I have a matrix like this:
fd =
x y z
2 5 10
2 6 10
3 5 11
3 9 11
4 3 11
4 9 12
5 4 12
5 7 13
6 1 13
6 5 13
I have two parts of my problem:
1) I want to calculate the difference of each two elements in a column.
So I tried the following code:
for i= 1:10
n=10-i;
for j=1:n
sdiff1 = diff([fd(i,1); fd(i+j,1)],1,1);
sdiff2 = diff([fd(i,2); fd(i+j,2)],1,1);
sdiff3 = diff([fd(i,3); fd(i+j,3)],1,1);
end
end
I want all the differences such as:
x1-x2, x1-x3, x1-x4....x1-x10
x2-x3, x2-x4.....x2-x10
.
.
.
.
.
x9-x10
same for y and z value differences
Then all the values should stored in sdiff1, sdiff2 and sdiff3
2) what I want next is for same z values, I want to keep the original data points. For different z values, I want to merge those points which are close to each other. By close I mean,
if abs(sdiff3)== 0
keep the original data
for abs(sdiff3) > 1
if abs(sdiff1) < 2 & abs(sdiff2) < 2
then I need mean x, mean y and mean z of the points.
So I tried the whole programme as:
for i= 1:10
n=10-i;
for j=1:n
sdiff1 = diff([fd(i,1); fd(i+j,1)],1,1);
sdiff2 = diff([fd(i,2); fd(i+j,2)],1,1);
sdiff3 = diff([fd(i,3); fd(i+j,3)],1,1);
if (abs(sdiff3(:,1)))> 1
continue
mask1 = (abs(sdiff1(:,1)) < 2) & (abs(sdiff2(:,1)) < 2) & (abs(sdiff3:,1)) > 1);
subs1 = cumsum(~mask1);
xmean1 = accumarray(subs1,fd(:,1),[],#mean);
ymean1 = accumarray(subs1,fd(:,2),[],#mean);
zmean1 = accumarray(subs1,fd(:,3),[],#mean);
fd = [xmean1(subs1) ymean1(subs1) zmean1(subs1)];
end
end
end
My final output should be:
2.5 5 10.5
3.5 9 11.5
5 4 12
5 7 13
6 1 13
where, (1,2,3),(4,6),(5,7,10) points are merged to their mean position (according to the threshold difference <2) whereas 8 and 9th point has their original data.
I am stuck in finding the differences for each two elements of a column and storing them. My code is not giving me the desired output.
Can somebody please help?
Thanks in advance.
This can be greatly simplified using vectorised notation. You can do for instance
fd(:,1) - fd(:,2)
to get the difference between columns 1 and 2 (or equivalently diff(fd(:,[1 2]), 1, 2)). You can make this more elegant/harder to read and debug with pdist but if you only have three columns it's probably more trouble than it's worth.
I suspect your first problem is with the third argument to diff. If you use diff(X, 1, 1) it will do the first order diff in direction 1, which is to say between adjacent rows (downwards). diff(X, 1, 2) will do it between adjacent columns (rightwards), which is what you want. Matlab uses the opposite convention to spreadsheets in that it indexes rows first then columns.
Once you have your diffs you can then test the elements:
thesame = find(sdiff3 < 2); % for example
this will yield a vector of the row indices of sdiff3 where the value is less than 2. Then you can use
fd(thesame,:)
to select the elements of fd at those indexes. To remove matching rows you would do the opposite test
notthesame = find(sdiff > 2);
to find the ones to keep, then extract those into a new array
keepers = fd(notthesame,:);
These won't give you the exact solution but it'll get you on the right track. For the syntax of these commands and lots of examples you can run e.g. doc diff in the command window.
Related
I am trying to reverse the elements of the matrix such that for given matrix order of the elements get reversed.
my code is as shown for the 3x3 matrix is working.
