I have a matrix train3.
1 2 3 4 5 6 7
2 12 13 14 15 16 17
3 62 53 44 35 26 17
4 52 13 24 15 26 37
I want to select only those rows of whose 1st columns contain specific values (in my case 1 and 2).
I have tried the following,
>> train3
train3 =
1 2 3 4 5 6 7
2 12 13 14 15 16 17
3 62 53 44 35 26 17
4 52 13 24 15 26 37
>> ind1 = train3(:,1) == 1
ind1 =
1
0
0
0
>> ind2 = train3(:,1) == 2
ind2 =
0
1
0
0
>> mat1 = train3(ind1, :)
mat1 =
1 2 3 4 5 6 7
>> mat2 = train3(ind2, :)
mat2 =
2 12 13 14 15 16 17
>> mat3 = [mat1 ; mat2]
mat3 =
1 2 3 4 5 6 7
2 12 13 14 15 16 17
>>
Is there any better way to do this?
Presumably you are trying to get mat3 in a single step which you can do with:
mat3 = train3(train3(:,1)==1 | train3(:,1)==2,:)
A more general way to do this would be to use ismember to get all of the rows that match the values in a list:
train3 =[
1 2 3 4 5 6 7
2 12 13 14 15 16 17
3 62 53 44 35 26 17
4 52 13 24 15 26 37];
chooseList = [1 2];
colIndex = ismember(train3(:, 1), chooseList);
subset = train3(colIndex, :);
subset =
1 2 3 4 5 6 7
2 12 13 14 15 16 17
Related
(For my problem, I use a matrix A 4x500000. And the values of A(4,k) varies between 1 and 200).
I give here an example for a case A 4x16 and A(4,k) varies between 1 and 10.
I want first to match a name to the value from 1 to 5 (=10/2):
1 = XXY;
2 = ABC;
3 = EFG;
4 = TXG;
5 = ZPF;
My goal is to find,for a vector X, a matrix M from the matrix A:
A = [20 52 70 20 52 20 52 20 20 10 52 20 11 1 52 20
32 24 91 44 60 32 24 32 32 12 11 32 2 5 24 32
40 37 24 30 11 40 37 40 40 5 10 40 40 3 37 40
2 4 1 3 4 5 2 1 3 3 8 6 7 9 6 10]
A(4,k) takes all values between 1 and 10. These values can be repeated and they all appear on the 4th line.
20
X= 32 =A(1:3,1)=A(1:3,6)=A(1:3,8)=A(1:3,9)=A(1:3,12)=A(1:3,16)
40
A(4,1) = 2;
A(4,6) = 5;
A(4,8) = 1;
A(4,9) = 3;
A(4,12) = 6;
A(4,16) = 10;
for A(4,k) corresponding to X, I associate 2 if A(4,k)<= 5, and 1 if A(4,k)> 5. For the rest of the value of A(4,k) which do not correspond to X, I associate 0:
[ 1 2 3 4 5 %% value of the fourth line of A between 1 and 5
2 2 2 0 2
ZX = 6 7 8 9 10 %% value of the fourth line of A between 6 and 10
1 0 0 0 1
2 2 2 0 2 ] %% = max(ZX(2,k),ZX(4,k))
the ultimate goal is to find the matrix M:
M = [ 1 2 3 4 5
XXY ABC EFG TXG ZPF
2 2 2 0 2 ] %% M(3,:)=ZX(5,:)
Code -
%// Assuming A, X and names to be given to the solution
A = [20 52 70 20 52 20 52 20 20 10 52 20 11 1 52 20
32 24 91 44 60 32 24 32 32 12 11 32 2 5 24 32
40 37 24 30 11 40 37 40 40 5 10 40 40 3 37 40
2 4 1 3 4 5 2 1 3 3 8 6 7 9 6 10];
X = [20 ; 32 ; 40];
names = {'XXY','ABC','EFG','TXG','ZPF'};
limit = 10; %// The maximum limit of A(4,:). Edit this to 200 for your actual case
%// Find matching 4th row elements
matches = A(4,ismember(A(1:3,:)',X','rows'));
%// Matches are compared against all possible numbers between 1 and limit
matches_pos = ismember(1:limit,matches);
%// Finally get the line 3 results of M
vals = max(2*matches_pos(1:limit/2),matches_pos( (limit/2)+1:end ));
Output -
vals =
2 2 2 0 2
For a better way to present the results, you can use a struct -
M_struct = cell2struct(num2cell(vals),names,2)
Output -
M_struct =
XXY: 2
ABC: 2
EFG: 2
TXG: 0
ZPF: 2
For writing the results to a text file -
output_file = 'results.txt'; %// Edit if needed to be saved to a different path
fid = fopen(output_file, 'w+');
for ii=1:numel(names)
fprintf(fid, '%d %s %d\n',ii, names{ii},vals(ii));
end
fclose(fid);
Text contents of the text file would be -
1 XXY 2
2 ABC 2
3 EFG 2
4 TXG 0
5 ZPF 2
A bsxfun() based approach.
