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I'm curious how the community uses the hold command when creating charts and graphics in MATLAB. In short, hold on sets up the axes to add new objects to the axes while hold off sets up the axes to reset when new objects are added.
When you use hold on do you always follow up with hold off? What's your reasoning on this decision?
Can't wait to discuss this here! I'd love to hear from newbies and experts alike!
Calling all students! New to MATLAB or need helpful resources? Check out our MATLAB GitHub for Students repository! Find MATLAB examples, videos, cheat sheets, and more!
Visit the repository here: MATLAB GitHub for Students
Imagine x is a large vector and you want the smallest 10 elements. How might you do it?
The way we've solved ODEs in MATLAB has been relatively unchanged at the user-level for decades. Indeed, I consider ode45 to be as iconic as backslash! There have been a few new solvers in recent years -- ode78 and ode89 for example -- and various things have gotten much faster but if you learned how to solve ODEs in MATLAB in 1997 then your knowledge is still applicable today.
In R2023b, there's a completely new framework for solving ODEs and I love it! You might argue that I'm contractually obliged to love it since I'm a MathWorker but I can assure you this is the real thing!
I wrote it up in a tutorial style on The MATLAB Blog https://blogs.mathworks.com/matlab/2023/10/03/the-new-solution-framework-for-ordinary-differential-equations-odes-in-matlab-r2023b/
The new interface makes a lot of things a much easier to do. Its also setting us up for a future where we'll be able to do some very cool algorithmic stuff behind the scenes.
Let me know what you think of the new functionality and what you think MathWorks should be doing next in the area of ODEs.
To solve the puzzle, first unscramble each of the words on the left. Then rearrange the letters in the yellow shaded boxes to complete the sentence on the right.
If you enjoyed this puzzle let me know with a like or in the comments below and I'll post more of them. Please don't post your answer, or any hints, and spoil it for those who come across this puzzle after you!! If you want to check your answer, you can messge me your guess through the link on my profile card (click on my name, Rena Berman, above and then on the envelope icon in the top right corner of the profile card that appears).
Thats the task:
Given a square cell array:
x = {'01', '56'; '234', '789'};
return a single character array:
y = '0123456789'
I wrote a code that passes Test 1 and 2 and one that passes Test 3 but I'm searching a condition so that the code for Test 3 runs when the cell array only contains letters and the one for Test 1 and 2 in every other case. Can somebody help me?
This is my code:
y = []
[a,b]=size(x)
%%TEST 3
delimiter=zeros(1,a)
delimiter(end)=1
delimiter=repmat(delimiter,1,b)
delimiter(end)=''
delimiter=string(delimiter)
y=[]
for i=1:a*b
y = string([y x(i)])
end
y=join(y,delimiter)
y=erase(y,'0')
y=regexprep(y,'1',' ')
%%TEST 1+2
for i=1:a*b
y = string([y x(i)])
y=join(y)
end
y=erase(y,' ' )
That's the question: Given four different positive numbers, a, b, c and d, provided in increasing order: a < b < c < d, find if any three of them comprise sides of a right-angled triangle. Return true if they do, otherwise return false .
I wrote this code but it doesn't pass test 7. I don't really understand why it isn't working. Can somebody help me?
function flag = isTherePythagoreanTriple(a, b, c, d)
a2=a^2
b2=b^2
c2=c^2
d2=d^2
format shortG
if a2+b2==c2
flag=true
else if a2+b2==d2
flag=true
else if a2+c2==d2
flag=true
else if c2+b2==d2
flag=true
else flag=false
end
end
end
end
end
That's the question:
The file cars.mat contains a table named cars with variables Model, MPG, Horsepower, Weight, and Acceleration for several classic cars.
Load the MAT-file. Given an integer N, calculate the output variable mpg.
Output mpg should contain the MPG of the top N lightest cars (by Weight) in a column vector.
I wrote this code and the resulting column vector has the right values but it doesn't pass the tests. What's wrong?
function mpg = sort_cars(N)
load cars.mat
sorted=sortrows(cars,4)
mpg = sorted(1:N,2)
end
I recently have found that I am no longer able to give my difficulty rating for questions on Cody after sucessfully completing a question. This is obviously not a big deal, I was just wondering if this was an issue on my end or if there was some change that I was not aware of.
The option to rate does not pop up after solving a problem, and the rating in general does not even show up anymore when answering questions (though it is visible from problem groups).
The MATLAB Answers community is an invaluable resource for all MATLAB users, providing selfless assistance and support. However, with the emergence of AI-based chatbots, like chatGPT, there may be concerns about the future relevance and utility of the MATLAB Answer community. What are your thoughts?
This is the 6th installment of the wish-list and bug report thread.
