I know how to check an 8-neighbourhood in matlab (i.e; nlfilter). But, I want to assign the value which is more repeated to the center value. So, say for instance that I have the following values in the 8-neighbourhood:
2-values = 56
3-values = 64
1-value = 70
1-value = 87
1-value = 65
In this case we would assign 64 to the center pixel.
How can we do that?
Thanks.
I think you want either the mode or the histc function.
M=mode(X) for vector X computes M as the sample mode, or most
frequently
occurring value in X.
Example with your data:
x = [56 56 64 64 64 70 87 65];
mode(x)
ans =
64
But this will only get you the most frequently occurring value.
If you want the count of each unique item in the array, you could do,
unqx = unique(x);
unqx =
56 64 65 70 87
valueCount = histc(x, unqx)
ans =
2 3 1 1 1
You could then sort this and take the first N values
valueCount = sort(valueCount, 'descend');
% Use unqx(valueCount(1:N))
Related
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)
Say I have the following columns vector Z
1 53 55 57 60 64 68 70 71 72 74 76 77 78 79 80 255
I want to use it to create a matrix such that each row would contain all the number between (and including) 2 adjacent elements in Z
So the output matrix should be something like this:
1 2 3 .... 53
53 54 55
55 56 57
57 58 60
....
80 81 ... 255
I've been searching for something similar but couldn't find it.
Thanks
See if this works for you -
lens = diff(Z)+1;
mask1 = bsxfun(#le,[1:max(lens)]',lens); %//'
array1 = zeros(size(mask1));
array1(mask1) = sort([1:255 Z(2:end-1)]);
out = array1.'; %//'# out is the desired output
Try this to break the monotony of bsxfun :) :
d = diff(Z);
N = max(d)+1;
R = zeros(length(Z)-1,N);
for i = 1:length(Z)-1
R(i,1:1+d(i)) = Z(i):Z(i+1);
end
EDIT:
I know that the general consensus is that one always should try to avoid loops in Matlab, but is this valid for this example? I know that this is a broad question, so lets focus on this particular problem and compare bsxfun to JIT loop. Comparing the two proposed solutions:
the code used for testing:
Z = [1 53 55 57 60 64 68 70 71 72 74 76 77 78 79 80 255];
%[1 3 4, 6];
nn = round(logspace(1,4,10));
tm1_nn = zeros(length(nn),1);
tm2_nn = zeros(length(nn),1);
for o = 1:length(nn)
tm1 = zeros(nn(o),1);
tm2 = zeros(nn(o),1);
% approach1
for k = 1:nn(o)+1
tic
d = diff(Z);
N = max(d)+1;
R = zeros(length(Z)-1,N);
for i = 1:length(Z)-1
R(i,1:1+d(i)) = Z(i):Z(i+1);
end
tm1(k) = toc;
end
%approach 2
for k = 1:nn(o)+1
tic
lens = diff(Z)+1;
mask1 = bsxfun(#le,[1:max(lens)]',lens); %//'
array1 = zeros(size(mask1));
array1(mask1) = sort([1:255 Z(2:end-1)]);
out = array1.';
tm2(k) = toc;
end
tm1_nn(o) = mean(tm1);%sum(tm1);%mean(tm1);%
tm2_nn(o) = mean(tm2);%sum(tm2);%mean(tm2);%
end
semilogx(nn,tm1_nn, '-ro', nn,tm2_nn, '-bo')
legend('JIT loop', 'bsxfun')
xlabel('log_1_0(Number of runs)')
%ylabel('Sum execution time')
ylabel('Mean execution time')
grid on
I encountered other tasks previously where the loop was faster. (or I mess up the comparison?)
Does anyone of you have a clue of why the following code is crashing with Index exceeds matrix dimensions. error for N_SUBJ = 17 or N_SUBJ = 14, but not for example for the values 13,15,16?
N_PICS = 7
COLR = hsv;
N_COLR = size(COLR,1);
COLR = COLR(1+[0:(N_PICS-1)]*round(N_COLR/N_PICS),:);
SUBJ_COLR = hsv;
N_SUBJ_COLR = size(SUBJ_COLR,1);
SUBJ_COLR = SUBJ_COLR(1+[0:(N_SUBJ-1)]*round(N_SUBJ_COLR/N_SUBJ),:);
And also, could somebody please explain me what it's doing exactly and how it's working?
