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806 个投票

.

MATLAB R2022a provides app developers more control over user navigation through app components using the keyboard's Tab key.

Part 1. The new focus function: programmatically set keyboard focus to a UI component

Part 2. Modify focus order of components

Today we'll review Part 1. Come back tomorrow for Part 2.

-------------------------------------------------------------------------------------------------

Programmatically set UI component focus

Did you know that you can save ~2 seconds every time you use a keyboard shortcut rather than reaching for your mouse [1,2]?

I need you to focus here: starting in MATLAB R2022a, use the new focus function to set keyboard focus to a specific UI component.

By specifying the component handle ( c ) in focus(c),

  1. The figure containing the component is displayed
  2. A blue frame appears around the component
  3. The user can directly interact with the component.

.

Which components are focusable?

Focusable components are those that a user can interact with using the keyboard. So an object set to Enable='off' or Visible='off' cannot be in focus. See the documentation for more details.

What will you do with all of that extra time saved?

-------------------------------------------------------------------------------------------------

Stay tuned

Tomorrow we'll learn how to apply the new focus function with control of tab order to create contextual flow of UI component focus. Follow Community Highlights to get notifications.

Let us know what interests you in the new MATLAB R2022a release in the comment section below.

See also

Footnotes

[1] Lane et. al. (2005). International Journal of Human-Computer Interaction, 18(2).

[2] Michels (2018). median.com

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2506 个投票

I saw this problem online recently.

Passenger distribution on a train

Not a terribly difficult problem to solve. But it was mildly interesting to find a solution using MATLAB. Perhaps just as interesting is the post analysis of the problem to understand what is happening, and why any unique solution exists at all for one specific car.

The question is, we have a passenger train with 11 cars in it. Feel free to number them 1 through 11. We know that 381 passengers boarded the train. Every passenger is in one of the cars, but all we know is there are exactly 99 passengers in every set of three consecutive cars. Now the question becomes, how many passengers are in car number 9?

One might say at first this is impossible to know. Surely there are many ways the passengers may be arranged, but is that true? Could it be impossible to solve?

First, before we go any further, a few tests seem important. Logically, we might think to distribute 33 passengers in every car. Would that work? So we would have a passenger distribution of

X1 = repmat(33,11,1)
X1 =
    33
    33
    33
    33
    33
    33
    33
    33
    33
    33
    33

While that satisfies the requirement of every 3 consecutive cars having 99 passengers, it fails the total count requirement, since we can see the sum of all passengers would be 363. This yields too few total passengers, with only a combined load of 363 passengers, and we need 381.

At the other end of the spectrum is another extreme case. We might have this distribution:

X2 = zeros(11,1);
X2(1:3:11) = 99
X2 =
    99
     0
     0
    99
     0
     0
    99
     0
     0
    99
     0

Again, it meets the requirement that the sum of passengers in any 3 consecutiuve cars will be 99. But that case yields too many total passengers at 396. Somewhere in the middle must/might/may be a solution, right? At least it is good to see that we can have more or less than 381 total passengers. But how can we find a solution using MATLAB?

There is one other problem with the X2 attempt at a solution, in that had I chosen a different first car to place the 99 passengers, we need not have a unique result in car number 9.

X3 = zeros(11,1);
X3(2:3:11) = 99;
X4 = zeros(11,1);
X4(3:3:11) = 99;

Each of those schemes would put 99 passengers in every set of 3 consecutive cars.

[X2,X3,X4]
ans =
    99     0     0
     0    99     0
     0     0    99
    99     0     0
     0    99     0
     0     0    99
    99     0     0
     0    99     0
     0     0    99
    99     0     0
     0    99     0

But car number 9 would have very different numbers of passengers, depending on the choice made, either 0 or 99 passengers.

An obvious solution is to look for a code that can solve such a problem for us. INTLINPROG stands out as the perfect tool, as this is a linear problem, with everything being in the form of a sum. The unknowns will be how many passengers are sitting in each car. There are 11 cars. So there are 11 unknowns. The bounds are simple.

lb = zeros(1,11); % There cannot be less than zero passengers in any car.
ub = repmat(99,1,11); % since the sum of any three consecutive cars is 99, we cannot have more than 99 people in any one car.

All of the unknown car counts must be integer. That is, we cannot have a fractional number of people in a car, unless this is part of an Agatha Christie murder mystery.

intcon = 1:11;

What constraints apply? First, the sum of all passengers on the train must be 381.

As well, we know that in every 3 consecutive cars, the sum must be 99. Both constraints will take the form of exact linear equality constraints. We can encode all of that into the matrix Aeq, and the vector Beq.

