I want to create a not-equally-spaced sequence of numbers in MATLAB starting from 24 and ending to 511.The Sequence uses 32 and 33 alternately as the increment. Thus, the sequence would be as below : [24 56 89 121 154 186 219 251 284 316 349 381 414 446 479 511] Notice that :
24+32=56
56+33=89
89+32=121
121+33=154
...
I just wonder how to modify my own codes or to write new codes to have the answer. My own codes are below:
t_3233=0;
for k=24:(32+t_3233):511
t_3233
k
if t_3233==1
t_3233=0;
else if t_3233==0
t_3233=1;
end
end
end
In this particular case you can use:
len = 16;
vector = round(linspace(24,511,len))
Related
I would like to partly load a PTX file by matlab (please see the following example)
I need to read and write the first two row (2 numbers) into 2 variables say a and b. And read and write the data from 5th row to the end into a matrix
Thanks for your help
114
221
1 0 0
1 0 0 0
-5.566405 -7.161944 -1.144557 0.197208 24 29 35
-5.560656 -7.154540 -1.137673 0.222400 29 32 39
-5.559846 -7.153491 -1.131895 0.254002 37 40 49
-5.560894 -7.154833 -1.126452 0.305013 51 54 63
-5.560084 -7.153783 -1.120633 0.290013 72 76 88
-5.561128 -7.155119 -1.115189 0.243214 105 113 134
-5.563203 -7.157782 -1.109926 0.227604 130 143 177
-5.569191 -7.165479 -1.105504 0.201602 121 140 173
-7.833616 -10.078705 -1.546952 0.130007 94 112 134
Look at the tdfread function in order to get the data into Matlab. It should be something like datafile = tdfread(filename, '\t'). Once you have that, index into the variable returned from that function like
a = datafile(1, 1);
b = datafile(2, 1);
data = datafile(5:end, :);
Just started MATLAB 2 days ago and I can't figure out a non-loop method (since I read they were slow/inefficient and MATLAB has better alternatives) to perform a simple task.
I have a matrix of 5 columns and 270 rows. What I want to do is:
if the value of an element in column 5 of matrix goodM is below 90, I want to take that element and and subtract it from 90.
So far I tried:
test = goodM(:,5) <= 90;
goodM(test) = 999;
It changes all goodM values within column 1 not 5 into 999, in addition this method doesn't allow me to perform operations on the elements below 90 in column 5. Any elegant solution to doing this?
edit:: goodM(:,5)(test) = 999; doesn't seem to work either so I have no idea to specify the target column.
I am assuming you are looking to operate on elements that have values below 90 as your text in the question reads, rather than 'below or equal to' as represented by '<=' as used in your code. So try this -
ind = find(goodM(:,5) < 90) %// Find indices in column 5 that have values less than 90
goodM(ind,5) = 90 - goodM(ind,5) %// Operate on those elements using indices obtained from previous step
Try this code:
b=90-a(a(:,5)<90,5);
For example:
a =
265 104 479 13 176
26 110 447 208 144
379 163 179 366 464
301 48 274 391 26
429 374 174 184 297
495 375 312 373 82
465 272 399 447 420
205 170 373 122 84
1 417 63 65 252
271 277 412 113 500
then,
b=90-a(a(:,5)<90,5);
b =
64
8
6
I am trying to compare multiple image using corr2 to see the similarity in correlation.
for i=1:2
first_img = imread(sprintf('%g.jpg',i));
first_size = size(first_img);
size_temp = size(first_size);
max_size = max(size_temp);
if max_size == 3
first_img = rgb2gray(first_img);
first_size = size(first_img);
end
for j=i+1:2
second_img = imread(sprintf('%g.jpg',j));
second_size = size(second_img);
size_temp = size(second_size);
max_size = max(size_temp);
if max_size == 3
second_img = rgb2gray(second_img);
second_size = size(second_img);
end
if i == j
continue;end
if first_size ~= second_size
continue;end
if first_size == second_size
correlation_fs = corr2(first_img,second_img);
if correlation_fs == 1
fprintf('%g is the same as %g\n',first_img,second_img);
end
end
end
end
now, the problem show up when the first image compared to the 3rd dummy image which is exactly the same as the first image.