When you say crashing, I assume you mean you are seeing the error, Index exceeds matrix dimensions.? If you are seeing this error then the matrix returned by hsv does not have enough rows for the sub-sample operation you are doing.
SUBJ_COLR = SUBJ_COLR(1+[0:(N_SUBJ-1)]*round(N_SUBJ_COLR/N_SUBJ),:);
selects a subset of the original matrix. 1+[0:(N_SUBJ-1)]*round(N_SUBJ_COLR/N_SUBJ) calculates which row to select, and : means all columns.
The matrix SUBJ_COLR is 64-by-3, thus N_SUBJ_COLR is equal to 64. You're indexing into the 64 rows of SUBJ_COLR and in some cases the particular index is greater than the number of row, resulting in a Index exceeds matrix dimensions. error. So the question is really why does this snippet
1+[0:(N_SUBJ-1)]*round(N_SUBJ_COLR/N_SUBJ)
evaluate to numbers greater than 64 for some values of N_SUBJ? This expression can be rewritten as:
1+(0:round(64/N_SUBJ):round(64/N_SUBJ)*(N_SUBJ-1))
or
1:round(64/N_SUBJ):round(64/N_SUBJ)*(N_SUBJ-1)+1
where I've replaced N_SUBJ_COLR by 64 for clarity. This latter expression more clearly shows what the largest index in the vector will be and how it depends on the value of N_SUBJ. You can print out this largest index as a function of N_SUBJ:
N_SUBJ = 1:30;
round(64./N_SUBJ).*(N_SUBJ-1)+1
which returns
ans =
Columns 1 through 13
1 33 43 49 53 56 55 57 57 55 61 56 61
Columns 14 through 26
66 57 61 65 69 55 58 61 64 67 70 73 51
Columns 27 through 30
53 55 57 59
As you can see, there are several values that exceed 64. This nonlinear behavior comes down to the use of round. The integers created by the round part don't appear to get small enough fast enough as they multiply (N_SUBJ-1) which is growing in order to keep the total term less than 64. One option might be to replace round with floor, but there are probably other ways.
I have a matrix like:
A=
10 31 32 22
32 35 52 77
68 42 84 32
I need a function like mode but with range, for example mymode(A,10) that return 30, find most frequent number in range 0-10, 10-20, 20-30, .... and return most number in range.
You can use histc to bin your data into the ranges of your desire and then find the bin with the most members using max on the output of histc
ranges = 0:10:50; % your desired ranges
[n, bins] = histc(A(:), ranges); % bin the data
[v,i] = max(n); % find the bin with most occurrences
[ranges(i) ranges(i+1)] % edges of the most frequent bin
For your specific example this returns
ans =
30 40
which matches with your required output, as the most values in A lay between 30 and 40.
[M,F] = mode( A((A>=2) & (A<=5)) ) %//only interested in range 2 to 5
...where M will give you the mode and F will give you frequency of occurence
> A = [10 31 32 22; 32 35 52 77; 68 42 84 32]
A =
10 31 32 22
32 35 52 77
68 42 84 32
> min = 10
min = 10
> max = 40
max = 40
> mode(A(A >= min & A <= max))
ans = 32
>
I guess by the number of different answers that we may be missing your goal. Here is my interpretation.
If you want to have many ranges and you want to output most frequent number for every range, create a cell containing all desired ranges (they could overlap) and use cellfun to run mode() for every range. You can also create a cell with desired ranges using arrayfun in a similar manner:
A = [10 31 32 22; 32 35 52 77; 68 42 84 32];
% create ranges
range_step = 10;
range_start=[0:range_step:40];
range=arrayfun(#(r)([r r+range_step]), range_start, 'UniformOutput', false)
% analyze ranges
o = cellfun(#(r)(mode(A(A>=r(1) & A<=r(2)))), range, 'UniformOutput', false)
o =
[10] [10] [22] [32] [42]
Quick MATLAB question.