Aeq = [ones(1,11);triu(tril(ones(9,11),2))]; % A tricky way to create the matrix Aeq
Beq = [381;repmat(99,9,1)];

If X is a potential solution that satisfies the bound constraints, it must satisfy the matrix equation Aeq*X==Beq.

We can see for example, the potential solutions I posed above as X1,...X4, all fail to satisfy the requirement on the total number of passengers, since while the sums for consecutive cars are correct, the total sum is not.

Aeq*[X1,X2,X3,X4]
ans =
   363   396   396   297
    99    99    99    99
    99    99    99    99
    99    99    99    99
    99    99    99    99
    99    99    99    99
    99    99    99    99
    99    99    99    99
    99    99    99    99
    99    99    99    99

Finally, there are no linear inequality constraints.

A = [];
B = [];

At this point, you might be wondering how we can formulate this as a linear programming problem at all. What would be the objective function? What could we hope to minimize? As it turns out, linear programming tools are pretty simple in that respect. We could pose just about any objective we want. For example, this is sufficient:

f = ones(1,11);

Effectively, we are just using intlinprog to see if a FEASIBLE integer solution exists that satisfies all of the bounds, as well as the equality constraints. This is why the objective can be the same as one of the equality constraints. Once INTLINPROG finds any solution, it will be done.

And now we can throw the problem into INTLINPROG, hoping something intelligent falls out.

[X,~,EXITFLAG] = intlinprog(f,intcon,A,B,Aeq,Beq,lb,ub)
P:                Optimal objective value is 381.000000.                                           
Optimal solution found.
Intlinprog stopped at the root node because the objective value is within a gap tolerance of the optimal value,
options.AbsoluteGapTolerance = 0 (the default value). The intcon variables are integer within tolerance,
options.IntegerTolerance = 1e-05 (the default value).
X =
     0
    84
    15
     0
    84
    15
     0
    84
    15
     0
    84
EXITFLAG =
     1

intlinprog has found a solution,

isequal(Aeq*X,Beq)
ans =
  logical
   1

The solution satisfies all of the constraints. Effectively, we see the repeating sub-sequence [0 84 15] in consecutive cars. And of course, as long as we repeat that sequence, it does satisfy all requirements. How many people are sitting in car number 9? 15 people.

Symmetry would suggest that car number 3 must have the same number of people, since we could as easily have numbered the cars starting from either end. And of course, X(3) was also 15.

Thankfully our intuition works there. It would seem we are done now. Or are we?

For some of you, you might be wondering if any other solutions can possibly exist. And some of you might be wondering if any of those solutions can have some other number of passengers than exactly 15 in cars number 3 and 9.

NULL is the MATLAB function to come to the rescue here.

This is essentially a linear algebra question. We wish to know the solutions of the problem Aeq*x==Beq. Here, Aeq is a 10x11 matrix, so it has rank at most 10. That means there is a vector Y, such that Aeq*Y == 0.

Y = double(null(sym(Aeq)))
Y =
    -1
     1
     0
    -1
     1
     0
    -1
     1
     0
    -1
     1

What does that tell us? If we have some particular solution (X) to the non-homogeneous problem Aeq*X==Beq, then the set of all possible solutions will be of the general form

syms t
X + t*Y
ans =
    -t
t + 84
    15
    -t
t + 84
    15
    -t
t + 84
    15
    -t
t + 84

You may see that this generates all solutions to the general problem. We can see a few of them in this array:

[X-1*Y, X - 2*Y, X - 3*Y, X - 84*Y]
ans =
     1     2     3    84
    83    82    81     0
    15    15    15    15
     1     2     3    84
    83    82    81     0
    15    15    15    15
     1     2     3    84
    83    82    81     0
    15    15    15    15
     1     2     3    84
    83    82    81     0

We see now there are 85 such possible integer solutions, all of the form X-k*Y, where k can be any positive integer from 0 to 84 inclusive. INTLINPROG found one of them. But as importantly, do you see that since the elements

Y([3 6 9])
ans =
     0
     0
     0

are all identically zero, that those elements in the solution can never change? Those cars must always contain exactly 15 passengers for all of the constraints to be satisfied. I'll be honest, it is not at all obvious as to why it works out that way, at least not initially in my eyes. That leaves my intuition wanting, just a bit.

How might we analyze this problem in a different way? Perhaps a different approach would yield a more satisfying solution. Suppose we chose a passenger partitioning that is strictly repetitive? For example, choose three non-negative integers u,v,w, such that u+v+w=99.