Suppose your inputs are (where N can be set to 200):
A = [20 52 70 20 52 20 52 20 20 10 52 20 11 1 52 20
32 24 91 44 60 32 24 32 32 12 11 32 2 5 24 32
40 37 24 30 11 40 37 40 40 5 10 40 40 3 37 40
2 4 1 3 4 5 2 1 3 3 8 6 7 9 6 10]
X = [20; 32; 40]
N = 10;
% Match first 3 rows and return 4th
idxA = all(bsxfun(#eq, X, A(1:3,:)));
Amatch = A(4,idxA);
% Match [1:5; 5:10] to 4th row
idxZX = ismember([1:N/2; N/2+1:N], Amatch)
idxZX =
1 1 1 0 1
1 0 0 0 1
% Return M3
M3 = max(bsxfun(#times, idxZX, [2;1]))
M3 =
2 2 2 0 2
I'm trying to implement the Baker map.
Is there a function that would allow one to divide a 8 x 8 matrix by providing, for example, a sequence of divisors 2, 4, 2 and rearranging pixels in the order as shown in the matrices below?
X = reshape(1:64,8,8);
After applying divisors 2,4,2 to the matrix X one should get a matrix like A shown below.
A=[31 23 15 7 32 24 16 8;
63 55 47 39 64 56 48 40;
11 3 12 4 13 5 14 6;
27 19 28 20 29 21 30 22;
43 35 44 36 45 37 46 38;
59 51 60 52 61 53 62 54;
25 17 9 1 26 18 10 2;
57 49 41 33 58 50 42 34]
The link to the document which I am working on is:
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.39.5132&rep=rep1&type=pdf
This is what I want to achieve:
Edit: a little more generic solution:
%function Z = bakermap(X,divisors)
function Z = bakermap()
X = reshape(1:64,8,8)'
divisors = [ 2 4 2 ];
[x,y] = size(X);
offsets = sum(divisors)-fliplr(cumsum(fliplr(divisors)));
if any(mod(y,divisors)) && ~(sum(divisors) == y)
disp('invalid divisor vector')
return
end
blocks = #(div) cell2mat( cellfun(#mtimes, repmat({ones(x/div,div)},div,1),...
num2cell(1:div)',...
'UniformOutput',false) );
%create index matrix
I = [];
for ii = 1:numel(divisors);
I = [I, blocks(divisors(ii))+offsets(ii)];
end
%create Baker map
Y = flipud(X);
Z = [];
for jj=1:I(end)
Z = [Z; Y(I==jj)'];
end
Z = flipud(Z);
end
returns:
index matrix:
I =
1 1 3 3 3 3 7 7
1 1 3 3 3 3 7 7
1 1 4 4 4 4 7 7
1 1 4 4 4 4 7 7
2 2 5 5 5 5 8 8
2 2 5 5 5 5 8 8
2 2 6 6 6 6 8 8
2 2 6 6 6 6 8 8
Baker map:
Z =
31 23 15 7 32 24 16 8
63 55 47 39 64 56 48 40
11 3 12 4 13 5 14 6
27 19 28 20 29 21 30 22
43 35 44 36 45 37 46 38
59 51 60 52 61 53 62 54
25 17 9 1 26 18 10 2
57 49 41 33 58 50 42 34
But have a look at the if-condition, it's just possible for these cases. I don't know if that's enough. I also tried something like divisors = [ 1 4 1 2 ] - and it worked. As long as the sum of all divisors is equal the row-length and the modulus as well, there shouldn't be problems.
Explanation:
% definition of anonymous function with input parameter: div: divisor vector
blocks = #(div) cell2mat( ... % converts final result into matrix
cellfun(#mtimes, ... % multiplies the next two inputs A,B
repmat(... % A...
{ones(x/div,div)},... % cell with a matrix of ones in size
of one subblock, e.g. [1,1,1,1;1,1,1,1]
div,1),... % which is replicated div-times according
to actual by cellfun processed divisor
num2cell(1:div)',... % creates a vector [1,2,3,4...] according
to the number of divisors, so so finally
every Block A gets an increasing factor
'UniformOutput',false...% necessary additional property of cellfun
));
Have also a look at this revision to have a simpler insight in what is happening. You requested a generic solution, thats the one above, the one linked was with more manual inputs.
I need a function that splits a vector in smaller frames with an overlap, like buffer, but instead of column-wise, it should be done row-wise.
This is how buffer works:
x = 1:20
x = buffer(x, 10, 5);
x = 0 1 6 11
0 2 7 12
0 3 8 13
0 4 9 14
0 5 10 15
1 6 11 16
2 7 12 17
3 8 13 18
4 9 14 19
5 10 15 20
What I want would be this though:
x = 0 0 1 2
1 2 3 4
3 4 5 6
5 6 7 8
7 8 9 10
9 10 11 12
11 12 13 14
13 14 15 16
15 16 17 18
17 18 19 20
Is there any function or way to achieve that? Maybe combination of buffer + some rearranging?
