How can I find the smallest number among three that is non zero.
I tried introducing a very small number eps = 1e-6 (my numbers are either zero or clearly larger than eps) and doing tests between min(x,eps), min(y,eps) etc. I didn't get anything. Is there a way to do that with a function?
If the numbers are all stored in a vector x you could do the following:
x = [1 0 2 0 3 0 4];
y = min(x(x>0));
This is based on your statement that
numbers are either zero or clearly larger than eps
If you mean larger in magnitude and you want to accept non-zero negative values you could use:
x = [1 0 -2 0 3 0 4];
y = min(x(x~=0));
Note that this will return the most negative number when negative numbers are present, rather than the number with the smallest non-zero magnitude. To get the number with the smallest non-zero magnitude, you could use:
x = [1 0 -2 0 3 0 4];
xnonzero = x(x~=0);
[~,idx] = min(abs(xnonzero));
y = xnonzero(idx);
It doesn't seem very elegant. There is probably a more direct way.
numbers = [1 3 4 -2 1 0];
answer = min(numbers(numbers>0));
answer == 1
Related
I am attempting to develop a Matlab program to balance chemical equations. I am able to balance them via solving a system of linear equations. Currently my output is a column vector with the coefficients.
My problem is that I need to return the smallest integer values of these coefficients. For example, if [10, 20, 30] was returned. I want [1, 2, 3] to be returned.
What is the best way to accomplish this?
I want this program to be fully autonomous once it is fed a matrix with the linear system. Thus I can not play around with the values, I need to automate this from the code. Thanks!
% Chemical Equation in Matrix Form
Chem = [1 0 0 -1 0 0 0; 1 0 1 0 0 -3 0; 0 2 0 0 -1 0 0; 0 10 0 0 0 -1 0; 0 35 4 -4 0 12 1; 0 0 2 -1 -3 0 2]
%set x4 = 1 then Chem(:, 4) = b and
b = Chem(:, 4); % Arbitrarily set x4 = 1 and set its column equal to b
Chem(:,4) = [] % Delete the x4 column from Chem and shift over
g = 1; % Initialize variable for LCM
x = Chem\b % This is equivalent to the reduced row echelon form of
% Chem | b
% Below is my sad attempt at factoring the values, I divide by the smallest decimal to raise all the values to numbers greater than or equal to 1
for n = 1:numel(x)
g = x(n)*g
M = -min(abs(x))
y = x./M
end
I want code that will take some vector with coefficients, and return an equivalent coefficient vector with the lowest possible integer coefficients. Thanks!
I was able to find a solution without using integer programming. I converted the non-integer values to rational expressions, and used a built-in matlab function to extract the denominator of each of these expressions. I then used a built in matlab function to find the least common multiples of these values. Finally, I multiplied the least common multiple by the matrix to find my answer coefficients.
% Chemical Equation in Matrix Form
clear, clc
% Enter chemical equation as a linear system in matrix form as Chem
Chem = [1 0 0 -1 0 0 0; 1 0 1 0 0 -3 0; 0 2 0 0 -1 0 0; 0 10 0 0 0 -1 0; 0 35 4 -4 0 -12 -1; 0 0 2 -1 -3 0 -2];
% row reduce the system
C = rref(Chem);
% parametrize the system by setting the last variable xend (e.g. x7) = 1
x = [C(:,end);1];
% extract numerator and denominator from the rational expressions of these
% values
[N,D] = rat(x);
% take the least common multiple of the first pair, set this to the
% variable least
least = lcm(D(1),D(2));
% loop through taking the lcm of the previous values with the next value
% through x
for n = 3:numel(x)
least = lcm(least,D(n));
end
% give answer as column vector with the coefficients (now factored to their
% lowest possible integers
coeff = abs(least.*x)
Say X is the given vector:
X=[1
2
4
2
3
1
4
5
2
4
5];
And Y is the given subset of elements from X:
Y=[3
4
5];
The required output is the number of times the elements in Y occur in X:
out=[1
3
2];
My solution to do this would be to use for loop:
for i=1:size(X,1)
temp = X(X(:,1)==Y(i,1),:);
out(i,1) = size(temp,1);
end
But when X and Y are large, this is inefficient. So, how to do it faster making use of vectorization? I know about hist and histc, but I can't think of how to use them in this case to get the desired output.