X = [ 1 2 3 ; 4 5 6 ; 7 8 9 ];
B = [fliplr(X(3,:));fliplr(X(2,:));fliplr(X(1,:))];
input X =
1 2 3
4 5 6
7 8 9
output:
B =
9 8 7
6 5 4
3 2 1
the above code I am trying to generalize for any matrix with the following code
[a,b]=size(X);
for i=0:a-1
A = [fliplr(X(a-i,:))];
end
but get only last row as output.
output A =
3 2 1
please help me to concatenate all the rows of the matrix one above to each other after it got reversed.
rot90 is the function made for this purpose.
B = rot90(A,2);
Your code doesn't work because you overwrite A in every loop iteration. Instead, you should index into A to save each of your rows.
However, fliplr can flip a whole matrix. You want to flip left/right and up/down:
B = flipud(fliplr(X));
This is the same as rotating the matrix (as Sardar posted while I was writing this):
B = rot90(X,2);
A totally different approach would work for arrays of any dimensionality:
X(:) = flipud(X(:));
suppose that we are determine peaks in vector as follow:
we have real values one dimensional vector with length m,or
x(1),x(2),.....x(m)
if x(1)>x(2) then clearly for first point peak(1)=x(1);else we are then comparing x(3) to x(2),if x(3)
[ indexes,peaks]=function(x,m);
c=[];
b=[];
if x(1)>x(2)
peaks(1)=x(1);
else
for i=2:m-1
if x(i+1)< x(i) & x(i)>x(i-1)
peak(i)=x(i);
end;
end
end
end
peaks are determined also using following picture:
sorry for the second picture,maybe it is not triangle,just A and C are on straight line,but here peak is B,so i can't continue my code for writing algorithm to find peak values in my vector.please help me to continue it
updated.numercial example given
x=[2 1 3 5 4 7 6 8 9]
here because first point is more then second,so it means that peak(1)=2,then we are comparing 1 to 3,because 3 is more then 1,we now want to compare 5 to 3,it is also more,compare 5 to 4,because 5 is more then 4,then it means that peak(2)=5,,so if we continue next peak is 7,and final peak would be 9
in case of first element is less then second,then we are comparing second element to third one,if second is more then third and first elements at the same time,then peak is second,and so on
You could try something like this:
function [peaks,peak_indices] = find_peaks(row_vector)
A = [min(row_vector)-1 row_vector min(row_vector)-1];
j = 1;
for i=1:length(A)-2
temp=A(i:i+2);
if(max(temp)==temp(2))
peaks(j) = row_vector(i);
peak_indices(j) = i;
j = j+1;
end
end
end
Save it as find_peaks.m
Now, you can use it as:
>> A = [2 1 3 5 4 7 6 8 9];
>> [peaks, peak_indices] = find_peaks(A)
peaks =
2 5 7 9
peak_indices =
1 4 6 9
This would however give you "plateaus" as well (adjacent and equal "peaks").
You can use diff to do the comparison and add two points in the beginning and end to cover the border cases:
B=[1 diff(A) -1];
peak_indices = find(B(1:end-1)>=0 & B(2:end)<=0);
peaks = A(peak_indices);
It returns
peak_indices =
1 4 6 9
peaks =
2 5 7 9
for your example.
findpeaks does it if you have a recent matlab version, but it's also a bit slow.
This proposed solution would be quite slow due to the for loop, and you also have a risk of rounding error due to the fact that you compare the maximal value to the central one instead of comparing the position of the maximum, which is better for your purpose.
You can stack the data so as to have three columns : the first one for the preceeding value, the second is the data and the third one is the next value, do a max, and your local maxima are the points for which the position of the max along columns is 2.
I've coded this as a subroutine of my own peak detection function, that adds a further level of iterative peak detection
http://www.mathworks.com/matlabcentral/fileexchange/42927-find-peaks-using-scale-space-approach
So if I have a matrix s;
s = [4;5;9;12;3]
and I want to calculate the difference between an entry and it's previous entry plus add the previous difference such that I'll get
s = [ 4 0; 5 1; 9 5; 12 8; 3 -1]
I'm quite new to matlab. I understand a for loop would be required to go through the original matrix
The second column of your result seems to be essentially cumsum(diff(s)). However, that's not "the difference between an entry and its previous entry plus the previous difference"; it's the cumulative sum of differences.
