Hi all I am working on checking for certain cases in a large loop, with in this large loop is this statement:
%Center Dialouge
timers_c = [start_dpc_sh_hi_timer_c, start_dpc_sh_lo_timer_c, start_ffm_sh_act_hi_timer_c,...
start_ffm_sh_act_lo_timer_c, start_hpot_eff_loss_timer_c, start_lfpt_dpsh_timer_c,...
start_pc_sh_time_c, start_post_tl_nl_c, start_pre_tl_nl_c, start_elec_lu_timer_c];
if perf_case_c ~= -1
for k = 1:10
if iscellstr(timers_c(k)) == 1
perf_case_timer_c = timers_c{k};
timer_set_c = 1;
end
end
end
To I identify the start time and the case type to be output in a dialogue message:
if timer_set_c == 1
pcase_c = msgbox([sprintf('%s'),perf_case_c,sprintf('\nMET:%s\n'),perf_case_timer_c],'PERFORMANCE CASE');
end
I can't get the if statement that determines which case has been determined to work. I am trying to use the change from -1 to a string somehow, but it doesn't quite work.
A minimal example would be
%Assume all timers initialized as zero
%Assume case ~= -1 (is a char array, string
timers_c = [timer1, timer2, timer3 ]
if case ~= -1
for k = 1:3
if iscellstr(timer_c(k)) ==1
case_time = timers(c{k});
time_set_flag = 1
end
end
end
...
and then outside of the loop would be the above msgbox
Related
I've been experimenting with a Sudoku solver using recursion. The problem I've encountered it that the recursive function after finding the right solution will not terminate, but goes on until every number is tested at every position.
How do I terminate such loop in Matlab? Error condition 'error('...') within the function can break the execution, but is by no means a good solution to my problem.
Here is a sample code for similar recursion, producing all possible 2 element vectors for numbers from 1 to 4. I would like it to stop when both numbers are equal to 2.
%possible moves at each position
moveMat = zeros([1,2])+3;
lineInput = zeros([1,2]);
%start recursion
recurNumbers(moveMat, 0, lineInput)
function recurNumbers(moveMat, position, lineVariable)
position = position + 1;
%if all numbers are equal to 2 then try to exit the function
if ~all(lineVariable == 2)
%if all numbers are not equal to 2, try other combination
if position < length(lineVariable)+1
for move = 0 : moveMat(position)
moveMat(position) = move;
lineVariable(position) = lineVariable(position) + 1;
recurNumbers(moveMat,position,lineVariable)
disp(lineVariable)
end
end
else
disp(lineVariable)
return
end
end
Now it will print the vector '[2 2]' twice, which indicates that it recognised the condition, but the 'return' will not do what I imagined it to do.
Although it is not very clear to me what you are trying to achieve, I assume that the following function meets your stopping criterion.
function exit_fl = recurNumbers(moveMat, position, lineVariable)
exit_fl = 0;
if (all(lineVariable == 2))
% Show the result and exit
disp(lineVariable)
exit_fl = 1;
else
position = position + 1;
%if all numbers are not equal to 2, try other combination
if position < length(lineVariable)+1
for move = 0 : moveMat(position)
moveMat(position) = move;
lineVariable(position) = lineVariable(position) + 1;
% Receive the exit status of your function
ex_fl = recurNumbers(moveMat,position,lineVariable);
if (ex_fl == 1)
% If the criterion was met, then stop
exit_fl = 1;
return
end
end
end
end
end
For this loop in MATLAB, after the 'if - end' I want to return to the same loop without executing next i. More specifically, I want to tell MATLAB to check until check(i) is different from 0.
for i = 1:length(numDate)
check(i)=any(Dates == numDate(i));
if check(i) == 0
numDate(i) = numDate(i)-1;
end
end
You cannot change the number of iterations of a for loop once it is decided. Use while loop for such a case.
k=1; %I replaced the loop variable with k because i (and j) are reserved for imag no.s
while k<=length(numDate)
if any(Dates == numDate(k)) == 0
numDate(k) = numDate(k)-1;
else k=k+1; %increment only if the condition is not satisfied
end
end
Use break
for i = 1:length(numDate)
check(i)=any(Dates == numDate(i));
if check(i) == 0
numDate(i) = numDate(i)-1;
else
break
end
end
I am new to matlab and I have couple of questions about it. First one, "Your function should terminate the sequence when either the value of ... or..." I use || in the code but it does not work as expected while && gives me the correct answer. Second question, how could the code be to display only the final answer?
Problem: calculate X which is represented by the sequence below
X = 1 - 1/2^2 + 1/3^2 - 1/4^2 +....
Requirement: Your function should terminate the sequence when either the value of 1/k^2 is less than 0.0001 or k is equal to k_max.
input k
Initialize x = 0
for loop i from 1 to k
if 1/i^2<0.0001 && i >= 100
break
end
Calculate X = (-1)^(i+1)./i^2 + X
end
You can use the break function as follows, where END_CONDITION is the condition you want to end your loop in.
if END_CONDITION
disp(X);
break;
end
To display the final answer, you can use the disp function. Eg. if your variable you want to print is called A then you use the following code.
disp(A)
Collectively this is your code. Since k_max terminates at the end of the for loop, we don't have to add any conditions to break out of the loop.
