I try to plot the orbital velocity with gnuplot, but somehow gnuplot gets completely different results than me. Now from experience I think my values are correct but I checked it with Google's calculator and get my results.
I use the formula from Wikipedia and Google gets a velocity at apoapsis of about 2.2 km/s. Now gnuplot itself gets a velocity of about 3.2 km/s.
set xlabel "Altitude above sea level (meters)"
set ylabel "Orbital velocity (meters per second)"
set title "Velocity of an 80×100 km orbit around Kebrin"
set terminal png size 800,640
set output "orbitv.png"
set xrange [80000:100000]
G=6.674*10**-11
M=5.2915793*10**22
R=600000
plot sqrt(G*M*(2/(x+R)-1/(90000+R))) title 'Orbital velocity' with lines
I'm wondering were did I make the mistake? I copied the formula directly to Google and replaced G, M and R with the constant values and x with 100000 and get the result linked above.
This problem has to do with how gnuplot handles integers when doing arithmetic. When you have an expression like 1/(90000 + R), if R is an integer, gnuplot evaluates 1/(690000) = 0, which is perfectly valid integer arithmetic. The solution is to add a period to a number so that gnuplot knows to cast it as a floating-point number:
R = 600000. # short option
R = 600000.0 # clearer option
Another solution is to use e-notation for big numbers:
R = 6e5
Gnuplot treats that as a float. This also helps prevent order-of-magnitude/number-of-zeroes errors.
Incidentally, python and other languages have the same problem with integer arithmetic--watch out!
Related
I'm making a c++ code which prints commands for gnuplot, in order to plot different things faster. The code plots the data already as the data fit as well, but now I'm adding some labels, and I want to print the fit equation, I mean something with this form
f(x) = (a +/- Δa)*x + (b +/- Δb)
I have the following line for printing it
set label 1 at screen 0.22, screen 0.75 sprintf('f(x) = %3.4f*x + %3.4f', a, b)
But, as you can see, there is only a and b values with no errors, I was thinking something like put there in the sprintf function any error related variables (FIT_something) and then have something like
set label 1 at screen 0.22, screen 0.75 sprintf('f(x) = (%3.4f +/- %3.4f)*x + (%3.4f + %3.4f)', a, deltaa, b, deltab)
But I can't find those, my answers are: does those exists? and if the answer is no, is there any way to print the variable errors further just writing it explicitly on the line?
Thanks for your help
Please read the statistical overview section of the gnuplot documentation (help statistical_overview). Keeping in mind the caveats described there, see also the documentation for set fit errorvariables, which I extract below:
If the `errorvariables` option is turned on, the error of each fitted
parameter computed by `fit` will be copied to a user-defined variable
whose name is formed by appending "_err" to the name of the parameter
itself. This is useful mainly to put the parameter and its error onto
a plot of the data and the fitted function, for reference, as in:
set fit errorvariables
fit f(x) 'datafile' using 1:2 via a, b
print "error of a is:", a_err
set label 1 sprintf("a=%6.2f +/- %6.2f", a, a_err)
plot 'datafile' using 1:2, f(x)
If the `errorscaling` option is specified, which is the default, the
calculated parameter errors are scaled with the reduced chi square. This is
equivalent to providing data errors equal to the calculated standard
deviation of the fit (FIT_STDFIT) resulting in a reduced chi square of one.
So I'm trying to plot three parametric functions using the gnuplot;
unfortunately, I cannot get around some garbage that is generated in the output plot. I tried to isolate the problem by splitting a function j into j1 and j2, just changing the position of the minus sign. Unexpectedly, the functions j1 and j2 jump strangely when close to the origin. I currently use version 4.6 of gnuplot, any suggestions?
