I have data that I want to create a barchart from. I require the x-axis to be log10.
[counts,xb]=hist(data(:,3),nbins); %IMHIST ONLY HANDLES 8 & 16 BIT IMAGES, NOT 12BIT
bar(log10(xb),counts,'r','EdgeColor','r');
grid on;
hold on
xlim([min(log10(xb)) max(log10(xb))])
I have read that I need to do:
set(gca,'Xtick',0:4); %// adjust manually; values in log scale
set(gca,'Xticklabel',10.^get(gca,'Xtick')); %// use labels with linear values
My data is typically 4000-6000
the histogram with log10 x axis looks like the same, and the ticks are not showing?

댓글 수: 8

dpb
dpb 2017년 2월 16일
Just using log10(xb) doesn't change anything excepting the scaling; you've already binned the values from the linear range; hence the shape is going to be the same. Would you not want to bin log10(x) instead?
Jason
Jason 2017년 2월 16일
편집: Jason 2017년 2월 16일
Hi, for my normal histogram I just plot:
bar(xb,counts,'b','EdgeColor','b');
dpb
dpb 2017년 2월 16일
Well, yes? The point is that's what you built by the hist call you made above; the binning is in linear scaling so the two are going to look the same.
If you mean that but still want to display x on a log scale, just
hBar=bar(xb,counts,'r','EdgeColor','r');
set(gca,'XScale','log')
then you can fixup range as want to make it look pretty if the autoscaling doesn't suit.
Jason
Jason 2017년 2월 16일
Unfortunately this doesn't work.
Star Strider
Star Strider 2017년 2월 16일
Note: Original data vector is now attached to the original Question.
dpb
dpb 2017년 2월 16일
For what specific definition of "doesn't"? See attached that pretty-much proves it does...
The above were generated by
>> subplot(2,1,1)
>> bar(bin,cnt),set(gca,'xscale','log')
>> xlim([2000 8000])
>> title('Semilog x')
>> subplot(2,1,2)
>> bar(bin,cnt),xlim([2000 8000])
>> title('Linear x')
If you use xlim([5000 8000]) the plots are superficially the same because the range of x-values is so limited the difference in scaling between the log and linear axes isn't much at all. Widening the scale as I did show how the log axis is foreshortened in the upper half significantly, IF there's a wide-enough range for the log to make any difference.
Of course, you still have the problem that the binning was done on the linear values first both ways so the actual counts/bin number are the same.
As noted, binning log(x) instead of x will change those counts some but again with such a narrow data range it won't make much difference either way.
IF (the proverbial "big if") the data range covered a couple of decades or more, then the difference between the two would be much more obvious and using the log would have some sense...in this case there just isn't enough variation to really matter much, either way within the range itself; only if you look at the range within a much larger scale.
Jason
Jason 2017년 2월 16일
Thankyou dpb. I see now that my spread isnt big enough to see as you suggested. I take back my comment that it doesnt work. Thanks again
dpb
dpb 2017년 2월 16일
편집: dpb 2017년 2월 22일
For confirmation
>> x=(5000:8000); lx=log10(x);
>> [(lx-lx(1))./(lx(end)-lx(1));(x-x(1))./(x(end)-x(1))]
ans =
0 0.3879 0.7159 1.0000
0 0.3333 0.6667 1.0000
>>
pretty same ratios; if don't study carefully wouldn't ever tell the difference...
I moved comment to Answer; please ACCEPT to close the subject if nothing else.

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dpb
dpb 2017년 2월 16일

3 개 추천

If you mean to still bin on linear range want to display x on a log scale, just use
hBar=bar(xb,counts,'r','EdgeColor','r');
set(gca,'XScale','log')
then you can fixup range as want to make it look pretty if the autoscaling doesn't suit.

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