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Log scale graphic with negative value

Asked by Camil Hamel on 22 Feb 2011
Latest activity Answered by Kilbock Lee on 11 Nov 2013

I want to plot a graphic with data that varies over many order (from 10e-4 to 10e-8) with some positive and negative values. Using 'semilog' or 'set xaxis log' can't plot my negative values (witch is normal). When I plot with a normal scale, we don't really see the variability of my datas, we only see the very high values and all the small one are not very visible because there are to close to zero. What would you suggest me to be able to show all my data (even the negative values) and be able to show the variability (like on a log-scale graph). Thank you

0 Comments

Camil Hamel

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7 Answers

Answer by Matt Tearle on 24 Feb 2011
Accepted answer

Oookay, I think I have it. If you're going to do this a lot, you may want to make yourself your own function. So, see if this does it for you:

function negsemilogx(x,y)
% Do log10 but keep sign
xlog = sign(x).*log10(abs(x));
% Just to get axis limits
plot(xlog,y,'o')
% Get limits
lims = xlim;
wdth = diff(lims);
% Wrap negative data around to positive side
xlog(xlog<0) = xlog(xlog<0) + wdth;
% Plot
plot(xlog,y,'o')
% Mess with ticks
tck = get(gca,'XTick')';
% Shift those that were wrapped from negative to positive (above) back 
% to their original values
tck(tck>lims(2)) = tck(tck>lims(2)) - wdth;
% Convert to string, then remove any midpoint
tcklbl = num2str(tck);
tcklbl(tck==lims(2),:) = ' ';
% Update tick labels
set(gca,'XTickLabel',tcklbl)

Then to try it out:

>> x = -100:100;
>> y = x.^3;
>> negsemilogx(x,y)

3 Comments

Camil Hamel on 24 Feb 2011

Excellent!
I had to work a little to get exactly what I wanted, but you put me on the right track.
Thanks a lot

Matt Tearle on 25 Feb 2011

Out of curiosity, what was still missing/wrong?

Camil Hamel on 25 Feb 2011

I changed the wdth selection by making it wdth=abs(min(negative data))+max(positive data). I put my zero at wdth/2. My graph look more smooth, there is not a so big area without data in the middle. I also customize all the tick part of the code making it automatic and adapted to my data. But I really just added a little to what you suggested me.

Matt Tearle
Answer by Matt Tearle on 22 Feb 2011

Richard pretty much beat me to it. Here's a slightly more automated way to get the tick marks. However, unfortunately there's no TeX markup for tick labels, so you'll have to choose how you want them to appear -- plain exponent values (2, 3, 4, etc), values written out in full (100, 1000, etc), or some scientific notation (but without TeX markup, eg "10^3" or "1e3")

% Make some fake data
n = 1:20;
y = factorial(n);
y(2:2:end) = -y(2:2:end);
% Bad plot
plot(n,y,'o')
% Transform
ylog = sign(y).*log10(abs(y));
figure
plot(n,ylog,'o')
% Do nothing else to get just exponents.  Otherwise:
yt = get(gca,'YTick')';
set(gca,'YTickLabel',num2str(sign(yt).*10.^abs(yt)))
% Or, for scientific notation
set(gca,'YTickLabel',strcat('10^',cellstr(num2str(abs(yt)))))

0 Comments

Matt Tearle
Answer by Matt Tearle on 22 Feb 2011

Dumb question, but is there any reason you can't just look at the absolute values? ( abs in MATLAB)

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Matt Tearle
Answer by Camil Hamel on 22 Feb 2011

I can't look at the absolute value because I have to show and upward (positive) or downward (negative) flux. I tried with the absolute value and plotting with a different colour for negative value. The problem was that my mean data are between a 2.5% and a 97,5% that are sometime positive sometime negative. By taking the absolute value, my mean line values where not between the 2,5% and 97,5% lines because sometime the 2,5% was a negative value (large value). This value in absolute value was higher than the 97,5% witch was positive (small value).

0 Comments

Camil Hamel
Answer by Richard on 22 Feb 2011

When you choose "log" as the scale for an axis, Matlab either chooses to use log10(x) or -log10(-x) as the transform, depending on whether your data is in the positive or negative domain. As you've found, you can't have both at the same time because both halves map data into the [-inf inf] space.

You can play with your own transforms by transforming the data yourself and at the same time manipulating the XTickLabel property of the axes so that you show untransformed data values. For example, this splices the normal negative and positive transforms, (with the singularity at 0 untransformed):

x = -100:100;
y = x.^3;
xt = x;
xt(xt>0) = log10(1+xt(xt>0));
xt(xt<0) = -log10(1-xt(xt<0));
figure;
ax = axes;
plot(xt,y);
set(ax, 'XTick', [-2 -1 0 1 2], ...
    'XTickLabel', {'-100', '-10', '0', '10', '100'});

0 Comments

Richard
Answer by Camil Hamel on 23 Feb 2011
The figure I have now
         x      |
            x   |
          x     |
        x       |
                |       x                  
                |    x    
                |      x
                |        x
        |    |  |   |    |
       -8   -2  0   2    8
meaning
  10^-8  10^-2  0  -10^-2   -10^-8
  positive data |negative data
  __________________________
  The figure I would like to have
               | x
               |   x
               |  x
               |x
              x|                  
            x  |  
          x    |
            x  |           
         |   | | |  |
         2   8  -8 -2
meaning
-10^-2 -10^-8  0 10^-8   10^-2
  negative data|positive data

Thank you for the answers. I did as Matt said giving the first figure.

I hope you understand the figure I made with my attempt at drawing.

All my data in between -10^-8 , -10^-2 and 10^-2 , 10^-8. So I have to kind of get the value of the zero outside and having my maximum value (10^-8) in the centre of the figure. After that I could change the xticklabel as I want it.

I want to get a figure looking like the second one. Would that be possible and how?

Thank you very much

0 Comments

Camil Hamel
Answer by Kilbock Lee on 11 Nov 2013

I think my way is easier using logical operation true false.

Assuming the matrix is P(x,y),

LogP=((P > 0).*log10(P))+(-(P<0).*log10(-P))

and then, plot LogP

First term converts positive values into log scale and Second term does the negative values

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Kilbock Lee

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