I need to compute the sum of this series
I need the output this way:
If n = 3;
x = function_name(n)
I need to get x = 11.
If n = 5;
x = function_name(n)
I need to get x = 45.
I believe I need a for-loop to iterate; but am finding it difficult to iterate the increment value itself.
I guess you want the sum of the cumsum of the differences d of the numbers:
d = 2;
n = 5;
s = d:d:d*(n-1)
cs = cumsum( [1 s] )
scs = sum(cs)
%// or as anonymous function
scsh = #(n,d) sum( cumsum( [1 d:d:d*(n-1)] ) )
scs =
45
scsh(5,2) =
45
No need for a loop!
inc=2;
sum=1;
next=1;
n=input('what is n?\n');
for i=2:n
next=next+inc;
sum=sum+next;
inc=inc+2;
end
disp('sum is ');
disp(sum);
function Total = abc(n)
nth_term=1;
Total = 1 ;
for d = 2:2:(2*(n-1))
nth_term = nth_term + d;
Total =Total + nth_term;
end
end
Related
I have some matlab code as follow, constructing KNN similarity weight matrix.
[D,I] = pdist2(X, X, 'squaredeuclidean', 'Smallest', k+1);
D = D < threshold;
W = zeros(n, n);
for i=1:size(I,2)
W(I(:,i), i) = D(:,i);
W(i, I(:,i)) = D(:,i)';
end
I want to vectorize the for loop. I have tried
W(I) = D;
but failed to get the correct value.
I add test case here:
n = 5;
D = [
1 1 1 1 1
0 1 1 1 1
0 0 0 0 0
];
I = [
1 2 3 4 5
5 4 5 2 3
3 1 1 1 1
];
There are some undefined variables that makes it hard to check what it is doing, but this should do the same as your for loop:
D,I] = pdist2(X, X, 'squaredeuclidean', 'Smallest', k+1);
D = D < threshold;
W = zeros(n);
% set the diagonal values
W(sub2ind(size(X), I(1, :), I(1, :))) = D(1,:);
% set the other values
W(sub2ind(size(W), I(2, :), 1:size(I, 2))) = D(2, :);
W(sub2ind(size(W), 1:size(I, 2), I(2, :))) = D(2, :).';
I splited the directions, it works now with your test case.
A possible solution:
idx1 = reshape(1:n*n,n,n).';
idx2 = bsxfun(#plus,I,0:n:n*size(I,2)-1);
W=zeros(n,n);
W(idx2) = D;
W(idx1(idx2)) = D;
Here assumed that you repeatedly want to compute D and I so compute idx only one time and use it repeatedly.
n = 5;
idx1 = reshape(1:n*n,n,n).';
%for k = 1 : 1000
%[D,I] = pdist2(X, X, 'squaredeuclidean', 'Smallest', k+1);
%D = D < threshold;
idx2 = bsxfun(#plus,I,0:n:n*size(I,2)-1);
W=zeros(n,n);
W(idx2) = D;
W(idx1(idx2)) = D;
%end
But if n isn't constant and it varies in each iteration it is better to change the way idx1 is computed:
n = 5;
%for k = 1 : 1000
%n = randi([2 10]);%n isn't constant
%[D,I] = pdist2(X, X, 'squaredeuclidean', 'Smallest', k+1);
%D = D < threshold;
idx1 = bsxfun(#plus,(0:n:n^2-1).',1:size(I,2));
idx2 = bsxfun(#plus,I,0:n:n*size(I,2)-1);
W=zeros(n,n);
W(idx2) = D;
W(idx1(idx2)) = D;
%end
You can cut some corners with linear indices but if your matrices are big then you should only take the nonzero components of D. Following copies all values of D
W = zeros(n);
W(reshape(sub2ind([n,n],I,[1;1;1]*[1:n]),1,[])) = reshape(D,1,[]);
How can I speed up the following MATLAB code, using vectorization? Right now the single line in the loop is taking hours to run for the case upper = 1e7.
Here is the commented code with sample output:
p = 8;
lower = 1;
upper = 1e1;
n = setdiff(lower:upper,primes(upper)); % contains composite numbers between lower + upper
x = ones(length(n),p); % Preallocated 2-D array of ones
% This loop stores the unique prime factors of each composite
% number from 1 to n, in each row of x. Since the rows will have
% varying lengths, the rows are padded with ones at the end.
for i = 1:length(n)
x(i,:) = [unique(factor(n(i))) ones(1,p-length(unique(factor(n(i)))))];
end
output:
x =
1 1 1 1 1 1 1 1
2 1 1 1 1 1 1 1
2 3 1 1 1 1 1 1
2 1 1 1 1 1 1 1
3 1 1 1 1 1 1 1
2 5 1 1 1 1 1 1
For example, the last row contains the prime factors of 10, if we ignore the ones. I have made the matrix 8 columns wide to account for the many prime factors of numbers up to 10 million.
