Knight (1966) has an algorithm to reduce it to O(n log(n)). Just wondering if there's any O(n*log(n)) algorithm for finding bivariate ecdf already out there.
Edit 2: This is the code I have in Fortran, as requested. This is called in R in the usual way, so the R code is omitted here. The code is meant for arbitrary dimensions, but for the specific thing I'm doing a bivariate one is good enough.
! Calculates multivariate empirical cdf for each point
! n: number of observations
! d: dimension (>=2)
! umat: data matrix
! outvec: vector of ecdf
subroutine mecdf(n,d,umat,outvec)
implicit none
integer :: n, d, i, j, k, tempsum
double precision, dimension(n) :: outvec
double precision, dimension(n,d) :: umat
logical :: flag
do i = 1,n
tempsum = 0
do j = 1,n
flag = .true.
do k = 1,d
if (umat(i,k) < umat(j,k)) then
flag = .false.
exit
end if
end do
if (flag) then
tempsum = tempsum + 1
end if
end do
outvec(i) = real(tempsum)/n
end do
return
end subroutine
I think my first effort was not really an ecdf, although it did map the points to the interval [0,1] The example, a 25 x 2 matrix generated with:
#M <- matrix(runif(100), ncol=2)
M <-
structure(c(0.0468267474789172, 0.296053855214268, 0.205678076483309,
0.467400068417192, 0.968577065737918, 0.435642971657217, 0.929023026255891,
0.038406387437135, 0.304360694251955, 0.964778139721602, 0.534192910650745,
0.741682186257094, 0.0848641532938927, 0.405901980120689, 0.957696850644425,
0.384813814423978, 0.639882878866047, 0.231505588628352, 0.271994129288942,
0.786155494628474, 0.349499785574153, 0.279077709652483, 0.206662984099239,
0.777465222170576, 0.705439242534339, 0.643429880728945, 0.887209519045427,
0.0794123203959316, 0.849177583120763, 0.704594585578889, 0.736909110797569,
0.503158083418384, 0.49449566937983, 0.408533290959895, 0.236613316927105,
0.297427259152755, 0.0677345870062709, 0.623845702270046, 0.139933609170839,
0.740499466424808, 0.628097783308476, 0.678438259987161, 0.186680511338636,
0.339367639739066, 0.373212536331266, 0.976724133593962, 0.94558056560345,
0.610417427960783, 0.887977657606825, 0.663434249348938, 0.447939050383866,
0.755168803501874, 0.478974275058135, 0.737040047068149, 0.429466919740662,
0.0021107573993504, 0.697435079608113, 0.444197302218527, 0.108997165458277,
0.856855363817886, 0.891898229718208, 0.93553287582472, 0.991948011796921,
0.630414301762357, 0.0604106825776398, 0.908968194155023, 0.0398679254576564,
0.251426834380254, 0.235532913124189, 0.392070295521989, 0.530511683085933,
0.319339724024758, 0.534880011575297, 0.92030712752603, 0.138276003766805,
0.213625695323572, 0.407931711757556, 0.605797187192366, 0.424798395251855,
0.471233424032107, 0.0105366336647421, 0.625802840106189, 0.524665891425684,
0.0375960320234299, 0.54812005511485, 0.0105806747451425, 0.438266788609326,
0.791981092421338, 0.363821814302355, 0.157931488472968, 0.47945317090489,
0.906797411618754, 0.762243523262441, 0.258681379957125, 0.308056800393388,
0.91944490163587, 0.412255838746205, 0.347220918396488, 0.68236422073096,
0.559149842709303), .Dim = c(50L, 2L))
So the task is to do a single summation of a two-part logical test on N items which I suspect is O(N*3). It might be marginally faster if implemented in Rcpp, but these are vectorized operations.
# Wrong: ecdf2d <- function(m,i,j) { ord <- rank(m[ , 1]^2+m[ , 2]^2)
# ord[i]/nrow(m)} # scales to [0,1] interval
ecdf2d.v2 <- function(obj, x, y) sum( obj[,1] < x & obj[,2] < y)/nrow(obj)
I'm trying to solve (for m_0) numerically the following ordinary differential equation:
dm0/dx=(((1-x)*(x*(2-x))**(1.5))/(k+x)**2)*(((x*(2-x))/3.0)*(dw/dx)**2 + ((8*(k+1))/(3*(k+x)))*w**2)
The values of w and dw/dx have been found already numerically using the Runge-Kutta 4th order and k is a factor that is fixed. I wrote a code where I call the values for w and dw/dx from an external file, then I organize them in an array, then I call the array in the function and then I run the integration. My outcome is not what it's expected :(, I don't know what is wrong. If anyone could give me a hand, it would be highly appreciated. Thank you!
from math import sqrt
from numpy import array,zeros,loadtxt
from printSoln import *
from run_kut4 import *
m = 1.15 # Just a constant.
k = 3.0*sqrt(1.0-(1.0/m))-1.0 # k in terms of m.
