help on graph (w/o common matrix plotting built-in functions)

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Patrick Laux
Patrick Laux on 20 May 2022
Edited: Voss on 21 May 2022
Dear all,
based on a matrix of zeros and ones, let's say 10 x 365 (10 years a 365 days), I would like to plot a '-' for a one and leave the position blank ' ' for a zero.
Thus, the graph should e.g. look like:
....
------ --- --- ---- ---- ------ ----- --
-------- ------ --------------- -----------
------ -- ---------- --------- ---------
----- --------------- ------ ---------- -
.....
Any hint appreciated.

Accepted Answer

Les Beckham
Les Beckham on 20 May 2022
In that case, maybe something like this will get you close.
A = randi(2,10,365)-1;
x = 1:size(A,2);
y = size(A,1):-1:1;
for i = 1:size(A,1)
A(i,:) = A(i,:)*y(i);
end
A(A==0) = NaN;
for i = 1:size(A,1)
line(x, A(i,:), 'Marker', '_', 'linestyle', 'none', 'MarkerSize', 0.1)
end

More Answers (5)

Les Beckham
Les Beckham on 20 May 2022
A = randi(2,10,365)-1
A = 10×365
1 1 1 1 1 1 1 1 0 1 0 1 1 0 1 1 0 0 0 0 0 0 1 1 0 1 1 0 1 0 1 0 0 1 0 1 1 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 1 1 1 1 0 1 1 0 1 0 1 1 1 1 0 1 1 0 0 0 0 1 1 0 0 0 1 1 1 1 0 0 0 1 0 0 1 0 1 1 1 1 0 1 1 0 0 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 0 0 0 1 1 0 1 0 0 0 1 1 0 1 1 0 0 0 0 1 1 1 0 0 1 1 1 0 0 0 1 0 0 1 0 1 0 1 1 0 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 0 1 1 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 1 0 1 0 1 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 0 0 1 0 1 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 0 0 0 0 0 0 1 0 1 1 0 0 1 1 1 1 1 0 0 0 0 0 1 0 0 1 0 0 1 1 1 0 1 1 1 0 0 1 0 1 0 1 0 0 1 0 0 0 1 1 1 0 1 0 0 1
str = repmat(' ', size(A));
str(A==1) = '-';
disp(str)
-------- - -- -- -- -- - - ------ - --- --- -- - - - - - -- --- - --- - -- - -- -- -- --- ---- -- --- ---- - - - - -- - ------- - - - - - ---- --- - -- --- - -- - --- - -- - - - ----- --- ---- ------- - - - - --- - ------- ---- -- -- - - - -- - -- - - - - -- - ---- -- -- - -- --- - - - -- - - -- ----- - - ---- ---- -- - -- ------- --- - - - - --- ---- - - - - - -- - - ---- - - - ------ ------ - --- --- -- -- - - - -- - - -- --- -- - ----- -- - - -- ---- ---- - -- - - - --- - - -- - - -- - -- - - - - - -- -- - --- -- --- - -- - - - --- - -- -- - ---- -- - ---- -- -- --- --- -- -- - - - ----- ---- - - ---- - - ---- - --- - -- --- -- - --- --- ----- -- -- --- - -- - --- - --- - --- --- ------ -- - -- - - ----- - --- ----- -- - - - ------ ---- - - -- --- ----- - - --- -- -- - - ----- - - - - ----- - - - ----- -- - - - - - ---- -- - ---- --- - ------ - - -- - -- ---- - - ---- --- ---- --- - --- - - - - - -- ------ ---- --- - --- - -- ------ --- -- -- - -- -- - ----- --- - - - --------- - - ---- - --- - - -- ----- --- --- - -------- ---- - - ------- -- - - ---- - - -- - - ---- --- - - - - - - -- - -- -- --- --- - -- - -- ---- - - -- - --- -- - - --- --- - - - - - ----- ----- -- - -- -- --- - -- -- - - - ---- - -- --- - -- - -- -- ----- ---------- - ---- -- -- - -- - - - ---- ----- - --- - -- - ------ - ---- -- - - - - - - -- - --- --- -- ---- - - - - - - - - - - - -- -- --- -- - ---- - --- - ---- - - - -- ---- - --- - - - --- - ----- -- ---- ---- - - ----- -- - - - -- - - --- - -- -- - - - --- -- - - -- - -- - - - -- -- - - -- --- - - - - - - - - ---- - -- - ----- - --- - - - - - -- - -- - --- -- - - - - -- --- --- - -- --- -- -- ---- --- - - - - - - -- - - -- - ----------- - - - ------ - - - --- -- - - - --- -- -- - - --- - - -- -- ---- - ---- - -- - ------- - - - - ---- ----- -- ---- - ---- ---- - -- - - - -- - - - - - - --- -- - --- ---- -- ------- -- -- - - - - - - - --- - -- --- --- - - - ----- - -- - -- - - --- ------- - -- -- - -- --- - - -- --- --- -- --- -- ------ -- - --- --- -- - ---- - - -- ---- -- - --- - -- - - - --- -- --------- -- - --- - - - --- - ---- -- - - --- - - -- ----- - - -- - --- --- - - - - - --- - -- ---- -- --- - -- -- - --- ------- - -- --- -- - ------- - - - - - - - -- ----- - - - - - -- - - - -- - - -- -- ---- -- -- -- -- - --- -- - -- - --- - ------ - - ---- -- -- - ----- - - - -- -- - -- - - ----- - -- - --- - --- - - - - -- -- --- - - - -- - --- --- --- - ---- - - - ---- - ---- -- - - -- - -- -- - -- ---- - - - - - - --- --- - - - - --- - -- -- - --- - --- -- - ---- ------- -- --- --- - - -- - - - - - -- - - ---- -- -- --- -- --- - - - --- - -- - -- --- - --- -- - - ---- - - - --- - - -- ---- --- - -- -- -- -- -- --- --- - - -- -- --- --- ---- -- -- ----- - -- - - -- ----- - -----
  3 Comments
Steven Lord
Steven Lord on 21 May 2022
The size of the index array doesn't matter for linear indexing. What matters are the values in that array.

