I try to plot 3D graph on Maple18. I plot 2 graphs on the same plane and I want it to show all intercepts. I actually want only integral intercepts if it's possible but I don't know the command.
Here is the graph I want it to show the intercepts
plot3d([x^2, 3^6*z-432], x = -50 .. 50, z = 0 .. 20)
If you are interested in just showing the intersection of the two expressions, you might try the plots:-intersectplot command. The following will show the intersection for the two surfaces:
p1 := plots:-intersectplot(y=x^2, y=3^6*z-432, x=-50..50, z=0..20, y=0..14000);
If you want to then superimpose this on your original plot:
p2 := plot3d([x^2, 3^6*z-432], x=-50..50, z=0..20):
plots:-display([p1,p2]);
Related
I have a 2 varaible data set that I have to plot (na and ob). I applied the kde2d kernel and plotted 1 to 4 sigma density curves (confidencebound). I need to select those points that are inside 2 sigma curves (letting out all those between 1 and 2 sigmas), but not just in the plot, I neet select them from the data set, put them in a new list. Could you please help me with this?
kde_BPT <- kde2d(na,ob, n=1000, lims=c(-2,2,-1.5,1.5))
confidencebound <- quantile(kde_BPT$z, probs=c(0.685,0.955,0.9975,0.99995), na.rm = TRUE)
The data are to large to paste here. I put here the plot if that helps, I need to know which data points (any colour) are in the area between the contour curves of 1 and 2 (sigmas).
The plot
Thanks for your help.
I'm looking for a way to plot histograms in 3d to produce something like this figure http://www.gnuplot.info/demo/surface1.17.png but where each series is a histogram.
I'm using the procedure given here https://stackoverflow.com/a/19596160 and http://www.gnuplotting.org/calculating-histograms/ to produce histograms, and it works perfectly in 2d.
Basically, the commands I use are
hist = 'u (binwidth*(floor(($2-binstart)/binwidth)+0.5)+binstart):(1) smooth freq w boxes
plot 'data.txt' #hist
Now I would just like to add multiple histograms in the same plot, but because they overlap in 2d, I would like to space them out in a 3d plot.
I have tried to do the following command (using above procedure)
hist = 'u (1):(binwidth*(floor(($2-binstart)/binwidth)+0.5)+binstart):(1) smooth freq w boxes
splot 'data.txt' #hist
But gnuplot complains that the z values are undefined.
I don't understand why this would not put a histogram along the value 1 on the x-axis with the bins along the y-axis, and plot the height on the z-axis.
My data is formatted simply in two columns:
Index angle
0 92.046
1 91.331
2 86.604
3 88.446
4 85.384
5 85.975
6 88.566
7 90.575
I have 10 files like this, and since the values in the files are close to each other, they will completely overlap if I plot them all in one 2d histogram. Therefore, I would like to see 10 histograms behind each other in a sort of 3d perspective.
This second answer is distinct from my first. Whereas the first addresses what the OP was trying to accomplish, this second provides an alternative approach which address the underlying problem the OP was trying to overcome.
I have posted an answer that addresses the ability to do this in 3d. However, this isn't usually the best way to do this with multiple histograms like this. A 3d graph like that will be difficult to compare.
We can address the overlap in 2D by stagnating the position of the boxes. With default settings, the boxes will spread out to touch. We can turn that off and adjust the position of the boxes to allow more than 1 histogram on a graph. Remember, that the coordinates you supply are the center of the boxes.
Suppose that I have the data you have provided and this additional data set
Index Angle
0 85.0804
1 92.2482
2 90.0384
3 99.2974
4 87.729
5 94.6049
6 86.703
7 97.9413
We can set the boxwidth to 2 units with set boxwidth 2 (your bins are 4 units wide). Additionally, we will turn on box filling with set style fill solid border lc black.
Then I can issue
plot datafile1 u (binwidth*(floor(($2-binstart)/binwidth)+0.5)+binstart):(1) smooth freq w boxes, \
datafile2 u (binwidth*(floor(($2-binstart)/binwidth)+0.5)+binstart+1):(1) smooth freq w boxes
The second plot command is identical to the first, except for the +1 after binstart. This will shift this box 1 unit to the right. This produces
Here, the two series are clear. Keeping track of which box is associated with each is easy because of the overlap, but it is not enough to mask the other series.
