Histogram using gnuplot?

Question

I know how to create a histogram (just use "with boxes") in gnuplot if my .dat file already has properly binned data. Is there a way to take a list of numbers and have gnuplot provide a histogram based on ranges and bin sizes the user provides?

Answer 1

I have found this discussion extremely useful, but I have experienced some "rounding off" problems.

More precisely, using a binwidth of 0.05, I have noticed that, with the techniques presented here above, data points which read 0.1 and 0.15 fall in the same bin. This (obviously unwanted behaviour) is most likely due to the "floor" function.

Hereafter is my small contribution to try to circumvent this.

bin(x,width,n)=x<=n*width? width*(n-1) + 0.5*binwidth:bin(x,width,n+1)
binwidth = 0.05
set boxwidth binwidth
plot "data.dat" u (bin($1,binwidth,1)):(1.0) smooth freq with boxes

This recursive method is for x >=0; one could generalise this with more conditional statements to obtain something even more general.

Answer 2

Do you want to plot a graph like this one? yes? Then you can have a look at my blog article: http://gnuplot-surprising.blogspot.com/2011/09/statistic-analysis-and-histogram.html

Key lines from the code:

n=100 #number of intervals
max=3. #max value
min=-3. #min value
width=(max-min)/n #interval width
#function used to map a value to the intervals
hist(x,width)=width*floor(x/width)+width/2.0
set boxwidth width*0.9
set style fill solid 0.5 # fill style

#count and plot
plot "data.dat" u (hist($1,width)):(1.0) smooth freq w boxes lc rgb"green" notitle

Answer 3

Be very careful: all of the answers on this page are implicitly taking the decision of where the binning starts - the left-hand edge of the left-most bin, if you like - out of the user's hands. If the user is combining any of these functions for binning data with his/her own decision about where binning starts (as is done on the blog which is linked to above) the functions above are all incorrect. With an arbitrary starting point for binning 'Min', the correct function is:

bin(x) = width*(floor((x-Min)/width)+0.5) + Min

You can see why this is correct sequentially (it helps to draw a few bins and a point somewhere in one of them). Subtract Min from your data point to see how far into the binning range it is. Then divide by binwidth so that you're effectively working in units of 'bins'. Then 'floor' the result to go to the left-hand edge of that bin, add 0.5 to go to the middle of the bin, multiply by the width so that you're no longer working in units of bins but in an absolute scale again, then finally add back on the Min offset you subtracted at the start.

Consider this function in action:

Min = 0.25 # where binning starts
Max = 2.25 # where binning ends
n = 2 # the number of bins
width = (Max-Min)/n # binwidth; evaluates to 1.0
bin(x) = width*(floor((x-Min)/width)+0.5) + Min

e.g. the value 1.1 truly falls in the left bin:

• this function correctly maps it to the centre of the left bin (0.75);
• Born2Smile's answer, bin(x)=width*floor(x/width), incorrectly maps it to 1;
• mas90's answer, bin(x)=width*floor(x/width) + binwidth/2.0, incorrectly maps it to 1.5.

Born2Smile's answer is only correct if the bin boundaries occur at (n+0.5)*binwidth (where n runs over integers). mas90's answer is only correct if the bin boundaries occur at n*binwidth.

Answer 4

As usual, Gnuplot is a fantastic tool for plotting sweet looking graphs and it can be made to perform all sorts of calculations. However, it is intended to plot data rather than to serve as a calculator and it is often easier to use an external programme (e.g. Octave) to do the more "complicated" calculations, save this data in a file, then use Gnuplot to produce the graph. For the above problem, check out the "hist" function is Octave using [freq,bins]=hist(data), then plot this in Gnuplot using

set style histogram rowstacked gap 0
set style fill solid 0.5 border lt -1
plot "./data.dat" smooth freq with boxes

Answer 5

We do not need to use recursive method, it may be slow. My solution is using a user-defined function rint instesd of instrinsic function int or floor.

rint(x)=(x-int(x)>0.9999)?int(x)+1:int(x)

This function will give rint(0.0003/0.0001)=3, while int(0.0003/0.0001)=floor(0.0003/0.0001)=2.

Why? Please look at Perl int function and padding zeros

Answer 6

yes, and its quick and simple though very hidden:

binwidth=5
bin(x,width)=width*floor(x/width)

plot 'datafile' using (bin($1,binwidth)):(1.0) smooth freq with boxes

check out help smooth freq to see why the above makes a histogram

to deal with ranges just set the xrange variable.