A Fast Option
You could use bsxfun combined with sum to compute this
sum(bsxfun(#eq, Y, X.'), 2)
Explanation
In this example, bsxfun performs a given operation on every combination of elements in X and Y. The operation we're gonig to use is eq (a check for equality). The result is a matrix that has a row for each element in Y and a column for each element in X. It will have a 1 value if the element in X equals the element in Y that corresponds to a given row.
bsxfun(#eq, Y, X.')
% 0 0 0 0 1 0 0 0 0 0 0
% 0 0 1 0 0 0 1 0 0 1 0
% 0 0 0 0 0 0 0 1 0 0 1
We can then sum across the columns to count the number of elements in X that were equal to a given value in Y.
sum(bsxfun(#eq, Y, X.'), 2)
% 1
% 3
% 2
On newer versions of MATLAB (since R2016b), you can omit the bsxfun since the equality operation will automatically broadcast.
sum(Y - X.', 2)
A Memory-Efficient Option
The first option isn't the most efficient since it requires creating a matrix that is [numel(Y), numel(X)] elements large. Another way which may be more memory efficient may be to use the second output of ismember combined with accumarray
[tf, ind] = ismember(X, Y);
counts = accumarray(ind(tf), ones(sum(tf), 1), [numel(Y), 1], #numel);
Explanation
ismember is used to determine if the values in one array are in another. The first input tells us if each element of the first input is in the second input and the second output tells you where in the second input each element of the first input was found.
[tf, ind] = ismember(X, Y);
% 0 0 1 0 1 0 1 1 0 1 1
% 0 0 2 0 1 0 2 3 0 2 3
We can use the second input to "group" the same values together. The accumarray function does exactly this, it uses the ind variable above to determine groups and then applies a given operation to each group. In our case, we want to simply determine the number of element within each group. So to do that we can pass a second input the size of the ind input (minus the ones that didn't match) of ones, and then use numel as the operation (counts the number in each group)
counts = accumarray(ind(tf), ones(sum(tf), 1), [numel(Y), 1], #numel);
% 1
% 3
% 2
I want to make a triadiagonal matrix with matlab, using
full(gallery('tridiag', 10, 1, -4, 6, -4, 1))
and i take that i have too many arguments in the function. Is there another way to do this?
I am trying to make the following matrix:
6 -4 1 0 0
-4 6 -4 1 0
1 -4 6 -4 1
0 1 -4 6 -4
0 0 1 -4 6
Since your matrix is pentadiagonal, I think the best solution is to use spdiags:
>> n = 5;
>> full(spdiags(ones(n,1)*[1,-4,6,-4,1],[-2,-1,0,1,2],n,n));
ans =
6 -4 1 0 0
-4 6 -4 1 0
1 -4 6 -4 1
0 1 -4 6 -4
0 0 1 -4 6
The full is optional and not recommended for large n.
Since there are 5 non-zero diagonals this is not a tridiagonal matrix so you cannot use the tridiag option. You have to manually generate such matrix by means of the diag() function, which allows you to create a matrix with a given diagonal and you can as well select which diagonal you want to write.
You can achieve this therefore by creating 5 different matrices, each of them will have a given non-zero diagonal:
n=5;
B=diag(6*ones(n,1),0)+diag(-4*ones(n-1,1),1)+diag(-4*ones(n-1,1),-1)+diag(1*ones(n-2,1),2)+diag(1*ones(n-2,1),-2);
In this code n=5 is the order of your matrix, then diag(6*ones(n,1),0) will create a vector (length n) with all 6 and such vector will be placed in the 0-th diagonal. Such matrix will have zero elsewhere.
Similarly diag(-4*ones(n-1,1),1) will create a vector (length n-1) with all -4 and such vector will be placed in the 1st superdiagonal. Such matrix will have zero elsewhere and we sum such matrix to the previous one.
And such "chain reaction" goes on until the matrix is fully generated.