So, if what you want in the second column is the cumulative sum of differences:
result = [s [0; cumsum(diff(s))]];
In matlab you have a lot of functions for working directly with matrix, the one that feeds here is diff and cumsum please visit the matlab documentation, and the functions for concatening like horzcat or vertcat int his case manually to get what you need work like this:
>> s = [4;5;9;12;3]
s =
4
5
9
12
3
Get the vector my_cum_diff which is the difference between elements in a vector
my_cum_diff = [0; cumsum(diff(s))]
my_cum_diff = [0; cumsum(diff(s))]
my_cum_diff =
0
1
5
8
-1
finally concat the two vectors
final_s=[s my_cum_diff]
final_s =
4 0
5 1
9 5
12 8
3 -1
As an example, I have a matrix [1,2,3,4,5]'. This matrix contains one column and 5 rows, and I have to generate a pair of points like (1,2),(1,3)(1,4)(1,5),(2,3)(2,4)(2,5),(3,4)(3,5)(4,5).
I have to store these values in 2 columns in a matrix. I have the following code, but it isn't quite giving me the right answer.
for s = 1:5;
for tb = (s+1):5;
if tb>s
in = sub2ind(size(pairpoints),(tb-1),1);
pairpoints(in) = s;
in = sub2ind(size(pairpoints),(tb-1),2);
pairpoints(in) = tb;
end
end
end
With this code, I got (1,2),(2,3),(3,4),(4,5). What should I do, and what is the general formula for the number of pairs?
One way, though is limited depending upon how many different elements there are to choose from, is to use nchoosek as follows
pairpoints = nchoosek([1:5],2)
pairpoints =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
See the limitations of this function in the provided link.
An alternative is to just iterate over each element and combine it with the remaining elements in the list (assumes that all are distinct)
pairpoints = [];
data = [1:5]';
len = length(data);
for k=1:len
pairpoints = [pairpoints ; [repmat(data(k),len-k,1) data(k+1:end)]];
end
This method just concatenates each element in data with the remaining elements in the list to get the desired pairs.
Try either of the above and see what happens!
Another suggestion I can add to the mix if you don't want to rely on nchoosek is to generate an upper triangular matrix full of ones, disregarding the diagonal, and use find to generate the rows and columns of where the matrix is equal to 1. You can then concatenate both of these into a single matrix. By generating an upper triangular matrix this way, the locations of the matrix where they're equal to 1 exactly correspond to the row and column pairs that you are seeking. As such:
%// Highest value in your data
N = 5;
[rows,cols] = find(triu(ones(N),1));
pairpoints = [rows,cols]
pairPoints =
1 2
1 3
2 3
1 4
2 4
3 4
1 5
2 5
3 5
4 5
Bear in mind that this will be unsorted (i.e. not in the order that you specified in your question). If order matters to you, then use the sortrows command in MATLAB so that we can get this into the proper order that you're expecting:
pairPoints = sortrows(pairPoints)
pairPoints =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
Take note that I specified an additional parameter to triu which denotes how much of an offset you want away from the diagonal. The default offset is 0, which includes the diagonal when you extract the upper triangular matrix. I specified 1 as the second parameter because I want to move away from the diagonal towards the right by 1 unit so I don't want to include the diagonal as part of the upper triangular decomposition.
for loop approach
If you truly desire the for loop approach, going with your model, you'll need two for loops and you need to keep track of the previous row we are at so that we can just skip over to the next column until the end using this. You can also use #GeoffHayes approach in using just a single for loop to generate your indices, but when you're new to a language, one key advice I will always give is to code for readability and not for efficiency. Once you get it working, if you have some way of measuring performance, you can then try and make the code faster and more efficient. This kind of programming is also endorsed by Jon Skeet, the resident StackOverflow ninja, and I got that from this post here.
As such, you can try this:
pairPoints = []; %// Initialize
N = 5; %// Highest value in your data
for row = 1 : N
for col = row + 1 : N
pairPoints = [pairPoints; [row col]]; %// Add row-column pair to matrix
end
end
We get the equivalent output:
pairPoints =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
Small caveat
This method will only work if your data is enumerated from 1 to N.
