I coded a function picircle() that estimates pi.
Now I would like to plot this function for N values.
function Plotpi()
p = 100 # precision of π
N = 5
for i in 1:N
picircle(p)
end
end
3.2238805970149254
3.044776119402985
3.1641791044776117
3.1243781094527363
3.084577114427861
Now I am not sure how to plot the function, I tried plot(PP()) but it didn't work
Here I defined picircle:
function picircle(n)
n = n
L = 2n+1
x = range(-1, 1, length=L)
y = rand(L)
center = (0,0)
radius = 1
n_in_circle = 0
for i in 1:L
if norm((x[i], y[i]) .- center) < radius
n_in_circle += 1
end
end
println(4 * n_in_circle / L)
end
Your problem is that your functions don't actually return anything:
julia> x = Plotpi()
3.263681592039801
3.0646766169154227
2.845771144278607
3.18407960199005
3.044776119402985
julia> x
julia> typeof(x)
Nothing
The numbers you see are just printed to the REPL, and print doesn't return any value:
julia> x = print(5)
5
julia> typeof(x)
Nothing
So you probably just want to change your function so that it returns what you want to plot:
julia> function picircle(n)
n = n
L = 2n+1
x = range(-1, 1, length=L)
y = rand(L)
center = (0,0)
radius = 1
n_in_circle = 0
for i in 1:L
if norm((x[i], y[i]) .- center) < radius
n_in_circle += 1
end
end
4 * n_in_circle / L
end
Then:
julia> x = picircle(100)
3.263681592039801
julia> x
3.263681592039801
So now the value of the function is actually returned (rather than just printed to the console). You don't really need a separate function if you just want to do this multiple times and plot the results, a comprehension will do. Here's an example comparing the variability of the estimate with 100 draws vs 50 draws:
julia> using Plots
julia> histogram([picircle(100) for _ ∈ 1:1_000], label = "100 draws", alpha = 0.5)
julia> histogram!([picircle(20) for _ ∈ 1:1_000], label = "20 draws", alpha = 0.5)
Related
Consider the following kernel, which reduces along the rows of a 2-D matrix
function row_sum!(x, ncol, out)
"""out = sum(x, dims=2)"""
row_idx = (blockIdx().x-1) * blockDim().x + threadIdx().x
for i = 1:ncol
#inbounds out[row_idx] += x[row_idx, i]
end
return
end
N = 1024
x = CUDA.rand(Float64, N, 2*N)
out = CUDA.zeros(Float64, N)
#cuda threads=256 blocks=4 row_sum!(x, size(x)[2], out)
isapprox(out, sum(x, dims=2)) # true
How do I write a similar kernel except for reducing along the columns (of a 2-D matrix)? In particular, how do I get the index of each column, similar to how we got the index of each row with row_idx?
Here is the code:
function col_sum!(x, nrow, out)
"""out = sum(x, dims=1)"""
col_idx = (blockIdx().x-1) * blockDim().x + threadIdx().x
for i = 1:nrow
#inbounds out[col_idx] += x[i, col_idx]
end
return
end
N = 1024
x = CUDA.rand(Float64, N, 2N)
out = CUDA.zeros(Float64, 2N)
#cuda threads=256 blocks=8 col_sum!(x, size(x, 1), out)
And here is the test:
julia> isapprox(out, vec(sum(x, dims=1)))
true
As you can see the size of the result vector is now 2N instead of N, hence we had to adapt the number of blocks accordingly (that is multiply by 2 and now we have 8 instead of 4)
More materials can be found here: https://juliagpu.gitlab.io/CUDA.jl/tutorials/introduction/
I am reading "Numerical Methods for Engineers" by Chapra and Canale. In it, they've provided pseudocode for the implementation of Euler's method (for solving ordinary differential equations). Here is the pseucode:
Pseucode for implementing Euler's method
I tried implementing this code in GNU Octave, but depending on the input values, I am getting one of two errors:
The program doesn't give any output at all. I have to press 'Ctrl + C' in order to break execution.
The program gives this message:
error: 'ynew' undefined near line 5 column 21
error: called from
Integrator at line 5 column 9
main at line 18 column 7
I would be very grateful if you could get this program to work for me. I am actually an amateur in GNU Octave. Thank you.
