I'm planning to do the following in matlab:
Create an empty set
Insert an element which specifies a certain criterion in it
So, as a quick run on the matlab command line I did the following:
>> e=78
e =
78
>> Ck=[]
Ck =
[]
>> Cn=Ck+e
Cn =
[]
But, I was expecting to get the following:
Cn = [78]
Why didn't I get the expected result? And, how can I solve this out?
Thanks.
The #plus operator is defined as an arithmetic operation, but not as a set operation.
To add to an existing (1D) set at specified locations, you perform catenation and/or indexing. For example, to add at the end of a set, you can write
Cn = [Ck,e];
or
Cn = Ck;
Cn(end+1) = e;
Probably I'll be wrong, but I would initialize an empty array by means of zeros, like:
C = zeros(m,n)
with m,n = 1,2,...,N
Then, you input your second array/matrix and treat it according to whichever algorithm you have in mind.
I hope this helps.
Related
If I wanted to create an array of specified class I would use an approach like this. So creating an array of int looks like this:
Aint = int16.empty(5,0);
Aint(1) = 3;
And it works fine. Now I want to create an array of tf class objects. My approach was similar:
L = tf.empty(5, 0);
s = tf('s');
L(1) = s;
This gives me an error:
Error using InputOutputModel/subsasgn (line 57)
Not enough input arguments.
Error in tf_array (line 6)
L(1) = s;
I also made sure to display class(s) and it correctly says it's tf. What do I do wrong here?
As usual, the MATLAB documentation has an example for how to do this sort of thing:
sys = tf(zeros(1,1,3));
s = tf('s');
for k = 1:3
sys(:,:,k) = k/(s^2+s+k);
end
So, the problem likely is that the indexing L(1) is wrong, it needs to be L(:,:,1).
Do note that tf.empty(5, 0) is instructing to create a 5x0 array (i.e. an empty array). There is no point to this. You might as well just skip this instruction. Because when you later do L(:,:,1), you'll be increasing the array size any way (it starts with 0 elements, you want to assign a new element, it needs to reallocate the array). You should always strive to create the arrays of the right size from the start.
I have an excel file and I need to read it based on string values in the 4th column. I have written the following but it does not work properly:
[num,txt,raw] = xlsread('Coordinates','Centerville');
zn={};
ctr=0;
for i = 3:size(raw,1)
tf = strcmp(char(raw{i,4}),char(raw{i-1,4}));
if tf == 0
ctr = ctr+1;
end
zn{ctr}=raw{i,4};
end
data=zeros(1,10); % 10 corresponds to the number of columns I want to read (herein, columns 'J' to 'S')
ctr=0;
for j = 1:length(zn)
for i=3:size(raw,1)
tf=strcmp(char(raw{i,4}),char(zn{j}));
if tf==1
ctr=ctr+1;
data(ctr,:,j)=num(i-2,10:19);
end
end
end
It gives me a "15129x10x22 double" thing and when I try to open it I get the message "Cannot display summaries of variables with more than 524288 elements". It might be obvious but what I am trying to get as the output is 'N = length(zn)' number of matrices which represent the data for different strings in the 4th column (so I probably need a struct; I just don't know how to make it work). Any ideas on how I could fix this? Thanks!
Did not test it, but this should help you get going:
EDIT: corrected wrong indexing into raw vector. Also, depending on the format you might want to restrict also the rows of the raw matrix. From your question, I assume something like selector = raw(3:end,4); and data = raw(3:end,10:19); should be correct.
