I have a matrix and I am interested in changing values that satisfy a certain condition inside that matrix differently, depending on where they are. Say I have a matrix smallPic. How do I obtain a matrix smallPicB with the same dimensions that changed all values that are above 50 in the first two columns to a 255, while those that are in the third and fourth column are changed to a 180?
I have this code which works, but it is pretty ugly and requires splitting the matrix and concatenating it again:
smallPic1=smallPic(:,1:2);smallPic1(smalllPic1>50)=255;
smallPic2=smallPic(:,3:4);smallPic2(smalllPic2>50)=180;
smallPicB = [smalllPic1 smalllPic2];
How would you combine the logical index with the scalar index in one command?
What doesn't work is this:
smallPic(:,smallPic(:,3:4)>50) = 180
Here, the second mention of smallPic inside the brackets does not allow indexing into the correct position of smallPic because it doesn't have the same dimensions as smallPic. So this command actually replaces values in the first two columns of smallPic that are in the same row as those values that are above 50 in the third and fourth column, instead of replacing the values in the third and fourth column themselves.
Any other suggestions?
It probably is not what you're looking for, but it can help if you have lots of assignments like that:
J = repmat(1:size(smallPic, 2), size(smallPic, 1), 1)
smallPic((J<3)&(smallPic>50))=255
smallPic((J>2)&(J<5)&(smallPic>50))=180
You can also call ismember function if column indices are not consecutive:
smallPic(ismember(J, [[1:2 5:6]])&(smallPic>50))=255
Related
I am having matrix with approx 3000 rows(changing) and 3 columns.
I have count of both rows and columns.
I am trying to plot the graph:
x=1:3000;
plot(matrix(x,1))
is there any way that I can include all rows in the plot instruction itself so that I can remove 'x=1:3000' ?
Also, I want to divide, 1st column of matrix which have 3000 rows into another matrix of 3 columns each with 1000 rows. Any specific instruction for this ?
I have made for loop for this and then i am placing individually the elements in the new array. But its taking long time.
As to the plotting issue, using the colon operator will plot all rows for your desired column:
plot(matrix(:,1));
EDIT: You mentioned you were a beginner. In case you haven't seen the colon operator used like this before, a colon operator all by itself when indexing into a matrix essentially means "all __", either "all rows" if in the first position or "all columns" if in the second position.
As for the second question, of splitting one column into a new matrix with multiple columns, you can use the reshape() function, which takes the input matrix to be reshaped and a number of output rows and columns. For example, to split the first column of matrix into 3 columns and put them into newMatrix, use the following:
newMatrix = reshape(matrix(:,1),[],3);
Note that the above code uses [] in the second argument (the number of rows argument) to mean "automatically determine number of rows".This is automatically determined based on the number of columns, which is defined in the third argument here as 3. The reshape function requires that the number of output rows * output columns be equal to input rows * input columns. So in the above case this will only work if the starting matrix has a number of rows which is divisible by 3.
My goal is to create a random, 20 by 5 array of integers, sort them by increasing order from top to bottom and from left to right, and then calculate the mean in each of the resulting 20 rows. This gives me a 1 by 20 array of the means. I then have to find the column whose mean is closest to 0. Here is my code so far:
RandomArray= randi([-100 100],20,5);
NewArray=reshape(sort(RandomArray(:)),20,5);
MeanArray= mean(transpose(NewArray(:,:)))
X=min(abs(x-0))
How can I store the column number whose mean is closest to 0 into a variable? I'm only about a month into coding so this probably seems like a very simple problem. Thanks
You're almost there. All you need is a find:
RandomArray= randi([-100 100],20,5);
NewArray=reshape(sort(RandomArray(:)),20,5);
% MeanArray= mean(transpose(NewArray(:,:))) %// gives means per row, not column
ColNum = find(abs(mean(NewArray,1))==min(abs(mean(NewArray,1)))); %// gives you the column number of the minimum
MeanColumn = RandomArray(:,ColNum);
find will give you the index of the entry where abs(mean(NewArray)), i.e. the absolute values of the mean per column equals the minimum of that same array, thus the index where the mean of the column is closest to 0.
Note that you don't need your MeanArray, as it transposes (which can be done by NewArray.', and then gives the mean per column, i.e. your old rows. I chucked everything in the find statement.