CODE BELOW:
set parametric
j1(x) = -((1.0/27.0*(1.+9.*x))/2.0)**(1./3.) #negative portion
j2(x) = (-(1.0/27.0*(1.+9.*x))/2.0)**(1./3.)
k(x) = ((-x/3.0)**(3./2.))**(1./3.)
l(x) = -((-x/3.0)**(3./2.))**(1.0/3.0)
tt(x) = sqrt(-x/3.)
set trange [-1.0/3.0:0]
set yrange [0:1.0/3.0]
set xrange [-1./6.:1./3]
plot j1(t),tt(t) w l ls 1, j2(t),tt(t) w l ls 1, k(t),tt(t) w l ls 2, l(t),tt(t) w l ls 3
The problem comes from selecting the cube root of a negative number. Gnuplot can work with complex numbers, and in the complex number system there are three cube roots of any number†. For a real number, one of these is real and two are complex. Gnuplot is selecting the first‡ one which is complex for a negative number (for a positive number, the first one is real).
print (-8)**(1/3.0) # prints {1.0, 1.73205080756888}
The solution is to construct our own cube root function
cuberoot(x) = sgn(x)*abs(x)**(1/3.0)
This will select take the cube root of the absolute value (always positive) and make the result have the same sign as the original.
We can then use it in our functions
j1(x) = -cuberoot((1.0/27.0*(1.+9.*x))/2.0) #negative portion
j2(x) = cuberoot(-(1.0/27.0*(1.+9.*x))/2.0)
leaving the rest of the code alone.
Without custom cuberoot function
With custom cuberoot function
† For the given example of -8, they are 1 + 1.7320508i, -2, and 1 - 1.7320508i.
‡ When ordered in increasing order by complex argument (restricted to the interval [0,2π) ).
I have a large set of data points from x = 1 to x = 10e13 (step size is fixed to about 3e8).
When I try to plot them using a logscale I certainly get an incredible huge point-density towards the end. Of course this affects my output plots since postscript and svg files (holding each and every data point) are getting really big.
Is there a way to tell gnuplot to decrease the data density dynamically?
Sample data here. Shows a straight line using logarithmic x-axis.
Usually, for this kind of plots, one can use a filter function which selects the desired points and discards all others (sets their value to 1/0:
Something like:
plot 'sample.dat' using (filter($1) ? $1 : 1/0):2
Now you must define an appropriate filter function to change the data density. Here is a proposal, with pseudo-data, although you might for sure find a better one, which doesn't show this typical logarithmic pattern:
set logscale x
reduce(x) = x/(10**(floor(log10(x))))
filterfunc(x) = abs(log10(sc)+(log10(x) - floor(log10(x))) - log10(floor(sc*reduce(x))))
filter(x) = filterfunc(x) < 1e-5 ? x : 1/0
set multiplot layout 1,2
sc = 1
plot 'sample.data' using (filter($1)):2 notitle
sc = 10
replot
The variable sc allows to change the density. The result is (with 4.6.5) is:
I did some work inspired by Christoph's answer and able to get equal spacing in log scale. I made a filtering, if you have numbers in the sequence you can simply use Greatest integer function and then find the nearest to it in log scale by comparing the fraction part. Precision is tuned by precision_parameter here.
precision_parameter=100
function(x)=(-floor(precision_parameter*log10(x))+(precision_parameter*log10(x)))
Now filter by using the filter function defined below
density_parameter = 3.5
filter(x)=(function(x) < 1/(log10(x))**density_parameter & function(x-1) > 1/(log10(x))**density_parameter ) ? x : 1/0
set datafile missing "NaN"
Last line helps in plotting with line point. I used x and x-1 assuming the xdata is in arithmetic progression with 1 as common difference, change it accordingly with your data. Just replace x by filter(x) in the plot command.
plot 'sample_data.dat' u (filter($1)):2 w lp
I'm using the following gnuplot script to plot a linear fit:
#!/usr/bin/gnuplot
set term cairolatex
set output "linear_fit.tex"
c = 299792458.
x(x) = c / x
y(x) = x
h(x) = a * x + b
fit h(x) "linear_fit.dat" u (x($1)):(y($2)) via a,b
plot "linear_fit.dat" u (x($1)):(y($2)) w points title "", \
(h(x)) with lines linecolor rgb "black" title "Linear Fit"
However, after the iterations converge, b is always 1.0: https://dpaste.de/ozReq/
How can I get gnuplot to adjust b as well as a?
