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?)
Related
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))
I have to make a matlab program, which should create a QR Code.
My problem is the Reed Solomon error correction
The user enters the word he wants. [...] I got a string of numbers I should be gone in a polynomial generator (Reed Solomon) (I found some sites that do this very well: http://www.pclviewer.com/rs2/calculator.html)
I would like it to happen: for example I input: 32 91 11 120 209 114 220 77 67 64 236 17 236
[Reed Solomon generator polynomial]
and I want to find out: 168 72 22 82 217 54 156 0 46 15 180 122 16
I found the functions rsenc comm.rsencoder gf ... But it is impossible to understand the operation of these functions. Functions are detailed: http://www.mathworks.fr/fr/help/comm...n.html#fp12225
I tried a code of this type :
n = 255; k = 13; % Codeword length and message length
m = 8; % Number of bits in each symbol
msg = [32 91 11 120 209 114 220 77 67 64 236 17 236]; % Message is a Galois array.
obj = comm.RSEncoder(n, k);
c1 = step(obj, msg(1,:)');
c = [c1].';
He produced a string of 255 while I want 13 output.
Thank you for your help.
I think that you are committing a mistake.
'n' is the length of final message with parity code.
'k' is the lenght of message (number of symbols)
I guess that this will help you:
clc, clear all;
M = 16; % Modulation Order || same that Max value, at your case: 256! 2^m
hEnc = comm.RSEncoder;
hEnc.CodewordLength = M - 1; % Max = M-1, Min = 4, Must be greater than MessageLenght
hEnc.MessageLength = 13; % Experiment change up and down value (using odd number)
hEnc.BitInput = false;
hEnc
t = hEnc.CodewordLength - hEnc.MessageLength;
frame = 2*hEnc.MessageLength; % multiple of MensagemLength
fprintf('\tError Detection (in Symbols): %d\n',t);
fprintf('\tError Correction: %.2f\n',t/2);
data = randi([0 M-1], frame, 1); % Create a frame with symbols range (0 to M-1)
encodedData = step(hEnc, data); % encod the frame
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]
If I had a matrix A such as:
63 55 85 21 71
80 65 85 48 53
55 60 93 71 66
21 65 40 33 21
61 90 80 48 50
... and so on how would I find the minimum values of each column and remove those numbers from the matrix completely, meaning essentially I would have one less row overall.
I though about using:
[C,I] = min(A);
A(I) = [];
but that wouldn't remove the necessary numbers, and also reshape would not work either. I would like for this to work with an arbitrary number of rows and columns.
A = [
63 55 85 21 71
80 65 85 48 53
55 60 93 71 66
21 65 40 33 21
61 90 80 48 50
];
B = zeros( size(A,1)-1, size(A,2));
for i=1:size(A,2)
x = A(:,i);
maxIndex = find(x==min(x(:)),1,'first');
x(maxIndex) = [];
B(:,i) = x;
end
disp(B);
Another vectorized solution:
M = mat2cell(A,5,ones(1,size(A,2)));
z = cellfun(#RemoveMin,M);
B = cell2mat(z);
disp(B);
function x = RemoveMin(x)
minIndex = find(x==min(x(:)),1,'first');
x(minIndex) = [];
x = {x};
end
Another solution:
[~,I] = min(A);
indexes = sub2ind(size(A),I,1:size(A,2));
B = A;
B(indexes) = [];
out = reshape(B,size(A)-[1 0]);
disp(out);
Personally I prefer the first because:
For loops aren't evil - many times they are actually faster (By using JIT optimizer)
The algorithm is clearer to the developer who reads your code.
But of course, its up to you.
Your original approach works if you convert the row indices resulting from min into linear indices:
[m, n] = size(A);
[~, row] = min(A,[],1);
A(row + (0:n-1)*m) = [];
A = reshape(A, m-1, n);
Assume we have the following data:
H_T = [36 66 21 65 52 67 73; 31 23 19 33 36 39 42]
P = [40 38 39 40 35 32 37]
Using MATLAB 7.0, I want to create three new matrices that have the following properties:
The matrix H (the first part in matrix H_T) will be divided to 3 intervals:
Matrix 1: the 1st interval contains the H values between 20 to 40
Matrix 2: the 2nd interval contains the H values between 40 to 60
Matrix 3: the 3rd interval contains the H values between 60 to 80
The important thing is that the corresponding T and P will also be included in their new matrices meaning that H will control the new matrices depending on the specifications defined above.
So, the resultant matrices will be:
H_T_1 = [36 21; 31 19]
P_1 = [40 39]
H_T_2 = [52; 36]
P_2 = [35]
H_T_3 = [66 65 67 73; 23 33 39 42]
P_3 = [38 40 32 37]
Actually, this is a simple example and it is easy by looking to create the new matrices depending on the specifications, BUT in my values I have thousands of numbers which makes it very difficult to do that.
Here's a quick solution
[~,bins] = histc(H_T(1,:), [20 40 60 80]);
outHT = cell(3,1);
outP = cell(3,1);
for i=1:3
idx = (bins == i);
outHT{i} = H_T(:,idx);
outP{i} = P(idx);
end
then you access the matrices as:
>> outHT{3}
ans =
66 65 67 73
23 33 39 42
>> outP{3}
ans =
38 40 32 37