Update: I've been looking around the gallery() help and there is indeed an option for a Toeplitz pentadiagonal. You might want to use then
full(gallery('toeppen',5,1,-4,6,-4,1))
I agree that for your huge case a sparse-based solution such as that of Troy Haskin is best. However, it's worth noting that you're precisely constructing a Toeplitz matrix (as Alessiox hinted), and you can use the built-in toeplitz() to do that. All that is needed is to figure out the number of zeros needed for padding the input nonzero elements of the first row and column (this is necessary since toeplitz asserts the size of the matrix to construct from the dimensions of the input vector):
n = 5; %// linear size of result
v = [1,-4,6,-4,1]; %// nonzero diagonal elements symmetrically
mid = ceil(length(v)/2); %// index of diagonal in the input vector
zerosvec = zeros(1,n-mid); %// zeros for padding the rest
colvec = [v(mid:-1:1), zerosvec]; %// first column of the result
rowvec = [v(mid:end), zerosvec]; %// first row of the result
toeplitz(colvec,rowvec) %// toeplitz does the heavy lifting
I want to calculate the cumulative sum of a vector, but stop summing up once the sum becomes negative, and start again at positive elements.
Example:
We have a vector:
[1 1 -1 -1 -1 -1 1 1 1 1]
The normal cumulative sum would then be:
[1 2 1 0 -1 -2 -1 0 1 2]
But i want:
[1 2 1 0 0 0 1 2 3 4]
The only solution i could come up with was to loop over the elements of the vector like this:
test = [1 1 -1 -1 -1 -1 1 1 1 1];
testCumsum = zeros(size(test));
for i=1:length(test)
if i==1
testCumsum(i) = test(i);
else
testCumsum(i) = testCumsum(i-1) + test(i);
end
if testCumsum(i)<0
testCumsum(i) = 0;
end
end
Is the a more matlab-ish solution?
(The sum can become negative an arbitrary number of times, the vectors can become pretty large, and the elements can be any number, not just 1 and -1)
You won't be able to vectorize it since you have to decide on each elemenet based on previous ones. You can find regions of positive and negative runs but it would be unnecessarily complex and I don't know if you can gain over your own solution.
Here is a simplification of your code for input A and output C:
C=A;
C(1) = max(C(1), 0);
for k=2:numel(C)
C(k) = max(C(k-1)+C(k), 0);
end
call your vector x,
y=x >0
z=x.*y
sum(z)
the y vector is 0 / 1 where the elemnts of x are greater than 0 the dot product to get z sets your negative values to 0, and then you can sum
_Ah i see more clearly what you want to do now, - looping is probably going to be quickest, you could break into block segments if the array is large and use parfor to speed it up
Given a vector of integers such as:
X = [1 2 3 4 5 1 2]
I would like to find a really fast way to count the number of unique combinations with 2-elements.
In this case the two-number combinations are:
[1 2] (occurs twice)
[2 3] (occurs once)
[3 4] (occurs once)
[4 5] (occurs once)
[5 1] (occurs once)
As it stands, I am currently doing this in MATLAB as follows
X = [1 2 3 4 5 1 2];
N = length(X)
X_max = max(X);
COUNTS = nan(X_max); %store as a X_max x X_max matrix
for i = 1:X_max
first_number_indices = find(X==1)
second_number_indices = first_number_indices + 1;
second_number_indices(second_number_indices>N) = [] %just in case last entry = 1
second_number_vals = X(second_number_indices);
for j = 1:X_max
COUNTS(i,j) = sum(second_number_vals==j)
end
end
Is there a faster/smarter way of doing this?
Here is a super fast way:
>> counts = sparse(x(1:end-1),x(2:end),1)
counts =
(5,1) 1
(1,2) 2
(2,3) 1
(3,4) 1
(4,5) 1
You could convert to a full matrix simply as: full(counts)
Here is an equivalent solution using accumarray:
>> counts = accumarray([x(1:end-1);x(2:end)]', 1)
counts =
0 2 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
1 0 0 0 0
EDIT: #Amro has provided a much better solution (well, better in the vast majority of cases, I suspect my method would work better if MaxX is very large and X contains zeros - this is because the presence of zeros will rule out the use of sparse while a large MaxX will slow down the accumarray approach as it creates a matrix of size MaxX by MaxX).
EDIT: Thanks to #EitanT for pointing out an improvement that can be made using accumarray.
Here is how I would solve it:
%Generate some random data
T = 20;
MaxX = 3;
X = randi(MaxX, T, 1);
%Get the unique combinations and an index. Note, I am assuming X is a column vector.
[UniqueComb, ~, Ind] = unique([X(1:end-1), X(2:end)], 'rows');
NumComb = size(UniqueComb, 1);
%Count the number of occurrences of each combination
Count = accumarray(Ind, 1);
All unique sequential two element combinations are now stored in UniqueComb, while the corresponding counts for each unique combination are stored in Count.