Edit 1: Here is my code. For main.m:
%prompt user
y = input('Initial value of y:');
xi = input('Initial value of x:');
xf = input('Final value of x:');
dx = input('Step size:');
xout = input('Output interval:');
x = xi;
m = 0;
xpm = x;
ypm = y;
while(1)
xend = x + xout;
if xend > xf
xend = xf;
h = dx;
Integrator(x,y,h,xend);
m = m + 1;
xpm = x;
ypm = y;
if x >= xf
break;
endif
endif
end
For Integrator.m:
function Integrator(x,y,h,xend)
while(1)
if xend - x < h
h = xend - x;
Euler(x,y,h,ynew);
y = ynew;
if x >= xend
break;
endif
endif
end
endfunction
For Euler.m:
function Euler(x,y,h,ynew)
Derivs(x,y,dydx);
ynew = y + dydx * h;
x = x + h;
endfunction
For Derivs.m:
function Derivs(x,y,dydx)
dydx = -2 * x^3 + 12 * x^2 - 20 * x + 8.5;
endfunction
Edit 2: I shoud mention that the differential equation which Chapra and Canale have given as an example is:
y'(x) = -2 * x^3 + 12 * x^2 - 20 * x + 8.5
That is why the 'Derivs.m' script shows dydx to be this particular polynomial.
Here is my final code. It has four different M-files:
main.m
%prompt the user
y = input('Initial value of y:');
x = input('Initial value of x:');
xf = input('Final value of x:');
dx = input('Step size:');
xout = dx;
%boring calculations
m = 1;
xp = [x];
yp = [y];
while x < xf
[x,y] = Integrator(x,y,dx,min(xf, x+xout));
m = m+1;
xp(m) = x;
yp(m) = y;
end
%plot the final result
plot(xp,yp);
title('Solution using Euler Method');
ylabel('Dependent variable (y)');
xlabel('Independent variable (x)');
grid on;
Integrator.m
%This function takes in 4 inputs (x,y,h,xend) and returns 2 outputs [x,y]
function [x,y] = Integrator(x,y,h,xend)
while x < xend
h = min(h, xend-x);
[x,y] = Euler(x,y,h);
end
endfunction
Euler.m
%This function takes in 3 inputs (x,y,h) and returns 2 outputs [x,ynew]
function [x,ynew] = Euler(x,y,h)
dydx = Derivs(x,y);
ynew = y + dydx * h;
x = x + h;
endfunction
Derivs.m
%This function takes in 2 inputs (x,y) and returns 1 output [dydx]
function [dydx] = Derivs(x,y)
dydx = -2 * x^3 + 12 * x^2 - 20 * x + 8.5;
endfunction
Your functions should look like
function [x, y] = Integrator(x,y,h,xend)
while x < xend
h = min(h, xend-x)
[x,y] = Euler(x,y,h);
end%while
end%function
as an example. Depending on what you want to do with the result, your main loop might need to collect all the results from the single steps. One variant for that is
m = 1;
xp = [x];
yp = [y];
while x < xf
[x,y] = Integrator(x,y,dx,min(xf, x+xout));
m = m+1;
xp(m) = x;
yp(m) = y;
end%while
I am trying to check the performance of k-means with the silhouette function but I am getting an error.
I am calling the function like this [out1,out2] = silhouette(normalized, idx); or [out1,out2] = silhouette(normalized, idx, 'cosine');
The definition of the function is function [si, h] = silhouette(X, clust, metric)
I expect to take a number between -1,+1 but instead of that I am getting error: element number 2 undefined in return list.
My code for the silhouette function:
function [si, h] = silhouette(X, clust, metric)
% Nan Zhou
% Code Matlab 'silhouette' into Octave function
% Sichuan University, Macquarie University
% zhnanx#gmail.com
% input parameters
% X, n-by-p data matrix
% - Rows of X correspond to points, columns correspond to coordinates.
% clust, clusters defined for X; n-by-1 vector
% metric, e.g. Euclidean, sqEuclidean, cosine
% return values
% si, silhouettte values, n-by-1 vector
% h, figure handle, waiting to be solved in the future
% algorithm reference
% - Peter J. Rousseeuw (1987)
% - Silhouettes: a Graphical Aid to the Interpretation and Validation of Cluster Analysis
% - doi:10.1016/0377-0427(87)90125-7
% check size
if (size(X, 1) != size(clust, 1))
error("First dimension of X <%d> doesn't match that of clust <%d>",...