As suggested in the comment by Matthias W. it's faster to use the second output of min directly instead of a find:
RandomArray= randi([-100 100],20,5);
NewArray=reshape(sort(RandomArray(:)),20,5);
% MeanArray= mean(transpose(NewArray(:,:))) %// gives means per row, not column
[~,ColNum] = min(abs(mean(NewArray,1)));
MeanColumn = RandomArray(:,ColNum);
So I'm working with sparse matrix and I have to find out different info about a very big one (10^6 size) and I need to find out the mean of the outlinks. Just to be sure I think mean is what you get from 3+4+5/3=4, 4 is the mean.
I thought of something like this:
[row,col] = find(A(:,2),1,'first')
and then I would do 1/numberInThatIndex or something similar, since it's a S-matrix (pretty sure it's called that).
And I would iterate column by column but for some reason it's not giving me the first number in each column, if I do find(A(:,1),1,'first') it does give me the first in the first column, but not in the second if I change it to A(:,2).
I'd also need something to store that index to access the value, I thought of a 2xN vector but I guess it's not the best idea. I mean, find is going to give me index, but I need the value in that index, and then store that or show it. Not sure if I'm explaining myself properly but I'm trying, sorry about that.
Just to be clear both when I input A(:,1) and A(:,2) it gives me index from the first column, and I do not want that, I want first element found from each column, so I can calculate the mean out of the number in that index.
edit: allright it seems like that indeed does work, but when I was checking the results I was putting 3817 instead of 3871 that was the given answer and so I found a 0 when I wanted something that's not a zero. Not sure if I should delete all of this.
To solve your problem, you can do the following:
numberNonZerosPerColumn = sum(S~=0,1);
meanValue = nanmean(1./numberNonZerosPerColumn);
Count the number of nonzero elements in every column n(i)
Compute the values v(i) that are stored there, which are defined by v(i) := 1/n(i)
Take the mean of those values where n(i) is not zero (i.e. summing all those values, where v(i) is not NaN and divide by the number of columns that contain at least one zero)
If you want to treat columns without any nonzero entry as v(i):= 0, but still use them in your mean, you can use:
numberNonZerosPerColumn = sum(S~=0,1);
meanValue = nansum(1./numberNonZerosPerColumn)/size(S,2);
i have two matrices
r=10,000x2
q=10,000x2
i have to find out those rows of q which are one value or both values(as it is a two column matrix) different then r and allocate them in another matrix, right now i am trying this.i cannot use isequal because i want to know those rows
which are not equal this code gives me the individual elements not the complete rows different
can anyone help please
if r(:,:)~=q(:,:)
IN= find(registeredPts(:,:)~=q(:,:))
end
You can probably do this using ismember. Is this what you want? Here you get the values from q in rows that are different from r.
q=[1,2;3,4;5,6]
r=[1,2;3,5;5,6]
x = q(sum(ismember(q,r),2) < 2,:)
x =
3 4
What this do:
ismember creates an array with 1's in the positions where q == r, and 0 in the remaining positions. sum(.., 2) takes the column sum of each of these rows. If the sum is less than 2, that row is included in the new array.
Update
If the values might differ some due to floating point arithmetic, check out ismemberf from the file exchange. I haven't tested it myself, but it looks good.
I have a 161*32 matrix (labelled "indpic") in MATLAB and I'm trying to find the frequency of a given number appearing in a row. So I think that I need to analyse each row separately for each value, but I'm incredibly unsure about how to go about this (I'm only new to MATLAB). This also means I'm incredibly useless with loops and whatnot as well.
Any help would be greatly appreciated!
If you want to count the number of times a specific number appears in each row, you can do this:
sum(indpic == val, 2)
where indpic is your matrix (e.g image) and val is the desired value to be counted.
Explanation: checking equality of each element with the value produces a boolean matrix with "1"s at the locations of the counted value. Summing each row (i.e summing along the 2nd dimension results in the desired column vector, where each element being equal to the number of times val is repeated in the corresponding row).
If you want to count how many times each value is repeated in your image, this is called a histogram, and you can use the histc command to achieve that. For example:
histc(indpic, 1:256)
counts how many times each value from 1 to 256 appears in image indpic.
Like this,
sum(indpic(rownum,:) == 7)
obviously change 7 to whatever.
You can just write
length(find(indpic(row_num,:)==some_value))
and it will give you the number of elements equal to "some_value" in the "row_num"th row in matrix "indpic"