This topic is the follow on to the first Wish-list for MATLAB Answer sections and second MATLAB Answers Wish-list #2 (and bug reports). The third started out as New design of the forum - grey on white and the fourth and fifth also grew so large they are slow to load and navigate.
Same idea as the previous ones: one wish (or bug report) per answer, so that people can vote their wishes.
What should you post where?
Next Gen threads (#1): features that would break compatibility with previous versions, but would be nice to have
@anyone posting a new thread when the last one gets too large (about 50 answers seems a reasonable limit per thread), please update this list in all last threads. (if you don't have editing privileges, just post a comment asking someone to do the edit)
When solving problems over on Cody, I can almost always view all solutions to a problem after submitting a correct solution of my own. Very rarely, however, this is not the case, and I instead get the following message:
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
You may solve another problem from Community group to unlock all the solutions to this problem.
If this happens, then again, I can almost always rectify this by submitting a (correct) solution to a different problem (I take it that the Community group is the implicit group of all problems on Cody --- is it?). But sometimes that, too, fails.
So my question is, why? What are the criteria that determine when all solutions are, in fact, unlocked?
Simple question: I noticed there's a Modeling & Simulation Challenge Master badge over on Cody, but I can't find the corresponding group. So: where is it? Does it still exist at all?
Error: The server timed out while running and assessing your solution in MATLAB CODY. How do I resolve this? My code is correct. I have run it on PC. But, when i submit in CODY the server throws an error.
I've now seen linear programming questions pop up on Answers recently, with some common failure modes for linprog that people seem not to understand.
One basic failure mode is an infeasible problem. What does this mean, and can it be resolved?
The most common failure mode seems to be a unbounded problem. What does this mean? How can it be avoided/solved/fixed? Is there some direction I can move where the objective obviously grows without bounds towards +/- inf?
Finally, I also see questions where someone wants the tool to produce all possible solutions.
A truly good exposition about linear programming would probably result in a complete course on the subject, and Aswers is limited in how much I can write (plus I'll only have a finite amount of energy to keep writing.) I'll try to answer each sub-question as separate answers, but if someone else would like to offer their own take, feel free to do so as an answer, since it has been many years for me since I learned linear programming.
Just in case, I have my license of MATLAB. I just have this question and I didn't find any information. I wouldn't like to create another account, for this reason I prefer to ask here.
I have submitted a problem in cody some days back. Now it is not showing in my profile. Initially it was accepted and some people submitted the solutions also, however It has been removed after that, are there some guidelines which I am not following?
This is not a question, but a point of discussion for the entire community. I am aware that every 1/2 months this theme comes out, but until this is not fixed it is totally necessary that this comes, indeed, out. And I said "fix" because Mathworks has to understand that a dark theme is not only a visual/aesthetic matter, it is a substantial part of the game. Most of the OS, GUIs, programs are actually in dark mode, and a vast majority of the users makes indeed use of a global dark mode. How much one does like it is personal, but the benefits to power savings and eye health is instead a fact. Mathworks being ignoring this for years is nothing but ridiculous. Of course it is not an easy task, but every minute of committment for it is worthy. And nope, Schemer is not helpful because it does not provide a real fix to this question.
I feel free to suggest something similar to the Spyder's dark theme, which came out like 2 years ago if I remember correctly.
Of course, my point is not being disrespectful (I am instead very respectful to the huge efforts of Mathworks for making this wonderful program run). But, form a user's point of view, the fact that not a single word has so far come out from Mathworks about a dark theme (meaning that for sure we will not see it in a timing of months) requires us to put a strong pressure on this.
Mathworks, please: it's time for a dark theme.
Hi everyone
I am a new of this community and I very interested in this forum and Matlab.I am trying to submit a soultion but as tiltle my code has a built-in function so the test systerm dont reconisie it.It run completely ok on my computer.
This is problem
This is my solution
function [boOut] = BoIfPointInPoly(PolyMatrix,p_test)
%Summary of this function goes here
%{
if we draw a line from test point to a central point of a side of The polygon
then we extend that line to the furthest point of the polygon ensure that
line go through all side of Polygon in 1 direction.I call that line is line_test
Next find number of intersert of line test and all sides w polyxpoly
function
num interset point is odd mean p_test inside
num interset point is even mean p_test outside
this solution go from the concept that if a line go in from a side it has go out
from other side.So if it go in but not go out that mean it start from
inside.