Now, fill the cars using the sequence

syms u v w
X = [u v w u v w u v w u v];

Surely you would agree that any subset of 3 consecutive cars adds to 99, as long as u+v+w=99. But then the sum of all 11 cars in that sequence must be 4*u+4v+3*w. And this leaves us with now two equations in three unknowns. We have

EQ1 = sum(X) == 381;
EQ2 = u + v + w == 99;

The rest is easy now, as we can do

EQ1 - 3*EQ2
ans =
u + v == 84

So if a solution in this form exists, we can see that u+v=84, and therefore w=99-u-v=15. (Remember that w was the number of passengers in car number 9, but also in cars numbered 3 and 6.) Any combination of non-negative integers that sums to 84 will work for u and v though.

This constructive approach does not insure it is the ONLY solution, since I built it from the sequence in the vector X. Perhaps a solution exists that is not simply repetitive as I created it. In fact, the previous analysis using null told us the whole story.

Starting in MATLAB R2022a, use the append option in exportgraphics to create GIF files from animated axes, figures, or other visualizations.

This basic template contains just two steps:

% 1. Create the initial image file
gifFile = 'myAnimation.gif';
exportgraphics(obj, gifFile);
% 2. Within a loop, append the gif image
for i = 1:20
      %   %   %   %   %   %    % 
      % Update the figure/axes %
      %   %   %   %   %   %    % 
      exportgraphics(obj, gifFile, Append=true);
  end

Note, exportgraphics will not capture UI components such as buttons and knobs and requires constant axis limits.

To create animations of images or more elaborate graphics, learn how to use imwrite to create animated GIFs .

Share your MATLAB animated GIFs in the comments below!

See Also

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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?

You've spent hours designing the perfect figure and now it's time to add it to a presentation or publication but the font sizes in the figure are too small to see for the people in the back of the room or too large for the figure space in the publication. You've got titles, subtitles, axis labels, legends, text objects, and other labels but their handles are inaccessible or scattered between several blocks of code. Making your figure readable no longer requires digging through your code and setting each text object's font size manually.

Starting in MATLAB R2022a, you have full control over a figure's font sizes and font units using the new fontsize function (see release notes ).

Use fontsize() to

  • Set FontSize and FontUnits properties for all text within specified graphics objects
  • Incrementally increase or decrease font sizes
  • Specify a scaling factor to maintain relative font sizes
  • Reset font sizes and font units to their default values . Note that the default font size and units may not be the same as the font sizes/units set directly with your code.

When specifying an object handle or an array of object handles, fontsize affects the font sizes and font units of text within all nested objects.

While you're at it, also check out the new fontname function that allows you to change the font name of objects in a figure!

Give the new fontsize function a test drive using the following demo figure in MATLAB R2022a or later and try the following commands:

