In my research, I have to identify row and column which has same value in matrix.
here for example the form of matrix:
A= [60 27 45 72 22 14 56 2 8 39 18 12;
72 27 60 45 11 7 3 23 41 17 56 39]
Then, I want to identify row 1 until 4 in column 1 and column 2.
here part of my code.
done =all(ismember(A(1,1:4),A(2,1:4))); %Code that I want to ask you.
Please Look at in row 1-4 and column 1-4;
if I use above code, done will always give true (1). but it is not my desire, because A(1,1) is not same with A(2,1), then A(1,3) is also not same with A(2,3), A(1,4) is also not same with A(2,4).
If you want to check whether the values in A(1,1:4) are the same as in A(2,1:4) you could just use
done =all(A(1,1:4) == A(2,1:4));
Related
How to add the value in specific range in a matrix?
I mean if i have a matrix
Columns 1 through 7
4 4 4 4 4 4 4
48 48 48 48 48 48 48
Columns 8 through 14
4 4 4 13 13 13 13
48 48 48 57 57 57 57
Columns 15 through 20
13 13 13 13 13 13
57 57 57 57 57 57
I want to sum all 4 values ,all 13 values,all 48 values,and all 57 values,so the result should be m=[40 130 480 570]
The easiest but stupid method is like this
a=sum(1,(1:10));
b=sum(1,(11:20));
c=sum(2,(1:10));
d=sum(2,(11:20));
m=[a b c d]
If i want to write a code with for-loop or while-loop to show the result i want.how do i write the code?
Or can i use the some method to write a code without loop to show this?
Though the solution of #phnx works fine, you can also use the other outputs of the unique function in combination with accumarray as described in the docs:
[C, ~, ic] = unique(a);
a_counts = accumarray(ic,1);
m = C.*a_counts
This will avoid the warning 'hist' is not recommended...
A simple two-line solution, with A as your original matrix, would be:
[a,b]=hist(A(:),unique(A(:)))
c = a .* b'
with a containing the number of occurances, b the unique elements and c the sums.
for a=1:50; %numbers 1 through 50
for b=1:50;
c=sqrt(a^2+b^2);
if c<=50&c(rem(c,1)==0);%if display only if c<=50 and c=c/1 has remainder of 0
pyth=[a,b,c];%pythagorean matrix
disp(pyth)
else c(rem(c,1)~=0);%if remainder doesn't equal to 0, omit output
end
end
end
answer=
3 4 5
4 3 5
5 12 13
6 8 10
7 24 25
8 6 10
8 15 17
9 12 15
9 40 41
10 24 26
12 5 13
12 9 15
12 16 20
12 35 37
14 48 50
15 8 17
15 20 25
15 36 39
16 12 20
16 30 34
18 24 30
20 15 25
20 21 29
21 20 29
21 28 35
24 7 25
24 10 26
24 18 30
24 32 40
27 36 45
28 21 35
30 16 34
30 40 50
32 24 40
35 12 37
36 15 39
36 27 45
40 9 41
40 30 50
48 14 50
This problem involves the Pythagorean theorem but we cannot use the built in function so I had to write one myself. The problem is for example columns 1 & 2 from the first two rows have the same numbers. How do I code it so it only deletes one of the rows if the columns 1 and 2 have the same number combination? I've tried unique function but it doesn't really delete the combinations. I have read about deleting duplicates from previous posts but those have confused me even more. Any help on how to go about this problem will help me immensely!
Thank you
welcome to StackOverflow.
The problem in your code seems to be, that pyth only contains 3 values, [a, b, c]. The unique() funcion used in the next line has no effect in that case, because only one row is contained in pyth. another issue is, that the values idx and out are calculated in each loop cycle. This should be placed after the loops. An example code could look like this:
pyth = zeros(0,3);
for a=1:50
for b=1:50
c = sqrt(a^2 + b^2);
if c<=50 && rem(c,1)==0
abc_sorted = sort([a,b,c]);
pyth = [pyth; abc_sorted];
end
end
end
% do final sorting outside of the loop
[~,idx] = unique(pyth, 'rows', 'stable');
out = pyth(idx,:);
disp(out)
a few other tips for writing MATLAB code:
You do not need to end for or if/else stements with a semicolon
else statements cover any other case not included before, so they do not need a condition.
Some performance reommendations:
Due to the symmetry of a and b (a^2 + b^2 = b^2 + a^2) the b loop could be constrained to for b=1:a, which would roughly save you half of the loop cycles.
if you use && for contencation of scalar values, the second part is not evaluated, if the first part already fails (source).