First figure out the answer in columns, then transpose the resulting matrix:
buffer(x, 4, 2).'
I have a pretty large matrix M and I am only interested in a few of the columns. I have a boolean vector V where a value of 1 represents a column that is of interest. Example:
-1 -1 -1 7 7 -1 -1 -1 7 7 7
M = -1 -1 7 7 7 -1 -1 7 7 7 7
-1 -1 7 7 7 -1 -1 -1 7 7 -1
V = 0 0 1 1 1 0 0 1 1 1 1
If multiple adjacent values of V are all 1, then I want the corresponding columns of M to be extracted into another matrix. Here's an example, using the matrices from before.
-1 7 7 -1 7 7 7
M1 = 7 7 7 M2 = 7 7 7 7
7 7 7 -1 7 7 -1
How might I do this efficiently? I would like all these portions of the matrix M to be stored in a cell array, or at least have an efficient way to generate them one after the other. Currently I'm doing this in a while loop and it is not as efficient as I'd like it to be.
(Note that my examples only include the values -1 and 7 just for clarity; this isn't the actual data I use.)
You can utilize the diff function for this, to break your V vector into blocks
% find where block differences exist
diffs = diff(V);
% move start index one value forward, as first value in
% diff represents diff between first and second in original vector
startPoints = find(diffs == 1) + 1;
endPoints = find(diffs == -1);
% if the first block begins with the first element diff won't have
% found start
if V(1) == 1
startPoints = [1 startPoints];
end
% if last block lasts until the end of the array, diff won't have found end
if length(startPoints) > length(endPoints)
endPoints(end+1) = length(V);
end
% subset original matrix into cell array with indices
results = cell(size(startPoints));
for c = 1:length(results)
results{c} = M(:,startPoints(c):endPoints(c));
end
The one thing I'm not sure of is if there's a better way to find the being_indices and end_indices.
Code:
X = [1 2 3 4 5 1 2 3 4 5
6 7 8 9 10 6 7 8 9 10
11 12 13 14 15 11 12 13 14 15
16 17 18 19 20 16 17 18 19 20
1 2 3 4 5 1 2 3 4 5
6 7 8 9 10 6 7 8 9 10
11 12 13 14 15 11 12 13 14 15
16 17 18 19 20 16 17 18 19 20];
V = logical([ 1 1 0 0 1 1 1 0 1 1]);
find_indices = find(V);
begin_indices = [find_indices(1) find_indices(find(diff(find_indices) ~= 1)+1)];
end_indices = [find_indices(find(diff(find_indices) ~= 1)) find_indices(end)];
X_truncated = mat2cell(X(:,V),size(X,1),[end_indices-begin_indices]+1);
X_truncated{:}
Output:
ans =
1 2
6 7
11 12
16 17
1 2
6 7
11 12
16 17
ans =
5 1 2
10 6 7
15 11 12
20 16 17
5 1 2
10 6 7
15 11 12
20 16 17
ans =
4 5
9 10
14 15
19 20
4 5
9 10
14 15
19 20
We have the following case:
Q = [idxcell{:,1}];
Sort = sort(Q,'descend')
Sort =
Columns 1 through 13
23 23 22 22 20 19 18 18 18 18 17 17 17
Columns 14 through 26
15 15 14 14 13 13 13 12 12 12 11 10 9
Columns 27 through 39
9 9 8 8 8 8 8 7 7 7 7 7 7
Columns 40 through 52
7 6 6 6 5 4 4 3 3 3 3 2 2
Columns 53 through 64
2 2 2 2 2 2 2 1 1 1 1 1
How can we sort matrix Sort according to how many times its values are repeated?
Awaiting result should be:
repeatedSort = 2(9) 7(7) 1(5) 8(5) 3(4) 18(4) 6(3) 9(3) 12(3) 13(3) 17(3) 4(2) 14(2) 15(2) 22(2) 23(2) 5(1) 10(1) 11(1) 19(1) 20(1)
or
repeatedSort = 2 7 1 8 3 18 6 9 12 13 17 4 14 15 22 23 5 10 11 19 20
Thank you in advance.
You can use the TABULATE function from the Statistics Toolbox, then call SORTROWS to sort by the frequency.
Example:
x = randi(10, [20 1]); %# random values
t = tabulate(x); %# unique values and counts
t = t(find(t(:,2)),1:2); %# get rid of entries with zero count
t = sortrows(t, -2) %# sort according to frequency
the result, where first column are the unique values, second is their count:
t =
2 4 %# value 2 appeared four times
5 4 %# etc...
1 3
8 3
7 2
9 2
4 1
6 1
Here's one way of doing it:
d = randi(10,1,30); %Some fake data
n = histc(d,1:10);
[y,ii] = sort(n,'descend');
disp(ii) % ii is now sorted according to frequency