X = 0;
for i = 1:k
if 1/i^2<0.0001 || i==100
break;
end
X = (-1)^(i+1)./i^2 + X;
end
disp(X);
I have a problem at the beginning of my function. The function is to combine several data column from some objects. Error happens at the beginning of function. It says as follows:
Error in find_by_coor (line 2)
for i = 1:length(obj_ac)
Here is only the declaration of variable and loop, but Matlab somehow returned error. I have no idea so would like someone to help me. I attached my code as follows. Thanks a lot in advance.
function arr = find_by_coor(obj_ac,obj_gps,obj_sen_dir,lat1,long1,lat2,long2)
for i = 1:length(obj_ac)
if eq(obj_sen_dir(i).sensor,4) && strcmp(obj_sen_dir(i).direction,'outbound')
ind = obj_gps(i).save_var_gps(:,1)>lat1;
if isempty(find(ind)) == 1
continue
end
temp = obj_gps(i).save_var_gps(ind,:);
ind = temp(:,1)<lat2;
if isempty(find(ind)) == 1
continue
end
temp2 = temp(ind,:);
ind = temp2(:,2)<long1;
if isempty(find(ind)) == 1
continue
end
temp3 = temp2(ind,:);
ind = temp3(:,2)>long2;
if isempty(find(ind)) == 1
continue
end
temp4 = temp3(ind,:);
mint = min(temp4(:,5))-min(obj_gps(i).save_var_gps(:,5));
maxt = max(temp4(:,5))-min(obj_gps(i).save_var_gps(:,5));
if isempty(mint) == 1 || isempty(maxt) == 1
continue
end
if floor(mint*(1.6516e+03)) == 0 || floor(maxt*(1.6516e+03)) == 0
continue
end
temp5 = obj_ac(i).save_var(floor(mint*(1.6516e+03)):floor(maxt*(1.6516e+03)));
temp6 = abs(fft(temp5));
arr(i,:) = [i objs(i).daten var(temp5) max(temp5) min(temp5) mean(temp5) std(temp5) mode(temp5) var(temp6) max(temp6) min(temp6) mean(temp6) std(temp6) mode(temp6)];
disp(i);
end
end
end
The problem is that when you run the function, the output variable arr is never assigned. In Matlab you must always assign a function output if you choose to have it in the definition. For example
function [a,b] = setAB()
err = 0; % Gives an error if err is true
a = 1;
if ~err
b = 1;
end
The reason is most certainly that for some inputs, all values fall into one of the if statements and you do never reach the point where arr is assigned. A good solution for this is to assign a default value for arr in the beginning. That may for example be nan or -1 or, in your case maybe an array arr = nan(wanted size) or arr = -1*ones(wanted size). If you do not preallocate arr you will likely get a "matrix out of bounds" error instead, should you solve the first issue.
It does not mean that you always need to have an output though.
function [] = noOutput()
disp('Hi, I am a void!');
You may also choose to return as many values as number of outputs.
function varargout = variableArgs()
a = 1;
b = 2;
c = 3;
if (nargout == 1)
varargout{1} = a;
elseif (nargout == 2)
varargout{1} = b;
varargout{2} = c;
else
error('Wrong number of output arguments!');
end
I am not saying which of the approaches you should use or that any of them are good. Normally I use varargout in case I write plotting functions. Then I may want to return nothing in case I do not have an output argument. Then I want to return handles or any extra information. Further as you may have understood there is also a varargin that may be of more use.
i want to clarify how to control returning of variables from function in matlab,for exmaple let us consider this code
function [x y z]=percentage(a)
n=length(a);
maximum=0;
minimum=0;
subst=0;
minus=0;
plus=0;
minus_perc=0;
plus_perc=0;
for i=1:1:n
if a(i)>0
plus=plus+1;
else
minus=minus+1;
end
end
minuc_perc=minus/n;
plus_perc=plus/n;
maximum=max(minus_perc,plus_perc);
minimum=min(minus_perc,plus_perc);
subst=maximum-minimum;
x=plus_perc;
y=minus_perc;
z=subst*100;
if plus_perc>minus_perc
disp('among the successful people,relevant propession was choosen by');
disp(z)
disp('% people');
else
disp('among the successful people,irrelevant propession was choosen by');
disp(z);
disp('% people');
end
end
what i want to return is plus_proc,min_proc and subst,but when i run following command,get result like this
[c d e]=percentage(a)
among the successful people,relevant propession was choosen by
58.3333
% people
c =
0.5833
d =
0
e =
58.3333
so i think something is wrong,array is like this
a =
1 -1 1 1 -1 1 -1 -1 1 1 1 -1
so ones again,i want to return plus_proc,minus_proc,and subst
To return a variable in matlab you just assign into one of the specified return parameters. For example: to return the number five I would use:
function [foo] = gimmeFive()
foo = 5;
end
Your code is not giving you the right answer because you have a typo:
minuc_perc=minus/n;
should be
minus_perc=minus/n;
You could greatly simplify the function by taking advantage of the find function, like so:
Find the indeces of any element of a > 0, count them.
plus = length(find(a > 0));
plus_perc = plus ./ length(a);
Or if you want to cut even more out:
a > 0 gives us a vector of 0 and 1, so sum up the 1's
plus = sum(a > 0);
plus_perc = plus ./ length(a);