Edit - August 20th, 2014
You wish to generalize this to any array of values. You also want to stick with the for loop approach. You can still keep the original for loop code there. You would simply have to add a couple more lines to index your new array. As such, supposing your data array was:
dat = [12, 45, 56, 44, 62];
You would use the pairPoints matrix and use each column to subset the data array to access your values. Also, you need to make sure your data is a column vector, or this won't work. If we didn't, we would be creating a 1D array and concatenating rows and that's not obviously what we're looking for. In other words:
dat = [12, 45, 56, 44, 62];
dat = dat(:); %// Make column vector - Important!
N = numel(dat); %// Total number of elements in your data array
pairPoints = []; %// Initialize
%// Skip if the array is empty
if (N ~= 0)
for row = 1 : N
for col = row + 1 : N
pairPoints = [pairPoints; [row col]]; %// Add row-column pair to matrix
end
end
vals = [dat(pairPoints(:,1)) dat(pairPoints(:,2))];
else
vals = [];
Take note that I have made a provision where if the array is empty, don't even bother doing any calculations. Just output an empty matrix.
We thus get:
vals =
12 45
12 56
12 44
12 62
45 56
45 44
45 62
56 44
56 62
44 62
suppose that we are determine peaks in vector as follow:
we have real values one dimensional vector with length m,or
x(1),x(2),.....x(m)
if x(1)>x(2) then clearly for first point peak(1)=x(1);else we are then comparing x(3) to x(2),if x(3)
[ indexes,peaks]=function(x,m);
c=[];
b=[];
if x(1)>x(2)
peaks(1)=x(1);
else
for i=2:m-1
if x(i+1)< x(i) & x(i)>x(i-1)
peak(i)=x(i);
end;
end
end
end
peaks are determined also using following picture:
sorry for the second picture,maybe it is not triangle,just A and C are on straight line,but here peak is B,so i can't continue my code for writing algorithm to find peak values in my vector.please help me to continue it
updated.numercial example given
x=[2 1 3 5 4 7 6 8 9]
here because first point is more then second,so it means that peak(1)=2,then we are comparing 1 to 3,because 3 is more then 1,we now want to compare 5 to 3,it is also more,compare 5 to 4,because 5 is more then 4,then it means that peak(2)=5,,so if we continue next peak is 7,and final peak would be 9
in case of first element is less then second,then we are comparing second element to third one,if second is more then third and first elements at the same time,then peak is second,and so on
You could try something like this:
function [peaks,peak_indices] = find_peaks(row_vector)
A = [min(row_vector)-1 row_vector min(row_vector)-1];
j = 1;
for i=1:length(A)-2
temp=A(i:i+2);
if(max(temp)==temp(2))
peaks(j) = row_vector(i);
peak_indices(j) = i;
j = j+1;
end
end
end
Save it as find_peaks.m
Now, you can use it as:
>> A = [2 1 3 5 4 7 6 8 9];
>> [peaks, peak_indices] = find_peaks(A)
peaks =
2 5 7 9
peak_indices =
1 4 6 9
This would however give you "plateaus" as well (adjacent and equal "peaks").
You can use diff to do the comparison and add two points in the beginning and end to cover the border cases:
B=[1 diff(A) -1];
peak_indices = find(B(1:end-1)>=0 & B(2:end)<=0);
peaks = A(peak_indices);
It returns
peak_indices =
1 4 6 9
peaks =
2 5 7 9
for your example.
findpeaks does it if you have a recent matlab version, but it's also a bit slow.
This proposed solution would be quite slow due to the for loop, and you also have a risk of rounding error due to the fact that you compare the maximal value to the central one instead of comparing the position of the maximum, which is better for your purpose.
You can stack the data so as to have three columns : the first one for the preceeding value, the second is the data and the third one is the next value, do a max, and your local maxima are the points for which the position of the max along columns is 2.
I've coded this as a subroutine of my own peak detection function, that adds a further level of iterative peak detection
http://www.mathworks.com/matlabcentral/fileexchange/42927-find-peaks-using-scale-space-approach