Thanks for any help!
This is not vectorization, but this version of the loop will save about half of the time:
for k = 1:numel(n)
tmp = unique(factor(n(k)));
x(k,1:numel(tmp)) = tmp;
end
Here is a quick benchmark for this:
function t = getPrimeTime
lower = 1;
upper = 2.^(1:8);
t = zeros(numel(upper),2);
for k = 1:numel(upper)
n = setdiff(lower:upper(k),primes(upper(k))); % contains composite numbers between lower to upper
t(k,1) = timeit(#() getPrime1(n));
t(k,2) = timeit(#() getPrime2(n));
disp(k)
end
p = plot(log2(upper),log10(t));
p(1).Marker = 'o';
p(2).Marker = '*';
xlabel('log_2(range of numbers)')
ylabel('log(time (sec))')
legend({'getPrime1','getPrime2'})
end
function x = getPrime1(n) % the originel function
p = 8;
x = ones(length(n),p); % Preallocated 2-D array of ones
for k = 1:length(n)
x(k,:) = [unique(factor(n(k))) ones(1,p-length(unique(factor(n(k)))))];
end
end
function x = getPrime2(n)
p = 8;
x = ones(numel(n),p); % Preallocated 2-D array of ones
for k = 1:numel(n)
tmp = unique(factor(n(k)));
x(k,1:numel(tmp)) = tmp;
end
end
Here's another approach:
p = 8;
lower = 1;
upper = 1e1;
p = 8;
q = primes(upper);
n = setdiff(lower:upper, q);
x = bsxfun(#times, q, ~bsxfun(#mod, n(:), q));
x(~x) = inf;
x = sort(x,2);
x(isinf(x)) = 1;
x = [x ones(size(x,1), p-size(x,2))];
This seems to be faster than the other two options (but is uses more memory). Borrowing EBH's benchmarking code:
function t = getPrimeTime
lower = 1;
upper = 2.^(1:12);
t = zeros(numel(upper),3);
for k = 1:numel(upper)
n = setdiff(lower:upper(k),primes(upper(k)));
t(k,1) = timeit(#() getPrime1(n));
t(k,2) = timeit(#() getPrime2(n));
t(k,3) = timeit(#() getPrime3(n));
disp(k)
end
p = plot(log2(upper),log10(t));
p(1).Marker = 'o';
p(2).Marker = '*';
p(3).Marker = '^';
xlabel('log_2(range of numbers)')
ylabel('log(time (sec))')
legend({'getPrime1','getPrime2','getPrime3'})
grid on
end
function x = getPrime1(n) % the originel function
p = 8;
x = ones(length(n),p); % Preallocated 2-D array of ones
for k = 1:length(n)
x(k,:) = [unique(factor(n(k))) ones(1,p-length(unique(factor(n(k)))))];
end
end
function x = getPrime2(n)
p = 8;
x = ones(numel(n),p); % Preallocated 2-D array of ones
for k = 1:numel(n)
tmp = unique(factor(n(k)));
x(k,1:numel(tmp)) = tmp;
end
end
function x = getPrime3(n) % Approach in this answer
p = 8;
q = primes(max(n));
x = bsxfun(#times, q, ~bsxfun(#mod, n(:), q));
x(~x) = inf;
x = sort(x,2);
x(isinf(x)) = 1;
x = [x ones(size(x,1), p-size(x,2))];
end
I'm trying to write my own program to sort vectors in matlab. I know you can use the sort(A) on a vector, but I'm trying to code this on my own. My goal is to also sort it in the fewest amount of swaps which is kept track of by the ctr variable. I find and sort the min and max elements first, and then have a loop that looks at the ii minimum vector value and swaps it accordingly.
