plot bar graph based on element type in matrix

I have a 20x100 (t,N) matrix with each element being either 1, 2, or 3. I want to create a bar graph showing the amount of each type of element. so at t=20, of the 100 columns, how many have 1, how many have 2, how many have 3.

Is that possible?

Answer 1

[Star Strider]

Do you want all of them, or just the last row (t=10)?

This does both —

A = randi(3, 20, 100)

A = 20×100

2 2 1 3 1 3 3 3 1 1 1 2 2 2 1 3 3 1 1 1 2 3 3 1 2 1 2 1 2 1 2 2 3 3 1 1 3 3 3 2 3 1 1 3 2 1 1 1 2 2 1 2 1 1 3 2 3 3 3 2 3 2 3 2 2 1 3 1 2 3 3 3 3 2 2 2 1 1 2 3 2 2 3 1 1 1 1 1 1 1 2 1 3 3 3 1 1 1 3 1 3 1 2 1 1 1 1 1 1 3 1 2 2 3 3 1 3 1 1 1 2 3 3 1 3 2 1 3 1 2 3 1 2 2 1 1 1 1 2 3 1 1 1 3 2 3 1 2 3 3 1 2 2 2 2 2 1 3 2 3 3 3 3 2 2 2 3 3 2 2 3 1 2 1 1 2 3 1 1 2 3 1 2 3 2 2 3 2 3 2 2 1 1 2 3 1 3 3 1 3 2 1 3 1 1 1 3 3 2 1 1 2 2 3 1 3 1 1 3 2 3 3 3 1 3 2 1 3 3 1 3 2 2 3 3 2 2 3 1 3 1 3 1 2 2 3 2 1 3 1 2 2 2 1 1 2 2 3 2 1 1 3 3 3 1 2 3 1 3 1 1 1 3 1 2 1 1 3 2 2 3 3 2 3 2 3 1 2 1 3 1 2 1 2 1 2 1 1 2 3

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TallyAll = accumarray(A(:), 1)

TallyAll = 3×1

675 645 680

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ResultAll = table(TallyAll, 'RowNames',compose('%d',1:3))

ResultAll = 3x1 table
         TallyAll
         ________

    1      675   
    2      645   
    3      680   

figure
bar(1:3, TallyAll)

Tally20 = accumarray(A(20,:).', 1)

Tally20 = 3×1

35 30 35

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Result20 = table(Tally20, 'RowNames',compose('%d',1:3))

Result20 = 3x1 table
         Tally20
         _______

    1      35   
    2      30   
    3      35   

figure
bar(1:3, Tally20)

EDIT — Forgot about the bar plots. Now added.

.

[Kitt]

In the end I'd like to see all of them, separated. 1, 2, and 3 all represent a decision an individual (N) can make, and I want to see the change in distribution of decisions over time (t). So like at time t = 1, 30 individuals chose 1, 40 individuals chose 2, and 30 individuals chose 3. And then do that for all 20 timesteps.

[Star Strider]

O.K. That was not initially obvious.

Two options —

A = randi(3, 20, 100)

A = 20×100

3 1 3 1 2 1 2 3 3 3 2 1 1 2 2 1 2 2 3 2 1 3 2 3 2 2 2 3 2 2 3 1 2 3 2 1 2 3 1 3 2 3 2 2 3 2 2 3 1 2 3 1 2 2 3 3 1 1 1 3 1 3 2 2 2 2 1 1 3 1 2 1 2 2 1 2 1 2 2 2 2 2 3 1 3 1 3 3 1 2 1 1 1 3 3 3 1 2 3 1 3 1 1 1 1 3 1 3 1 3 1 3 1 3 1 3 1 1 2 3 2 3 3 3 1 1 1 3 1 2 2 3 1 2 2 1 1 2 1 2 2 2 2 2 2 2 2 2 1 2 1 1 2 1 2 3 3 2 2 3 3 3 1 2 2 2 1 2 3 2 3 1 1 2 3 2 2 3 3 2 2 3 3 3 1 1 2 2 3 3 3 1 1 2 2 1 2 1 1 1 1 1 2 3 1 1 2 2 3 1 2 1 3 2 3 3 3 2 1 3 2 3 2 2 3 1 2 3 2 1 3 2 2 2 3 2 3 3 2 2 2 1 2 1 1 3 3 3 1 3 3 3 1 2 1 1 3 2 3 1 3 2 1 3 3 1 1 1 3 3 1 2 3 3 1 3 3 2 1 3 3 3 2 3 3 1 2 1 3 1 2 3 1 2 3 2 2 1 1 1

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for k = 1:size(A,1)
    Tally(:,k) = accumarray(A(k,:).', 1);
end

figure
tiledlayout(5,4, TileIndexing='columnmajor')
for k = 1:size(Tally,2)
    nexttile
    bar(1:3, Tally(:,k))
    grid
    title("Row "+k)
end

figure
bar3(1:3, Tally)
xlabel('Rows')
ylabel('Responses')

The first option presents each result in a separate plot, the second option presents all of them in a single 3D plot.