219 is the same as 219
220 is the same as 220
221 is the same as 221
221 is the same as 222
224 is the same as 223
222 is the same as 221
221 is the same as 222
223 is the same as 224
218 is the same as 236
242 is the same as 232
217 is the same as 219
226 is the same as 228
220 is the same as 229
241 is the same as 251
254 is the same as 253
250 is the same as 247
253 is the same as 253
252 is the same as 248
237 is the same as 224
217 is the same as 218
225 is the same as 219
219 is the same as 223
219 is the same as 214
222 is the same as 237
I don't know why this is showing up, it should print that image 1 is the same as image 3, at least is what i want it to.
You are printing out the entire image matrices instead of the image number. Try:
fprintf('%g is the same as %g\n',i,j)
Think about what first_img is, it's a matrix of pixel intensities. So you're printing out all the pixel values.
fprintf('%g is the same as %g\n',first_img,second_img);
Here you are passing two images as arguments. You should have passed the image numbers instead.
fprintf('%g is the same as %g\n', i, j);
I would like to add a constant value of 360 to a vector of values after the maximum value is reached. That is, if H=[12 26 67 92 167 178 112 98 76 85], how do I write a matlab code so that 180 is added to all values after 178? The answer should be H=[12 26 67 92 167 178 292 278 256 265].
This should work on earlier Matlab versions as well:
H=[12 26 67 92 167 178 112 98 76 85]
[n, n] = max(H);
H(n+1:end) = H(n+1:end) + 180
Try following:
n=find(H==max(H));
H(n+1:end)=H(n+1:end)+180;
Since desired vector values are in increasing order, idea here is to find the index of maximum value and increment all the subsequent elements with 180.
EDIT
Better approach for finding max index, as suggested by #LeonidBeschastny
[~,n]=max(H);
I would appreciate your help with the following problem in matlab:
I have a vector and I would like to select parts of it based on the following two vector of start and end indices of parts:
aa = [1 22 41 64 83 105 127 147 170 190 212 233]
bb = [21 40 63 82 104 126 146 169 189 211 232 252]
Basically I would like to perform some function on V(1:21), V(22:40),... V(233:252).
I have tried V(aa:bb) or V(aa(t):bb(t)) where t = 1:12 but I get only V(1:21), probably because V(22:40) has 19 elements compared to V(1:21) which has 22 elements.
Is there a fast way of programming this?
Put your selection in a cell array, and apply your function to each cell:
aa = [1 22 41 64 83 105 127 147 170 190 212 233]
bb = [21 40 63 82 104 126 146 169 189 211 232 252]
V = rand(252,1); % some sample data
selV = arrayfun(#(t) V(aa(t):bb(t)), 1:12,'uniformoutput',false);
result = cellfun(#yourfunction,selV)
% or
result = cellfun(#(selVi) yourfunction(selVi), selV);
If the function you want to apply has scalar output to every vector input, this should give you an 1x12 array. If the function gives vector output, you'll have to include the uniformoutput parameter:
result = cellfun(#(selVi) yourfunction(selVi), selV,'uniformoutput',false);
which gives you a 1x12 cell array.
If you want to run this in a highly condensed form, you can write (in two lines, for clarity)
aa = [1 22 41 64 83 105 127 147 170 190 212 233]
bb = [21 40 63 82 104 126 146 169 189 211 232 252]
V = rand(252,1); % some sample data borrowed from #Gunther
%# create an anonymous function that accepts start/end of range as input
myFunctionHandle = #(low,high)someFunction(V(low:high));
%# calculate result
%# if "someFunction" returns a scalar, you can drop the 'Uni',false part
%# from arrayfun
result = arrayfun(myFunctionHandle(low,high),aa,bb,'uni',false)
Note that this may run more slowly than an explicit loop at the moment, but arrayfun is likely to be multithreaded in a future release.