This is where I start to run into problems, I'm trying to remove all the ii minimum values from my starting vector but I'm not sure how to use the ~= on a vector. Is there a way do this this with a loop? Thanks!
clc;
a = [8 9 13 3 2 8 74 3 1] %random vector, will be function a once I get this to work
[one, len] = size(a);
[mx, posmx] = max(a);
ctr = 0; %counter set to zero to start
%setting min and max at first and last elements
if a(1,len) ~= mx
b = mx;
c = a(1,len);
a(1,len) = b;
a(1,posmx) = c;
ctr = ctr + 1;
end
[mn, posmn] = min(a);
if a(1,1) ~= mn
b = mn;
c = a(1,1);
a(1,1) = b;
a(1,posmn) = c;
ctr = ctr + 1;
end
ii = 2; %starting at 2 since first element already sorted
mini = [mn];
posmini = [];
while a(1,ii) < mx
[mini(ii), posmini(ii - 1)] = min(a(a~=mini))
if a(1,ii) ~= mini(ii)
b = mini(ii)
c = a(1,ii)
a(1,ii) = b
a(1,ii) = c
ctr = ctr + 1;
end
ii = ii + 1;
end
I have two matrices of big sizes, which are something similar to the following matrices.
m; with size 1000 by 10
n; with size 1 by 10.
I would like to subtract each element of n from all elements of m to get ten different matrices, each has size of 1000 by 10.
I started as follows
clc;clear;
nrow = 10000;
ncol = 10;
t = length(n)
for i = 1:nrow;
for j = 1:ncol;
for t = 1:length(n);
m1(i,j) = m(i,j)-n(1);
m2(i,j) = m(i,j)-n(2);
m3(i,j) = m(i,j)-n(3);
m4(i,j) = m(i,j)-n(4);
m5(i,j) = m(i,j)-n(5);
m6(i,j) = m(i,j)-n(6);
m7(i,j) = m(i,j)-n(7);
m8(i,j) = m(i,j)-n(8);
m9(i,j) = m(i,j)-n(9);
m10(i,j) = m(i,j)-n(10);
end
end
end
can any one help me how can I do it without writing the ten equations inside the loop? Or can suggest me any convenient way especially when the two matrices has many columns.
Why can't you just do this:
m01 = m - n(1);
...
m10 = m - n(10);
What do you need the loop for?
Even better:
N = length(n);
m2 = cell(N, 1);
for k = 1:N
m2{k} = m - n(k);
end
Here we go loopless:
nrow = 10000;
ncol = 10;
%example data
m = ones(nrow,ncol);
n = 1:ncol;
M = repmat(m,1,1,ncol);
N = permute( repmat(n,nrow,1,ncol) , [1 3 2] );
result = bsxfun(#minus, M, N );
%or just
result = M-N;
Elapsed time is 0.018499 seconds.
or as recommended by Luis Mendo:
M = repmat(m,1,1,ncol);
result = bsxfun(#minus, m, permute(n, [1 3 2]) );
Elapsed time is 0.000094 seconds.
please make sure that your input vectors have the same orientation like in my example, otherwise you could get in trouble. You should be able to obtain that by transposements or you have to modify this line:
permute( repmat(n,nrow,1,ncol) , [1 3 2] )
according to your needs.
You mentioned in a comment that you want to count the negative elements in each of the obtained columns:
A = result; %backup results
A(A > 0) = 0; %set non-negative elements to zero
D = sum( logical(A),3 );
which will return the desired 10000x10 matrix with quantities of negative elements. (Please verify it, I may got a little confused with the dimensions ;))
Create the three dimensional result matrix. Store your results, for example, in third dimension.
clc;clear;
nrow = 10000;
ncol = 10;
N = length(n);
resultMatrix = zeros(nrow, ncol, N);
neg = zeros(ncol, N); % amount of negative values
for j = 1:ncol
for i = 1:nrow
for t = 1:N
resultMatrix(i,j,t) = m(i,j) - n(t);
end
end
for t = 1:N
neg(j,t) = length( find(resultMatrix(:,j,t) < 0) );
end
end
I wrote a code in Matlab which I predefine the variable "a" and then set up a for loop of 5 iterations where the variable "a" goes through some basic operations. However, the for loop output only saves the fifth iteration of "a." How do I save all 5 iterations in a 1x5 array?
The code is as follows:
a = 10;
k = 0.5;
n = 2;
for m = 1:5
a = a + (a*k) + n;
end
Edit:
I just found it that I have to create a new variable.
a = 10;
k = 0.5;
n = 2;
a_n = zeros(1,5);
for m = 1:5
a = a + (a*k) + n;
a_n(m) = a;
end
You may need to store value of a after each iteration into an another variable x
a = 10;
k = 0.5;
n = 2;
for m = 1:5
a = a + (a*k) + n;
x(m) = a;
end
x
Output:
x =
17.000 27.500 43.250 66.875 102.312
You would need to use a different variable to store the 5 iterations as an array.
Code would look something like this:
a = 10;
k = 0.5;
n = 2;
b = [];
for m = 1:5
a = (a + (a*k) + n)
b = [b a];
end
You can now print b for all 5 iteration values.
Here is an alternate way to update values into the 1-D matrix.