Update: Repeating the fit command a few hundred times with alternating via a/via b does give pretty good results, but that just can't be how it's supposed to be done.
Update 2: Here's the data in linear_fit.dat:
# lambda, V
360e-9 1.119
360e-9 1.148
360e-9 1.145
400e-9 0.949
400e-9 0.993
400e-9 0.971
440e-9 0.883
440e-9 0.875
440e-9 0.863
490e-9 0.737
490e-9 0.728
490e-9 0.755
540e-9 0.575
540e-9 0.571
540e-9 0.592
590e-9 0.457
590e-9 0.455
590e-9 0.482
I think your troubles stem from the fact that your x-values are very large (on the order of 10e14).
If you do not provide gnuplot with an initial guess for a and b, it will assume a=1 and b=1 as starting points for the fit. However, this is a poor initial guess:
Please note the log scale on both the x- and y-axis.
From the gnuplot documentation:
fit may, and often will get "lost" if started far from a solution, where SSR is large and changing slowly as the parameters are varied, or it may reach a numerically unstable region (e.g., too large a number causing a floating point overflow) which results in an "undefined value" message or gnuplot halting.
To improve the chances of finding the global optimum, you should set the starting values at least roughly in the vicinity of the solution, e.g., within an order of magnitude, if possible. The closer your starting values are to the solution, the less chance of stopping at another minimum. One way to find starting values is to plot data and the fitting function on the same graph and change parameter values and replot until reasonable similarity is reached. The same plot is also useful to check whether the fit stopped at a minimum with a poor fit.
In your case, such starting values could be:
a = 1e-15
b = -0.5
I obtained these values by eye-balling your range of values.
With those starting values, the linear fit results in:
Final set of parameters Asymptotic Standard Error
======================= ==========================
a = 1.97355e-015 +/- 6.237e-017 (3.161%)
b = -0.5 +/- 0.04153 (8.306%)
Which looks like this:
You can play with the control setting of fit (such as setting FIT_LIMIT = 1.e-35) or the starting values to achieve a better fit than this.
EDIT
While I still have not been able to coax gnuplot into modifying both parameters a, b at the same time, I found an alternate approach using R. I am aware that there are many other (scripting) languages that can perform a linear fit and this question was about gnuplot. However, the required effort with R appeared to be minimal.
Here's an example, which, when saved as linear_fit.R and called with
R CMD BATCH linear_fit.R
will provide the two coefficients of the linear fit, that gnuplot failed to provide.
y <- c(1.119, 1.148, 1.145, 0.949, 0.993, 0.971, 0.883, 0.875, 0.863,
0.737, 0.728, 0.755, 0.575, 0.571, 0.592, 0.457, 0.455, 0.482)
x <- c(3.60E-007, 3.60E-007, 3.60E-007, 4.00E-007, 4.00E-007,
4.00E-007, 4.40E-007, 4.40E-007, 4.40E-007, 4.90E-007,
4.90E-007, 4.90E-007, 5.40E-007, 5.40E-007, 5.40E-007,
5.90E-007, 5.90E-007, 5.90E-007)
c = 299792458.
x <- c/x
lm.out <- lm(y ~ x)
svg("linear_fit.svg")
plot(x,y)
abline(lm.out,col="red")
summary(lm.out)
You will end up with an svg-file that contains the plot and a linear_fit.Rout text file. In there you'll find the following coefficients:
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -5.429e-01 4.012e-02 -13.53 3.55e-10 ***
x 2.037e-15 6.026e-17 33.80 2.61e-16 ***
So, in the terminology of the original question, we obtain:
a = 2.037e-15
b = -5.429e-01
These values are very close to the values you quoted from alternating the fit.
In case the comments get purged, these questions were identified as related:
What is gnuplot's internal representation of floating point numbers?
Gnuplot behaves oddly in polynomial fit. Why is that?
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?
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.
I have a couple corrections/additions to Born2Smile's very useful answer:
Empty bins caused the box for the adjacent bin to incorrectly extend into its space; avoid this using set boxwidth binwidth
In Born2Smile's version, bins are rendered as centered on their lower bound. Strictly they ought to extend from the lower bound to the upper bound. This can be corrected by modifying the bin function: bin(x,width)=width*floor(x/width) + width/2.0
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.