We can even move them next to each other, with no overlap, by subtracting 1 from the first plot command:
plot datafile1 u (binwidth*(floor(($2-binstart)/binwidth)+0.5)+binstart-1):(1) smooth freq w boxes, \
datafile2 u (binwidth*(floor(($2-binstart)/binwidth)+0.5)+binstart+1):(1) smooth freq w boxes
producing
This first answer is distinct from my second. This answer address what the OP was trying to accomplish whereas the second addresses the underlying problem the OP was trying to overcome.
Gnuplot isn't going to be able to do this on it's own, as the relevant styles (boxes and histograms) only work in 2D. You would have to do it using an external program.
For example, using your data and your 2d command (your first command), we get (using your data and the linked values of -100 and 4 for binstart and binwidth)
To draw these boxes on the 3d grid, we will need to use the line style and have four points for each: lower left, upper left, upper right, and lower right. We can use the previous command and capture to a table, but this will only gives the upper center point. We can use an external program to pre-process, however. The following python program, makehist.py, does just that.
from sys import argv
import re
from math import floor
pat = re.compile("\s+")
fname = argv[1]
binstart = float(argv[2])
binwidth = float(argv[3])
data = [tuple(map(float,pat.split(x.strip()))) for x in open(fname,"r").readlines()[1:]]
counts = {}
for x in data:
bn = binwidth*(floor((x[-1]-binstart)/binwidth)+0.5)+binstart
if not bn in counts: counts[bn] = 0
counts[bn]+=1
for x in sorted(counts.keys()):
count = counts[x]
print(x-binwidth/2,0)
print(x-binwidth/2,count)
print(x+binwidth/2,count)
print(x+binwidth/2,0)
print(max(counts.keys())+binwidth/2,0)
print(min(counts.keys())-binwidth/2,0)
Essentially, this program does the same thing as the smooth frequency option does, but instead of getting the upper center of each box, we get the four previously mentioned points along with two points to draw a line along the bottom of all the boxes.
Running the following command,
plot "< makehist.py data.txt -100 4" u 1:2 with lines
produces
which looks very similar to the original graph. We can use this in a 3d plot
splot "< makehist.py data.txt -100 4" u (1):1:2 with lines
which produces
This isn't all that pretty, but does lay the histogram out on a 3d plot. The same technique can be used to add multiple data files onto it spread out. For example, with the additional data
Index Angle
0 85.0804
1 92.2482
2 90.0384
3 99.2974
4 87.729
5 94.6049
6 86.703
7 97.9413
We can use
splot "< makehist.py data.txt -100 4" u (1):1:2 with lines, \
"< makehist.py data2.txt -100 4" u (2):1:2 with lines
to produce
I am back at college learning maths and I want to try and use some this knowledge to create some svg with d3.js.
If I have a function f(x) = x^3 - 3x^2 + 3x - 1
I would take the following steps:
Find the x intercepts for when y = 0
Find the y intercept when x = 0
Find the stationary points when dy\dx = 0
I would then have 2 x values from point 3 to plug into the original equation.
I would then draw a nature table do judge the flow of the graph or curve.
Plot the known points from the above and sketch the graph.
Translating what I would do on pen and paper into code instructions is what I really could do with any sort of advice on the following:
How can I programmatically factorise point 1 of the above to find the x-intercepts for when y = 0. I honestly do not know where to even start.
How would I programmatically find dy/dx and the values for the stationary points.
If I actually get this far then what should I use in d3 to join the points on the graph.
Your other "steps" have nothing to do with d3 or plotting.
Find the x intercepts for when y = 0
This is root finding. Look for algorithms to help with this.
Find the y intercept when x = 0
Easy: substitute to get y = 1.
Find the stationary points when dy\dx = 0
Take the first derivative to get 3x^2 - 12x + 9 and repeat the root finding step. Easy to get using quadratic equation.
I would then have 2 x values from point 3 to plug into the original
equation. I would then draw a nature table do judge the flow of the
graph or curve. Plot the known points from the above and sketch the
graph.
I would just draw the curve. Pick a range for x and go.
It's great to learn d3. You'll end up with something like this:
https://maurizzzio.github.io/function-plot/
For a cubic polynomial, there are closed formulas available to find all the particular points that you want (https://en.wikipedia.org/wiki/Cubic_function), and it is a sound approach to determine them.
Anyway, you will have to plot the smooth curve, which means that you will need to compute close enough points and draw a polyline that joins them.
Doing this, you are actually performing the first steps of numerical root isolation, with such an accuracy that the approximate and exact roots will be practically undistinguishable.