% Increase all font sizes within the figure by a factor of 1.5
fontsize(fig, scale=1.5)
% Set all font sizes in the uipanel to 16
fontsize(uip, 16, "pixels")
% Incrementally increase the font sizes of the left two axes (x1.1)
% and incrementally decrease the font size of the legend (x0.9)
fontsize([ax1, ax2], "increase")
fontsize(leg, "decrease")
% Reset the font sizes within the entire figure to default values
fontsize(fig, "default")
% Create fake behavioral data
rng('default')
fy = @(a,x)a*exp(-(((x-8).^2)/(2*3.^2)));
x = 1 : 0.5 : 20;
y = fy(32,x);
ynoise = y+8*rand(size(y))-4;
selectedTrial = 13;
% Plot behavioral data
fig = figure('Units','normalized','Position',[0.1, 0.1, 0.4, 0.5]);
movegui(fig, 'center')
tcl = tiledlayout(fig,2,2); 
ax1 = nexttile(tcl); 
hold(ax1,'on')
h1 = plot(ax1, x, ynoise, 'bo', 'DisplayName', 'Response');
h2 = plot(ax1, x, y, 'r-', 'DisplayName', 'Expected');
grid(ax1, 'on')
title(ax1, 'Behavioral Results')
subtitle(ax1, sprintf('Trial %d', selectedTrial))
xlabel(ax1, 'Time (seconds)','Interpreter','Latex')
ylabel(ax1, 'Responds ($\frac{deg}{sec}$)','Interpreter','Latex')
leg = legend([h1,h2]);
% Plot behavioral error
ax2 = nexttile(tcl,3);
behavioralError = ynoise-y; 
stem(ax2, x, behavioralError)
yline(ax2, mean(behavioralError), 'r--', 'Mean', ...
    'LabelVerticalAlignment','bottom')
grid(ax2, 'on')
title(ax2, 'Behavioral Error')
subtitle(ax2, ax1.Subtitle.String)
xlabel(ax2, ax1.XLabel.String,'Interpreter','Latex')
ylabel(ax2, 'Response - Expected ($\frac{deg}{sec}$)','Interpreter','Latex')
% Simulate spike train data
ntrials = 25; 
nSamplesPerSecond = 3; 
nSeconds = max(x) - min(x); 
nSamples = ceil(nSeconds*nSamplesPerSecond);
xTime = linspace(min(x),max(x), nSamples);
spiketrain = round(fy(1, xTime)+(rand(ntrials,nSamples)-.5));
[trial, sample] = find(spiketrain);
time = xTime(sample);
% Spike raster plot
axTemp = nexttile(tcl, 2, [2,1]);
uip = uipanel(fig, 'Units', axTemp.Units, ...
    'Position', axTemp.Position, ...
    'Title', 'Neural activity', ...
    'BackgroundColor', 'W');
delete(axTemp)
tcl2 = tiledlayout(uip, 3, 1);
pax1 = nexttile(tcl2); 
plot(pax1, time, trial, 'b.', 'MarkerSize', 4)
yline(pax1, selectedTrial-0.5, 'r-', ...
    ['\leftarrow Trial ',num2str(selectedTrial)], ...
    'LabelHorizontalAlignment','right', ...
    'FontSize', 8); 
linkaxes([ax1, ax2, pax1], 'x')
pax1.YLimitMethod = 'tight';
title(pax1, 'Spike train')
xlabel(pax1, ax1.XLabel.String)
ylabel(pax1, 'Trial #')
% Show MRI
pax2 = nexttile(tcl2,2,[2,1]); 
[I, cmap] = imread('mri.tif');
imshow(I,cmap,'Parent',pax2)
hold(pax2, 'on')
th = 0:0.1:2*pi; 
plot(pax2, 7*sin(th)+84, 5*cos(th)+90, 'r-','LineWidth',2)
text(pax2, pax2.XLim(2), pax2.YLim(1), 'ML22a',...
    'FontWeight', 'bold', ...
    'Color','r', ...
    'VerticalAlignment', 'top', ...
    'HorizontalAlignment', 'right', ...
    'BackgroundColor',[1 0.95 0.95])
title(pax2, 'Area of activation')
% Overall figure title
title(tcl, 'Single trial responses')

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Antonello Zito
Antonello Zito
上次活动时间: 2024-12-27,18:49

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.
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6178 个投票
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:
1) from a field of a structure array using dot-indexing:
struct_array.field % all elements
struct_array(idx).field % selected elements
2) from a cell array using curly-braces:
cell_array{:} % all elements
cell_array{idx} % selected elements
3) from a string array using curly-braces:
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)
Nested Structures (why you get an error trying to index into a comma-separated list)
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)
Yes, I'm it.
45%
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55%
6563 个投票

Several major updates have been introduced to Answers’ reputation system! The updates include a new User Levels system, a new Editor indicator, and updated badges series.

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Becoming an Editor (upon earning 3,000 points) is a huge milestone in Answers. Therefore, we introduced the Editor indicator to show our appreciation. From the screenshot below, you will notice a user can have both a User Level and an Editor indicator.

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Based on our analysis of existing badges, we decided to introduce 2 new badges into existing series and retire an entire badge series.

  • The Knowledgeable badge series and the Thankful badge series now have 5 levels.
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You will find more information on Answers help page . If you have any questions, comments or feedback, free feel to leave a comment below.

Cody is a useful tool to practice MATLAB skills not only by solving the problems but also learn from each other’s solutions. Sometimes you see subpar solutions that are cheats and hacks. With the flagging feature we released recently, you can help us identify solutions that administrators, including Community Advisory Board members, can review and delete.

How to flag?

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640 个投票
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Other not mentioned, or mixture
4%
10577 个投票
Ambient | Atmospheric | Nature
9%
Classical | Jazz | Musicals
10%
Electronic | Dubstep | House | VGM
15%
Lo-fi | Chill | Coffee House
22%
Rock | Metal | Pop | Punk | Hip Hop
27%
Other | Podcasts | Nothing
17%
753 个投票
I only use it for homework problems
22%
1
5%
2-10
24%
10-100
28%
More than 100
12%
None yet (just started learning)
9%
6646 个投票
1 (just me)
8%
2-10
9%
11-100
8%
More than 100 (e.g. University)
65%
I can't even guess.
9%
8483 个投票
1 monitor/screen
42%
2
46%
3
9%
4 or more
3%
12173 个投票
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