size(X, 1), size(clust, 1));
endif
% check metric
if (! exist('metric', 'var'))
metric = 'sqEuclidean';
endif
%%%% function set
function [dist] = EuclideanDist(x, y)
dist = sqrt((x - y) * (x - y)');
endfunction
function [dist] = sqEuclideanDist(x, y)
dist = (x - y) * (x - y)';
endfunction
function [dist] = cosineDist(x, y)
cosineValue = dot(x,y)/(norm(x,2)*norm(y,2));
dist = 1 - cosineValue;
endfunction
%%% end function set
% calculating
si = zeros(size(X, 1), 1);
%h
%calculate values of si one by one
for iii = 1:length(si)
%%% distance of iii to all others
iii2all = zeros(size(X, 1), 1);
for jjj = 1:size(X, 1)
switch (metric)
case 'Euclidean'
iii2all(jjj) = EuclideanDist(X(iii, :), X(jjj, :));
case 'sqEuclidean'
iii2all(jjj) = sqEuclideanDist(X(iii, :), X(jjj, :));
case 'cosine'
iii2all(jjj) = cosineDist(X(iii, :), X(jjj, :));
otherwise
error('Invalid metric.');
endswitch
endfor
%%% end distance to all
%%% allocate values to clusters
clusterIDs = unique(clust); % eg [1; 2; 3; 4]
groupedValues = {};
for jjj = 1:length(clusterIDs)
groupedValues{clusterIDs(jjj)} = [iii2all(clust == clusterIDs(jjj))];
endfor
%%% end allocation
%%% calculate a(i)
% dist of object iii to all other objects in the same cluster
a_iii = groupedValues{clust(iii)};
% average distance of iii to all other objects in the same cluster
a_i = sum(a_iii) / (size(a_iii, 1) - 1);
%disp(a_i);pause;
%%% end a(i)
%%% calculate b(i)
clusterIDs_new = clusterIDs;
% remove the cluster iii in
clusterIDs_new(find(clusterIDs_new == clust(iii))) = [];
% average distance of iii to all objects of another cluster
a_iii_2others = zeros(length(clusterIDs_new), 1);
for jjj = 1:length(clusterIDs_new)
values_another = groupedValues{clusterIDs_new(jjj)};
a_iii_2others(jjj) = mean(values_another);
endfor
b_i = min(a_iii_2others);
%disp(b_i);disp('---');pause;
%%% end b(i)
%%% calculate s(i)
si(iii) = (b_i - a_i) / max([a_i; b_i]);
%%% end s(i)
endfor
end
I'm trying to plot a piecewise function in R that is equal to -3 if x < 0, 1/3 * x^3 if x is between 0 and 5 (inclusive), and 4x otherwise. The function I've been running seems to work, but the plot returns an error message. The code I wrote is:
g <- function(x) {
if (x < 0)
-3 # first component
else if (x >= 0 & x <= 5)
(1/3) * x #second component
else
4*x # third component
}
Then when I try to plot it with
plot(g, -20, 20)
I get
Error in curve(expr = x, from = from, to = to, xlim = xlim, ylab = ylab, :
'expr' did not evaluate to an object of length 'n'
In addition: Warning message:
In if (x < 0) -3 else if (x >= 0 & x <= 5) (1/3) * x else 4 * x :
the condition has length > 1 and only the first element will be used
I can't quite figure out how to troubleshoot this, and the 'ifelse' function returned a dark black line when I attempted to plot it, which I'd prefer to avoid, so this seems to be my best bet.
I would greatly appreciate any insights.
For the function
function function_maker(N)
if N == 1
x = 1.0
else
x = 1
end
f(y) = x+y
end
I want the output of this to not be type-stable, but I want it to generate an f that is type-stable, i.e. uses the type of x determined by the value of N to generate a function dependent on N. Basically, I want the functions that come out of this to be performant, but the function_maker itself doesn't need to be performant because it's only used in the global scope or above a function barrier.
f = function_maker(1)
#code_warntype f(1)
Variables:
#self#::#f#9
y::Int64
Body:
begin
return ((Core.getfield)((Core.getfield)(#self#::#f#9, :x)::ANY, :contents)::ANY + y::Int64)::ANY
end::ANY
This doesn't happen by default. I tried f(y) = x::typeof(x)+y but that didn't work either. Is there a simple way to do this?
There's:
julia> function function_maker2(N)
if N == 1
let x = 1.0
return f(y) = x + y
end
else
let x = 1
return f(y) = x + y
end
end
end
function_maker2 (generic function with 1 method)
julia> f2 = function_maker2(1)
(::f) (generic function with 1 method)
julia> #code_warntype f2(1)
Variables:
#self#::#f#5{Float64}
y::Int64
Body:
begin
return (Base.add_float)((Core.getfield)(#self#::#f#5{Float64}, :x)::Float64, (Base.sitofp)(Float64, y::Int64)::Float64)::Float64
end::Float64
This version separates the x in each branch inside let blocks. Otherwise the compiler seems to get confused.