[~,~,raw] = xlsread('Coordinates','Centerville');
selector = raw(:,4);
data = raw(:,10:19);
[selector,~,grpidx] = unique(selector);
nGrp = numel(selector);
out = cell(nGrp,1);
for i=1:nGrp
idx = grpidx==i;
out{i} = cell2mat(data(idx,:));
end
out is the output variable. The key here is the variable grpidx that is an output of the unique function and allows you to trace back the unique values to their position in the original vector. Note that unique as I used it may change the order of the string values. If that is an issue for you, use the setOrderparameter of the unique function and set it to 'stable'
Is it possible to refer back to/access the names of variables (say nx1 arrays) that make up a matrix? I wish to access them to insert there names into a plot or figure (as a text) that I have created. Here is an example:
A = [supdamp, clgvlv,redamp,extfanstat,htgvlv,occupied,supfanspd]
%lots of code here but not changing A, just using A(:,:)'s
%drawn figure
text(1,1,'supdamp')
...
text(1,n,'supfanspd')
I have failed in an attempt create a string named a with their names in so that I could loop through a(i,1), then use something like text(1,n,'a(i,1)')
Depending on your problem, it might make sense to use structures with dynamical field names.
Especially if your data in the array have some meaning other than just entries of a matrix in linear algebra sense.
# name your variables so that your grandma could understand what they store
A.('supdamp') = supdamp
A.('clgvlv') = clgvlv
...
fieldsOfA = fieldnames(a)
for n = 1 : numel(fieldsOfA )
text(1, n, fieldsOfA{n})
end
temp(i,1) = rand(1)*(pb(1,num).pos(i,1) - pw(1,num).pos(i,1));
This line gives the following error:
Error using ==> minus
Not enough input arguments.
The following are the definitions of pb and pw.
pw=struct('fitness',[],'pos',{});
pb=struct('fitness',[],'pos',{});
pos is a 2 x 1 array.
When tracking down errors like this, I break the problem up into smaller bits. Especially when the logic isn't readily apparent. Not only does it provide a path that can be used to step through your function using the debugger, but it also makes it more readable.
I've taken liberty with the intermediate variable names.
thisPb = pb(1,num);
thisPw = pw(1,num);
initialPos= pw.pos(i,1);
finalPos = pb.pos(i,1);
whos initialPos finalPos
temp(i,1) = rand(1) * (finalPos - initialPos);
The line with whos will print out the values. Make sure that finalPos and initialPos are both numbers.
One way that you can get this error is when num is an empty matrix.
The expression
>> s(x).a
can return a variable number of outputs, depending on the size of x.
If x = [1,2,3] for example, it will return three values (as long as s has at least three elements).
If x = [] on the other hand, then s(x).a will return no outputs, so the expression
>> disp(s(x).a)
will give you a Not enough input arguments error, which is almost certainly what you're seeing. You should check that num is not empty.
Are you sure, that all values are really initialised? Try to check this before your codeline.
disp(pb(1,num).pos(i,1))
disp(pw(1,num).pos(i,1))
temp(i,1) = rand(1)*(pb(1,num).pos(i,1) - pw(1,num).pos(i,1));
I have a function that returns one or more variables, but as it changes (depending on whether the function is successful or not), the following does NOT work:
[resultA, resultB, resultC, resultD, resultE, resultF] = func(somevars);
This will sometimes return an error, varargout{2} not defined, since only the first variable resultA is actually given a value when the function fails. Instead I put all the output in one variable:
output = func(somevars);
However, the variables are defined as properties of a struct, meaning I have to access them with output.A. This is not a problem in itself, but I need to count the number of properties to determine if I got the proper result.
I tried length(output), numel(output) and size(output) to no avail, so if anyone has a clever way of doing this I would be very grateful.
length(fieldnames(output))
There's probably a better way, but I can't think of it.
It looks like Matthews answer is the best for your problem:
nFields = numel(fieldnames(output));
There's one caveat which probably doesn't apply for your situation but may be interesting to know nonetheless: even if a structure field is empty, FIELDNAMES will still return the name of that field. For example:
>> s.a = 5;
>> s.b = [1 2 3];
>> s.c = [];
>> fieldnames(s)
ans =
'a'
'b'
'c'
If you are interested in knowing the number of fields that are not empty, you could use either STRUCTFUN:
nFields = sum(~structfun(#isempty,s));
or a combination of STRUCT2CELL and CELLFUN:
nFields = sum(~cellfun('isempty',struct2cell(s)));
Both of the above return an answer of 2, whereas:
nFields = numel(fieldnames(s));
returns 3.