omegas = loadtxt("omega.txt",float) # Import values of w
domegas = loadtxt("domega.txt",float) # Import values of dw/dx
w = [] # Defines the array w to store the values w^2
s = 0.0
for s in omegas:
w.append(s**2) # Calculates the values w**2
omeg = array(w,float) # Array to store the value of w**2
dw = [] # Defines the array dw to store the values dw**2
t = 0.0
for t in domegas:
dw.append(t**2) # Calculates the values for dw**2
domeg = array(dw,float) # Array to store the values of dw**2
x = 1.0e-12 # Starting point of integration
xStop = (2.0 - k)/3.0 # Final point of integration
def F(x,y): # Define function to be integrated
F = zeros(1)
for i in domeg: # Loop to call w^2, (dw/dx)^2
for j in omeg:
F[0] = (((1.0-x)*(x*(2.0-x))**(1.5))/(k+x)**2)*((1.0/3.0)*x* (2.0-x)*domeg[i] + (8.0*(k+1.0)*omeg[j])/(3.0*(k+x)))
return F
y = array([((32.0*sqrt(2.0)*(k+1.0)*(x**2.5))/(15.0*(k**3)))]) # Initial condition for m_{0}
h = 1.0e-5 # Integration step
freq = 0 # Prints only initial and final values
X,Y = integrate(F,x,y,xStop,h) # Calls Runge-Kutta 4
printSoln(X,Y,freq) # Prints solution
Interpreting your verbal description, there is an ODE for omega, w'=F(x,w), and a coupled ODE for m0, m'=G(x,m,w,w'). The almost always optimal way to solve this is to treat it as system of ODE,
def ODEfunc(x,y)
w,m = y
dw = F(x,w)
dm = G(x,m,w,dw)
return np.array([dw, dm])
which you can then insert in the ODE solver of your choice, e.g., the fictitious
ODEintegrate(ODEfunc, xsamples, y0)
I'm using Julia 0.3.4
I'm trying to write LU-decomposition using Gaussian elimination. So I have to swap rows. And here's my problem:
If I'm using a,b = b,a I get an error,
but if I'm using:
function swapRows(row1, row2)
temp = row1
row1 = row2
row2 = temp
end
then everything works just fine.
Am I doing something wrong or it's a bug?
Here's my source code:
function lu_t(A::Matrix)
# input value: (A), where A is a matrix
# return value: (L,U), where L,U are matrices
function swapRows(row1, row2)
temp = row1
row1 = row2
row2 = temp
return null
end
if size(A)[1] != size(A)[2]
throw(DimException())
end
n = size(A)[1] # matrix dimension
U = copy(A) # upper triangular matrix
L = eye(n) # lower triangular matrix
for k = 1:n-1 # direct Gaussian elimination for each column `k`
(val,id) = findmax(U[k:end,k]) # find max pivot element and it's row `id`
if val == 0 # check matrix for singularity
throw(SingularException())
end
swapRows(U[k,k:end],U[id,k:end]) # swap row `k` and `id`
# U[k,k:end],U[id,k:end] = U[id,k:end],U[k,k:end] - error
for i = k+1:n # for each row `i` > `k`
μ = U[i,k] / U[k,k] # find elimination coefficient `μ`
L[i,k] = μ # save to an appropriate position in lower triangular matrix `L`
for j = k:n # update each value of the row `i`
U[i,j] = U[i,j] - μ⋅U[k,j]
end
end
end
return (L,U)
end
###### main code ######
A = rand(4,4)
#time (L,U) = lu_t(A)
#test_approx_eq(L*U, A)
The swapRows function is a no-op and has no effect whatsoever – all it does is swap around some local variable names. See various discussions of the difference between assignment and mutation:
https://groups.google.com/d/msg/julia-users/oSW5hH8vxAo/llAHRvvFVhMJ
http://julia.readthedocs.org/en/latest/manual/faq/#i-passed-an-argument-x-to-a-function-modified-it-inside-that-function-but-on-the-outside-the-variable-x-is-still-unchanged-why
http://julia.readthedocs.org/en/latest/manual/faq/#why-does-x-y-allocate-memory-when-x-and-y-are-arrays
The constant null doesn't mean what you think it does – in Julia v0.3 it's a function that computes the null space of a linear transformation; in Julia v0.4 it still means this but has been deprecated and renamed to nullspace. The "uninteresting" value in Julia is called nothing.
I'm not sure what's wrong with your commented out row swapping code, but this general approach does work:
julia> X = rand(3,4)
3x4 Array{Float64,2}:
0.149066 0.706264 0.983477 0.203822
0.478816 0.0901912 0.810107 0.675179
0.73195 0.756805 0.345936 0.821917
julia> X[1,:], X[2,:] = X[2,:], X[1,:]
(
1x4 Array{Float64,2}:
0.478816 0.0901912 0.810107 0.675179,
1x4 Array{Float64,2}:
0.149066 0.706264 0.983477 0.203822)
julia> X
3x4 Array{Float64,2}:
0.478816 0.0901912 0.810107 0.675179
0.149066 0.706264 0.983477 0.203822
0.73195 0.756805 0.345936 0.821917
Since this creates a pair of temporary arrays that we can't yet eliminate the allocation of, this isn't the most efficient approach. If you want the most efficient code here, looping over the two rows and swapping pairs of scalar values will be faster:
function swapRows!(X, i, j)
for k = 1:size(X,2)
X[i,k], X[j,k] = X[j,k], X[i,k]
end
end
Note that it is conventional in Julia to name functions that mutate one or more of their arguments with a trailing !. Currently, closures (i.e. inner functions) have some performance issues, so you'll want such a helper function to be defined at the top-level scope instead of inside of another function the way you've got it.
Finally, I assume this is an exercise since Julia ships with carefully tuned generic (i.e. it works for arbitrary numeric types) LU decomposition: http://docs.julialang.org/en/release-0.3/stdlib/linalg/#Base.lu.
-
It's quite simple
julia> A = rand(3,4)
3×4 Array{Float64,2}:
0.241426 0.283391 0.201864 0.116797
0.457109 0.138233 0.346372 0.458742
0.0940065 0.358259 0.260923 0.578814
julia> A[[1,2],:] = A[[2,1],:]
2×4 Array{Float64,2}:
0.457109 0.138233 0.346372 0.458742
0.241426 0.283391 0.201864 0.116797
julia> A
3×4 Array{Float64,2}:
0.457109 0.138233 0.346372 0.458742
0.241426 0.283391 0.201864 0.116797
0.0940065 0.358259 0.260923 0.578814