What would be the best/most efficient way to select a certain number of elements, 'n' in windows of 'm'. In other words, I want to select the first 50 elements of a sequence, then elements 10-60, then elements 20-70 ect.
Right now, my sequence is in vector format(but this can easily be changed).
EDIT:
The sequences that I am dealing with are too long to be stored in my RAM. I need to be able to create the windows, and then call upon the window that I want to analyze/preform another command on.
Do you have enough RAM to store a 50-by-nWindow array in memory? In that case, you can generate your windows in one go, and then apply your processing on each column
%# idxMatrix has 1:50 in first col, 11:60 in second col etc
idxMatrix = bsxfun(#plus,(1:50)',0:10:length(yourVector)-50); %'#
%# reshapedData is a 50-by-numberOfWindows array
reshapedData = yourVector(idxMatrix);
%# now you can do processing on each column, e.g.
maximumOfEachWindow = max(reshapedData,[],1);
To complement Kerrek's answer: if you want to do it in a loop, you can use something like
n = 50
m = 10;
for i=1:m:length(v)
w = v(i:i+n);
% Do something with w
end
There's a slight issue with the description of your problem. You say that you want "to select the first 50 elements of a sequence, then elements 10-60..."; however, this would translate to selecting elements:
1-50
10-60
20-70
etc.
That first sequence should be 0-10 to fit the pattern which of course in MATLAB would not make sense since arrays use one-indexing. To address this, the algorithm below uses a variable called startIndex to indicate which element to start the sequence sampling from.
You could accomplish this in a vectorized way by constructing an index array. Create a vector consisting of the starting indices of each sequence. For reuse sake, I put the length of the sequence, the step size between sequence starts, and the start of the last sequence as variables. In the example you describe, the length of the sequence should be 50, the step size should be 10 and the start of the last sequence depends on the size of the input data and your needs.
>> startIndex = 10;
>> sequenceSize = 5;
>> finalSequenceStart = 20;
Create some sample data:
>> sampleData = randi(100, 1, 28)
sampleData =
Columns 1 through 18
8 53 10 82 82 73 15 66 52 98 65 81 46 44 83 9 14 18
Columns 19 through 28
40 84 81 7 40 53 42 66 63 30
Create a vector of the start indices of the sequences:
>> sequenceStart = startIndex:sequenceSize:finalSequenceStart
sequenceStart =
10 15 20
Create an array of indices to index into the data array:
>> index = cumsum(ones(sequenceSize, length(sequenceStart)))
index =
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
>> index = index + repmat(sequenceStart, sequenceSize, 1) - 1
index =
10 15 20
11 16 21
12 17 22
13 18 23
14 19 24
Finally, use this index array to reference the data array:
>> sampleData(index)
ans =
98 83 84
65 9 81
81 14 7
46 18 40
44 40 53
Use (start : step : end) indexing: v(1:1:50), v(10:1:60), etc. If the step is 1, you can omit it: v(1:50).
Consider the following vectorized code:
x = 1:100; %# an example sequence of numbers
nwind = 50; %# window size
noverlap = 40; %# number of overlapping elements
nx = length(x); %# length of sequence
ncol = fix((nx-noverlap)/(nwind-noverlap)); %# number of sliding windows
colindex = 1 + (0:(ncol-1))*(nwind-noverlap); %# starting index of each
%# indices to put sequence into columns with the proper offset
idx = bsxfun(#plus, (1:nwind)', colindex)-1; %'
%# apply the indices on the sequence
slidingWindows = x(idx)
The result (truncated for brevity):
slidingWindows =
1 11 21 31 41 51
2 12 22 32 42 52
3 13 23 33 43 53
...
48 58 68 78 88 98
49 59 69 79 89 99
50 60 70 80 90 100
In fact, the code was adapted from the now deprecated SPECGRAM function from the Signal Processing Toolbox (just do edit specgram.m to see the code).
I omitted parts that zero-pad the sequence in case the sliding windows do not evenly divide the entire sequence (for example x=1:105), but you can easily add them again if you need that functionality...