If I have a matrix
F=[ 24 3 17 1;
28 31 19 1;
24 13 25 2;
47 43 39 1;
56 41 39 2];
in the first three columns I have feature values a forth column is for class labels. my problem is to get rid of same feature values when class label is different for that particular values.
like for F matrix I have to remove the rows 1,3,4 and 5 ,because for first column there are 2 different values in column four and same is for third column (39 and 39)as class label again got changed.
so output should look like
F=[28 31 19 1];
The straightforward approach would be iterating over the columns, counting the number of different classes for each value, and removing the rows for values associated to more than one class.
Example
F = [24 3 17 1; 28 31 19 1; 24 13 25 2; 47 43 39 1; 56 41 39 2];
%// Iterate over columns
for col = 1:size(F, 2) - 1
%// Count number of different classes for each value
[vals, k, idx] = unique(F(:, col));
count = arrayfun(#(x)length(unique(F(F(:, col) == x, end))), vals);
%// Remove values associated to more than one class
F(count(idx) > 1, :) = [];
end
This results in:
F =
28 31 19 1
Another take at the problem, without arrayfun (edited)
F = [24 3 17 1; 28 31 19 1; 24 13 25 2; 47 43 39 1; 56 41 39 2];
Separate both classes:
A1 = F(F(:,4)==1,1:3);
A2 = F(F(:,4)==2,1:3);
Replicate them to a 3D matrix to compare each line of class1 with each line of class2:
B2 = repmat(shiftdim(A2',-1),size(A1,1),1);
B1 = repmat(A1,[1,1,size(A2,1)]);
D4 = squeeze(sum(B1 == B2,2));
remove rows duplicated rows
A1(logical(sum(D4,2)),:) = [];
A2(logical(sum(D4,1)),:) = [];
reconstruct original matrix
R = [A1 ones(size(A1,1),1);A2 2*ones(size(A2,1),1)];
Related
Suppose I have two rows as follows with the data
R1 = 12 13 15 17 200 23
R2 = 32 22 43 67 21 74
I would like to know how to shift the values of the 2nd index and 3rd index of R1 (e.g, 13 15) into the second row of R2 so it becomes
R2 = 32 13 15 67 21 74
It's very simple: R2(2:3) = R1(2:3);
Code sample:
R1 = [12 13 15 17 200 23];
R2 = [32 22 43 67 21 74];
R2(2:3) = R1(2:3);
You can also use the following: R2([2,3]) = R1([2,3]);, in case indexes are not sequential.
In case R1 and R2 are two rows in a matrix, you can use the following sample:
% Create the input matrix A:
R1 = [12 13 15 17 200 23];
R2 = [32 22 43 67 21 74];
A = [R1; R2];
%Copy values from index 2 and 3 of first row to index 2 and 3 of second row:
A(2, [2,3]) = A(1, [2,3]);
In case there are more rows, and you need to "shift" all down, you can use the following example:
%Create sample matrix A (6x6 elements).
A = magic(6);
%"Shift" values of index 2,3 of all rows, one row down:
A(2:end, [2,3]) = A(1:end-1, [2,3]);
Refer here: http://www.mathworks.com/company/newsletters/articles/matrix-indexing-in-matlab.html
So say I have two arrays:
A:14 63 13
38 44 23
11 12 13
38 44 23
B:38 44 23
I am trying to use ismember to return the index of every location where B is found in A. All examples I have found online only list the first or last occurrence of a match, I am trying to have a list indices for all values that match, even repeating ones. Thanks
Use ismember with the 'rows' arugment:
ismember(A, B, 'rows')
which results in a logical array [0 1 0 1] which is often better than an array of indices but if you want the indices specifically then just use find:
find(ismember(A,B,'rows'))
to return [2,4]
Note that this method will still work if B has multiple rows e.g. B = [38 44 23; 11 12 13], it will return [0; 1; 1; 1]
You can use bsxfun for the comarison:
idx = find( all( bsxfun(#eq, A, B), 2 )); %// only where all line matches
Results with
idx =
2
4
You can look into pdist2 if you have A and B as Nx3 sized arrays -
[indA,indB] = ind2sub([size(A,1) size(B,1)],find(pdist2(A,B)==0));
ind = [indA,indB]
Thus, in ind you would have the pairwise indices for the matches with the first column representing the indices for A and the second one for B.
Sample run -
A =
14 63 13
38 44 23
11 12 13
14 63 13
38 44 23
B =
38 44 23
14 63 13
ind =
2 1
5 1
1 2
4 2
This is just an improved version of shai's answer for handling multiple rows of B
idx = find(any(all( bsxfun(#eq, A, permute(B,[3 2 1])), 2 ),3));
Sample Run:
A = [14 63 13;
38 44 23;
11 12 13;
38 44 23];
B = [38 44 23;
11 12 13];
idx = find(any(all( bsxfun(#eq, A, permute(B,[3 2 1])), 2 ),3));
>> idx
idx =
2
3
4
If I have a matrix
F=[ 24 3 17 1;
28 31 19 1;
24 13 25 2;
47 43 39 1;
56 41 39 2];
in the first three columns I have feature values a forth column is for class labels. my problem is to get rid of same feature values when class label is different for that particular values.
like for F matrix I have to remove the rows 1,3,4 and 5 ,because for first column there are 2 different values in column four and same is for third column (39 and 39)as class label again got changed.
so output should look like
F=[28 31 19 1];
The straightforward approach would be iterating over the columns, counting the number of different classes for each value, and removing the rows for values associated to more than one class.
Example
F = [24 3 17 1; 28 31 19 1; 24 13 25 2; 47 43 39 1; 56 41 39 2];
%// Iterate over columns
for col = 1:size(F, 2) - 1
%// Count number of different classes for each value
[vals, k, idx] = unique(F(:, col));
count = arrayfun(#(x)length(unique(F(F(:, col) == x, end))), vals);
%// Remove values associated to more than one class
F(count(idx) > 1, :) = [];
end
This results in:
F =
28 31 19 1
Another take at the problem, without arrayfun (edited)
F = [24 3 17 1; 28 31 19 1; 24 13 25 2; 47 43 39 1; 56 41 39 2];
Separate both classes:
A1 = F(F(:,4)==1,1:3);
A2 = F(F(:,4)==2,1:3);
Replicate them to a 3D matrix to compare each line of class1 with each line of class2:
B2 = repmat(shiftdim(A2',-1),size(A1,1),1);
B1 = repmat(A1,[1,1,size(A2,1)]);
D4 = squeeze(sum(B1 == B2,2));
remove rows duplicated rows
A1(logical(sum(D4,2)),:) = [];
A2(logical(sum(D4,1)),:) = [];
reconstruct original matrix
R = [A1 ones(size(A1,1),1);A2 2*ones(size(A2,1),1)];
e=[40 19 18 20 30 34 65 97 155 160];
If there is a minimum difference between two consecutive values (for e.g. (19,18), (30, 34) and (155,160)) then merge these values..
Similar values also...Whatever condition can be used to solve this..Kindly help to solve this..
Iteratively,
e = [ 40 19 18 20 30 34 65 97 155 160];
current = e + 1; % init
prev = e;
while ~isequal( current, prev )
prev = current;
d = [ diff( prev ) < 5 true]; % always keep the last one
current = prev( d );
end
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...