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Patrick Laux
Patrick Laux on 20 May 2022
excellent already, but actually I thought about having a real figure with xaxis and yaxis. Sorry that I was not very clear here.
  1 Comment
Steven Lord
Steven Lord on 21 May 2022
A = randi([0 1], 5, 5)
A = 5×5
1 0 0 1 1 1 1 1 1 1 1 0 1 0 1 0 1 1 1 0 0 1 0 1 1
spy(A)
If you really want dashes:
figure
spy(A, '_')

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DGM
DGM on 20 May 2022
Edited: DGM on 20 May 2022
In the spirit of ridiculousness, here is my attempt:
% random binary data
data = randi([0 1],10,10)
data =
1 1 0 0 1 1 1 1 0 0
0 0 0 1 0 1 1 1 1 1
0 0 0 1 1 0 0 0 1 0
1 1 1 1 0 1 0 0 1 0
0 1 0 1 0 1 1 0 1 1
0 1 0 1 0 0 1 1 1 0
0 0 1 0 1 0 1 0 0 0
0 0 0 0 0 1 1 0 0 1
1 0 0 0 1 1 0 1 0 1
0 1 1 0 0 1 0 1 0 1
% convert to char data (space and full-width crossbar)
charmap = char([32 196]);
chardata = charmap(data+1)
chardata =
10×10 char array
'ÄÄ ÄÄÄÄ '
' Ä ÄÄÄÄÄ'
' ÄÄ Ä '
'ÄÄÄÄ Ä Ä '
' Ä Ä ÄÄ ÄÄ'
' Ä Ä ÄÄÄ '
' Ä Ä Ä '
' ÄÄ Ä'
'Ä ÄÄ Ä Ä'
' ÄÄ Ä Ä Ä'
% construct an image by flowing char glyphs into a rectangular box
charsize = [8 8]; % depends on the selected font
datasize = size(chardata);
tsize = [NaN datasize(2)*charsize(2)+charsize(2)/2];
outpict = textblock(chardata(:)',tsize,'font','compaq-8x8','tightwidth',true,'hardbreak',true);
% display the image with character-centered tickmarks
xrange = 0.5:datasize(1)+0.5;
yrange = 0.5:datasize(2)+0.5;
imagesc(xrange,yrange,1-outpict)
colormap(gray)
Alternatively, instead of using a full crossbar, you can just use
charmap = ' -';
and get
Bear in mind those are different random data. The point is that the bars don't touch.
This uses a few tools from MIMT, namely textblock() and imstacker().

Patrick Laux
Patrick Laux on 21 May 2022
great answers! Thank you very much, very helpful

Voss
Voss on 21 May 2022
Edited: Voss on 21 May 2022
data = randi([0 1],10,30)
data = 10×30
0 0 1 1 0 1 1 1 1 0 1 0 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 0 0 0 0 1 1 0 1 0 1 1 1 1 1 0 0 1 1 0 0 1 0 1 1 1 1 0 1 0 0 0 1 0 1 1 1 1 0 1 1 0 0 0 0 0 1 1 0 1 1 0 0 1 0 1 0 0 1 0 0 1 1 1 0 1 1 0 1 0 0 1 0 0 0 0 1 1 0 1 0 1 0 0 0 1 1 1 1 1 1 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 1 0 1 1 1 1 0 0 1 0 1 0 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 0 1 1 0 0 1 0 1 1 1 0 0 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 1 1 1 0 1 1 0 1 0 0 0 1 1 0 1 0 1 1 1 0 0 0 0 0 0 0 1 1 0 1 0 1 1 0 0 0 0 0 0 0 1 0 1 0 1 0 1 1 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 0 1 1 0 1 1 0 1 1 1 1 1 0 0 0 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 0 1 0 0 1 1 0 1
imshow(1-data)

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