plot bar graph based on element type in matrix
显示 更早的评论
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?
采纳的回答
Do you want all of them, or just the last row (t=10)?
This does both —
A = randi(3, 20, 100)
TallyAll = accumarray(A(:), 1)
ResultAll = table(TallyAll, 'RowNames',compose('%d',1:3))
figure
bar(1:3, TallyAll)
Tally20 = accumarray(A(20,:).', 1)
Result20 = table(Tally20, 'RowNames',compose('%d',1:3))
figure
bar(1:3, Tally20)
.
8 个评论
Kitt
2024-10-7
O.K. That was not initially obvious.
Two options —
A = randi(3, 20, 100)
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
2024-10-7
Star Strider
2024-10-7
As always, my pleasure!
No worries!
Kitt
2024-10-7
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(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
2024-10-7
Star Strider
2024-10-7
As always, my pleasure!
更多回答（1 个）
M=randi([1 3],20,100);
whos t
[min(M(:)) max(M(:))]
histogram(M(20,:))
xticks(1:3)
xlabel('Bin'), ylabel('Count')
title('Counts for t=20')
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