Regards,
Chris
You can also linearize your algorithm (but we're still using bruteforce):
[X,Y] = meshgrid(1:50,1:50); %generate all the combination
C = (X(:).^2+Y(:).^2).^0.5; %sums of two square for every combination
ind = find(rem(C,1)==0 & C<=50); %get the index
res = unique([sort([X(ind),Y(ind)],2),C(ind)],'rows'); %check for uniqueness
Now you could really optimized your algorithm using math, you should read this question. It will be useful if n>>50.
I have a few multidimensional matrices of dimensions mxnxt, where each element in mxn is an individual sensor input, and t is time. What I want to do is analyse only the peak values for each element in mxn over t, so I would end up with a single 2D matrix of mxn containing only max values.
I know there are are ways to get a single overall max value, but is there a way to combine this with element-by-element operations like bsxfun so that it examines each individual element over t?
I'd be grateful for any help you can give because I'm really stuck at the moment. Thanks in advance!
Is this what you want?
out = max(A,[],3); %// checking maximum values in 3rd dimension
Example:
A = randi(50,3,3,3); %// Random 3x3x3 dim matrix
out = max(A,[],3);
Results:
A(:,:,1) =
35 5 8
38 12 42
23 46 27
A(:,:,2) =
50 6 39
4 49 41
23 1 44
A(:,:,3) =
5 41 10
20 22 14
13 46 8
>> out
out =
50 41 39
38 49 42
23 46 44
You can call max() with the matrix and select the dimension (look the documentation) on which the operation will be calculated, e.g
M = max(A,[],3)
I have an MxN matrix and I want a column vector v, using the vector s that tells me for each row in the matrix what column I will take.
Here's an example:
Matrix =
[ 4 13 93 20 42;
31 18 94 64 02;
7 44 24 91 15;
11 20 43 38 31;
21 42 72 60 99;
13 81 31 87 50;
32 22 83 24 04]
s = [4 4 5 4 4 4 3].'
And the desired output is:
v = [20 64 15 38 60 87 83].'
I thought using the expression
Matrix(:,s)
would've work but it doesn't. Is there a solution without using for loops to access the rows separately?
It's not pretty, and there might be better solutions, but you can use the function sub2ind like this:
M(sub2ind(size(M),1:numel(s),s'))
You can also do it with linear indexing, here is an example:
M=M'; s=s';
M([0:size(M,1):numel(M)-1]+s)
I have a large number of entries arranged in three columns. Sample of the data is:
A=[1 3 2 3 5 4 1 5 ;
22 25 27 20 22 21 23 27;
17 15 15 17 12 19 11 18]'
I want the first column (hours) to control the entire matrix to create new matrix as follows:
Anew=[1 2 3 4 5 ; 22.5 27 22.5 21 24.5; 14 15 16 19 15]'
Where the 2nd column of Anew is the average value of each corresponding hour for example:
from matrix A:
at hour 1, we have 2 values in 2nd column correspond to hour 1
which are 22 and 23 so the average is 22.5
Also the 3rd column: at hour 1 we have 17 and 11 and the
average is 14 and this continues to the hour 5 I am using Matlab
You can use ACCUMARRAY for this:
Anew = [unique(A(:,1)),...
cell2mat(accumarray(A(:,1),1:size(A,1),[],#(x){mean(A(x,2:3),2)}))]
This uses the first column A(:,1) as indices (x) to pick the values in columns 2 and 3 for averaging (mean(A(x,2:3),1)). The curly brackets and the call to cell2mat allow you to work on both columns at once. Otherwise, you could do each column individually, like this
Anew = [unique(A(:,1)), ...
accumarray(A(:,1),A(:,2),[],#mean), ...
accumarray(A(:,1),A(:,3),[],#mean)]
which may actually be a bit more readable.
EDIT
The above assumes that there's no missing entry for any of the hours. It will result in an error otherwise. Thus, a more robust way to calculate Anew is to allow for missing values. For easy identification of the missing values, we use the fillval input argument to accumarray and set it to NaN.
Anew = [(1:max(A(:,1)))', ...
accumarray(A(:,1),A(:,2),[],#mean,NaN), ...
accumarray(A(:,1),A(:,3),[],#mean,NaN)]
You can use consolidator to do the work for you.
[Afinal(:,1),Afinal(:,2:3)] = consolidator(A(:,1),A(:,2:3),#mean);
Afinal
Afinal =
1 22.5 14
2 27 15
3 22.5 16
4 21 19
5 24.5 15