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
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
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.
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
I have a little modification to Born2Smile's solution.
I know that doesn't make much sense, but you may want it just in case. If your data is integer and you need a float bin size (maybe for comparison with another set of data, or plot density in finer grid), you will need to add a random number between 0 and 1 inside floor. Otherwise, there will be spikes due to round up error. floor(x/width+0.5) will not do because it will create pattern that's not true to original data.
binwidth=0.3
bin(x,width)=width*floor(x/width+rand(0))
With respect to binning functions, I didn't expect the result of the functions offered so far. Namely, if my binwidth is 0.001, these functions were centering the bins on 0.0005 points, whereas I feel it's more intuitive to have the bins centered on 0.001 boundaries.
In other words, I'd like to have
Bin 0.001 contain data from 0.0005 to 0.0014
Bin 0.002 contain data from 0.0015 to 0.0024
...
The binning function I came up with is
my_bin(x,width) = width*(floor(x/width+0.5))
Here's a script to compare some of the offered bin functions to this one:
rint(x) = (x-int(x)>0.9999)?int(x)+1:int(x)
bin(x,width) = width*rint(x/width) + width/2.0
binc(x,width) = width*(int(x/width)+0.5)
mitar_bin(x,width) = width*floor(x/width) + width/2.0
my_bin(x,width) = width*(floor(x/width+0.5))
binwidth = 0.001
data_list = "-0.1386 -0.1383 -0.1375 -0.0015 -0.0005 0.0005 0.0015 0.1375 0.1383 0.1386"
my_line = sprintf("%7s %7s %7s %7s %7s","data","bin()","binc()","mitar()","my_bin()")
print my_line
do for [i in data_list] {
iN = i + 0
my_line = sprintf("%+.4f %+.4f %+.4f %+.4f %+.4f",iN,bin(iN,binwidth),binc(iN,binwidth),mitar_bin(iN,binwidth),my_bin(iN,binwidth))
print my_line
}
and here's the output
data bin() binc() mitar() my_bin()
-0.1386 -0.1375 -0.1375 -0.1385 -0.1390
-0.1383 -0.1375 -0.1375 -0.1385 -0.1380
-0.1375 -0.1365 -0.1365 -0.1375 -0.1380
-0.0015 -0.0005 -0.0005 -0.0015 -0.0010
-0.0005 +0.0005 +0.0005 -0.0005 +0.0000
+0.0005 +0.0005 +0.0005 +0.0005 +0.0010
+0.0015 +0.0015 +0.0015 +0.0015 +0.0020
+0.1375 +0.1375 +0.1375 +0.1375 +0.1380
+0.1383 +0.1385 +0.1385 +0.1385 +0.1380
+0.1386 +0.1385 +0.1385 +0.1385 +0.1390
Different number of bins on the same dataset can reveal different features of the data.
Unfortunately, there is no universal best method that can determine the number of bins.
One of the powerful methods is the Freedman–Diaconis rule, which automatically determines the number of bins based on statistics of a given dataset, among many other alternatives.
Accordingly, the following can be used to utilise the Freedman–Diaconis rule in a gnuplot script:
Say you have a file containing a single column of samples, samplesFile:
# samples
0.12345
1.23232
...
The following (which is based on ChrisW's answer) may be embed into an existing gnuplot script:
...
## preceeding gnuplot commands
...
#
samples="$samplesFile"
stats samples nooutput
N = floor(STATS_records)
samplesMin = STATS_min
samplesMax = STATS_max
# Freedman–Diaconis formula for bin-width size estimation
lowQuartile = STATS_lo_quartile
upQuartile = STATS_up_quartile
IQR = upQuartile - lowQuartile
width = 2*IQR/(N**(1.0/3.0))
bin(x) = width*(floor((x-samplesMin)/width)+0.5) + samplesMin
plot \
samples u (bin(\$1)):(1.0/(N*width)) t "Output" w l lw 1 smooth freq