%}
% Detailed explanation goes here
%line from p test throuh central of a side to furthest point of polygon
%find vector
V = ((PolyMatrix(1,:) + PolyMatrix(2,:)) /2) - p_test ;
%draw that vector to furtest point
pend = p_test + V * max(PolyMatrix(:));
%with multi of V and biggest element I assume that line will go all out the
%polygon which ensure out logic will right
line_test = [p_test ; pend];
disp('Our line test\n');
disp(line_test);
%find interst point
p_inter = polyxpoly(PolyMatrix(:,1),PolyMatrix(:,2),line_test(:,1),line_test(:,2));
%find number of interset (row)
[numIntere,trash] = size(p_inter);
disp('Number of interest point:');
disp(numIntere);
%determine in or out
if (rem(numIntere,2) == 0)
boOut = 0;
else
boOut = 1;
end
end
Can anyone has solution.
Introduction
Comma-separated lists are really very simple. You use them all the time. Here is one:
a,b,c,d
That is a comma-separated list containing four variables, the variables a, b, c, and d. Every time you write a list separated by commas then you are writing a comma-separated list. Most commonly you would write a comma-separated list as inputs when calling a function:
fun(a,b,c,d)
or as arguments to the concatenation operator or cell construction operator:
[a,b,c,d]
{a,b,c,d}
or as function outputs:
[a,b,c,d] = fun();
It is very important to understand that in general a comma-separated list is NOT one variable (but it could be). However, sometimes it is useful to create a comma-separated list from one variable (or define one variable from a comma-separated list), and MATLAB has several ways of doing this from various container array types:
struct_array.field % all elements
struct_array(idx).field % selected elements
cell_array{:} % all elements
cell_array{idx} % selected elements
string_array{:} % all elements
string_array{idx} % selected elements
Note that in all cases, the comma-separated list consists of the content of the container array, not subsets (or "slices") of the container array itself (use parentheses to "slice" any array). In other words, they will be equivalent to writing this comma-separated list of the container array content:
content1, content2, content3, .. , contentN
and will return as many content arrays as the original container array has elements (or that you select using indexing, in the requested order). A comma-separated list of one element is just one array, but in general there can be any number of separate arrays in the comma-separated list (zero, one, two, three, four, or more). Here is an example showing that a comma-separated list generated from the content of a cell array is the same as a comma-separated list written explicitly:
>> C = {1,0,Inf};
>> C{:}
ans =
1
ans =
0
ans =
Inf
>> 1,0,Inf
ans =
1
ans =
0
ans =
Inf
How to Use Comma-Separated Lists
Function Inputs: Remember that every time you call a function with multiple input arguments you are using a comma-separated list:
fun(a,b,c,d)
and this is exactly why they are useful: because you can specify the arguments for a function or operator without knowing anything about the arguments (even how many there are). Using the example cell array from above:
>> vertcat(C{:})
ans =
1
0
Inf
which, as we should know by now, is exactly equivalent to writing the same comma-separated list directly into the function call:
>> vertcat(1,0,Inf)
ans =
1
0
Inf
How can we use this? Commonly these are used to generate vectors of values from a structure or cell array, e.g. to concatenate the filenames which are in the output structure of dir:
S = dir(..);
F = {S.name}
which is simply equivalent to
F = {S(1).name, S(2).name, S(3).name, .. , S(end).name}
Or, consider a function with multiple optional input arguments:
opt = {'HeaderLines',2, 'Delimiter',',', 'CollectOutputs',true);
fid = fopen(..);
C = textscan(fid,'%f%f',opt{:});
fclose(fid);
Note how we can pass the optional arguments as a comma-separated list. Remember how a comma-separated list is equivalent to writing var1,var2,var3,..., then the above example is really just this:
C = textscan(fid,'%f%f', 'HeaderLines',2, 'Delimiter',',', 'CollectOutputs',true)
with the added advantage that we can specify all of the optional arguments elsewhere and handle them as one cell array (e.g. as a function input, or at the top of the file). Or we could select which options we want simply by using indexing on that cell array. Note that varargin and varargout can also be useful here.
Function Outputs: In much the same way that the input arguments can be specified, so can an arbitrary number of output arguments. This is commonly used for functions which return a variable number of output arguments, specifically ind2sub and gradient and ndgrid. For example we can easily get all outputs of ndgrid, for any number of inputs (in this example three inputs and three outputs, determined by the number of elements in the cell array):
C = {1:3,4:7,8:9};
[C{:}] = ndgrid(C{:});
which is thus equivalent to:
[C{1},C{2},C{3}] = ndgrid(C{1},C{2},C{3});
Further Topics:
MATLAB documentation:
Click on these links to jump to relevant comments below:
Dynamic Indexing (indexing into arrays with arbitrary numbers of dimensions)
Summary
Just remember that in general a comma-separated list is not one variable (although they can be), and that they are exactly what they say: a list (of arrays) separated with commas. You use them all the time without even realizing it, every time you write this:
fun(a,b,c,d)
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