.

[Kitt]

Apologies for not making that more clear in the original message. Thank you so much!

[Star Strider]

As always, my pleasure!

No worries!

[Kitt]

@Star Strider

Sorry to call on you again, but I keep getting an error

"Unable to perform assignment because the size of the left side is 2-by-1 and the size of the right side is

3-by-1.

Tally(:,k) = accumarray(optchosen(k,:).', 1);"

I have k as 1:size(optchosen,1)

I'm not sure how I'm getting this error.

It might be because I realized the last timestep is actually all 0, so elements can be either 0, 1, 2, or 3. Though I don't see how that influences this first line of code.

[Star Strider]

No worries.

Your ‘optchosen’ variable needs to be a (20x100) matrix with all three choices present in each row to work wiith my original code. There may be only two choices in some rows, and that would definitely not work with my code, since it depends on every row having all three choices.

One possibiility is to preallocate the ‘Tally’ matrix, and additionally be certain that the choices are allocated correctly in each column —

A = randi(3, 20, 100)

A = 20×100

2 2 1 1 3 3 3 2 2 1 2 1 1 3 3 3 3 3 2 3 3 3 1 1 2 1 3 1 2 3 3 2 1 2 2 3 1 3 1 1 1 1 2 1 3 2 3 1 1 2 3 3 1 3 2 1 2 1 3 1 2 1 1 1 3 2 3 2 3 3 1 2 1 1 3 2 3 3 1 3 2 3 2 3 2 2 1 2 3 2 3 3 1 3 3 1 1 2 3 1 3 2 1 2 3 3 1 1 2 1 2 3 2 2 3 1 2 3 1 3 1 2 3 1 3 1 2 2 3 2 1 3 3 3 3 3 1 2 1 2 3 3 2 1 3 3 2 1 2 2 2 1 1 2 1 3 1 3 2 1 1 3 2 3 1 2 2 3 2 2 2 2 1 2 3 1 3 2 2 3 2 1 1 3 2 1 1 2 1 2 3 3 2 2 1 1 1 2 3 3 3 3 2 3 2 1 1 2 2 3 2 2 1 3 2 3 3 3 1 1 2 1 2 1 3 1 3 1 3 1 1 1 2 2 1 1 3 2 3 3 1 1 3 1 3 1 3 3 2 1 3 2 2 3 3 2 2 1 2 3 3 3 3 2 3 3 2 2 3 3 3 1 3 1 1 1 3 3 3 3 2 3 3 2 3 3 3 3 3 3 1 3 3 1 3 2 1 2 3 3

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A(5,:) = randi(2, 1, 100);                                  % Test #1 (No 3’s)
A(10,:) = randi([2 3], 1, 100);                             % Test #2 (No 1’s)
A(15,A(15,:) == 2) = 3;                                     % Test #3 (No 2’s)

Tally = zeros(3, size(A,1));
for k = 1:size(A,1)
    Uchoice = unique(A(k,:));
    Count = accumarray(A(k,:).', 1);                        % Raw Count Veector
    Tally(Uchoice,k) = Count(Count ~= 0);                   % Allocate ‘Tally’ Results To Non-Zero Rows
end

figure
tiledlayout(5,4, TileIndexing='columnmajor')
for k = 1:size(Tally,2)
    nexttile
    bar(1:3, Tally(:,k))
    grid
    title("Row "+k)
end

figure
bar3(1:3, Tally)
xlabel('Rows')
ylabel('Responses')

That should allocate the choices correctly, and be certain that every column has three rows, even if some contain zero counts. (This worked when I tested it.)

.

[Kitt]

That worked PERFECTLY haha!!

Thank you so much!

[Star Strider]

As always, my pleasure!

Answer 2

[dpb]

M=randi([1 3],20,100);
whos t
[min(M(:)) max(M(:))]

ans = 1×2

1 3

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histogram(M(20,:))
xticks(1:3)
xlabel('Bin'), ylabel('Count')
title('Counts for t=20')

[Kitt]

I've tried the histogram but the problem is I want to see the distribution over time, and when I try to plot multiple histograms they are just on top of each other and I can't really see the change