So an easy combined solution is to draw the curve as a polyline and find the intersections with the X axis as well as extrema using this polyline representation, rather than by means of more sophisticated methods.
This approach works for any continuous curve and is very easy to implement. So you actually draw the curve to find particular points rather than conversely as is done by analytical methods.
For best results on complicated curves, you can adapt the point density based on the local curvature, but this is another story.
I am trying to generate a plot which uses arrows as markers in Gnuplot. These arrows I want to turn in a specific angle which I know. So I have value triples of x1 ... xn, y1...yn, alpha1...alphan. Sorry, I wasn't able to include a pic from my hard drive to illustrate what I want to achieve.
Basically, for every (15th or so) x-y pair, the marker should be an arrow which uses a certain angle.
The measured data is tightly packed so I suppose I will have to define an increment between the markers. The length of the arrow can be the same all over.
I would appreciate your ideas.
Gnuplot has a plot mode with vectors that is what you want
Given that your file has the following format, x y angle and assuming that
your angle is in radians, you have to take into account that
with vectors requires 4 parameters, namely x y dx dy where dx
and dy are the projections of the lenght of the arrow.
this draws only the arrows, if you want a line you have to make
two passes on the data.
you want to draw an arrow for a data point over, say, 10 points.
That said, I'd proceed like this
dx(a) = 0.2*cos(a) # 0.2 is an arbitrary scaling factor
dy(a) = 0.2*sin(a)
# this draws the arrows
plot 'mydata.dat' every 10 using 1:2:(dx(a)):(dy(a)) with vectors
# this draws the line
plot 'mydata.dat'
You may want to use help plot to find the detailed explanation of all the parameters that you can apply to a with vectors plot.
Credits: An article on the gnuplotting site
I apologize in advance if my code looks very amateurish.
I'm trying to assign quadrants to 4 measurement stations approximately located on the edges of a town.
I have the coordinates of these 4 stations:
a <- c(13.2975,52.6556)
b <- c(14.0083,52.5583)
c <- c(13.3722,52.3997)
d <- c(12.7417,52.6917)
Now my idea was to create lines connecting the north-south and east-west stations:
line.1 <- matrix(c(d[1],b[1],d[2],b[2]),ncol=2)
line.2 <- matrix(c(a[1],c[1],a[2],c[2]),ncol=2)
Plotting all the stations the connecting lines looks allright, however not very helpful for analyzing it on a computer.
So I calculated the eucledian vectors for the two lines:
vec.1 <- as.vector(c((b[1]-d[1]),(b[2]-d[2])))
vec.2 <- as.vector(c((c[1]-a[1]),(c[2]-a[2])))
which allowed me to calculate the angle between the two lines in degrees:
alpha <- acos((vec.1%*%vec.2) / (sqrt(vec.1[1]^2+vec.1[2]^2)*
sqrt(vec.2[1]^2+vec.2[2]^2)))) * 180/pi
The angle I get for alpha is 67.7146°. This looks fairly good. From this angle I can easily calculate the other 3 angles of the intersection, however I need values relative to the grid so I can assign values from 0°-360° for the wind directions.
Now my next planned step was to find the point where the two lines intersect, add a horizontal and vertical abline through that point and then calculate the angle relative to the grid. However I can't find a proper example that does that and I don't think I have a nice linear equation system I could solve.
Is my code way off? Or maybe anyone knows of a package which could help me? It feels like my whole approach is a bit wrong.
Okay I managed to calculate the intersection point, using line equations. Here is how.
The basic equation for two points is like this:
y - y_1 = (y_2-y_1/x_2-x_1) * (x-x_1)
If you make one for each of the two lines, you can just substitute the fractions.
k.1 <- ((c[2]-a[2])/(c[1]-a[1]))
k.2 <- ((b[2]-d[2])/(b[1]-d[1]))
Reshaping the two functions you get a final form for y:
y <- (((-k.1/k.2)*d[2]+k.1*d[1]-k.1*c[1]+d[2])/(1-k.1/k.2))
This one you can now use to calculate the x-value:
x <- ((y-d[2])+d[1]*k.2)/k.2
In my case I get
y = 52.62319
x = 13.3922
I'm starting to really enjoy this program!
Wikipedia has a good article on finding the intersection between two line segments with an explicit formula. However, you don't need to know the point of intersection to calculate the angle to the grid (or axes of coordinate system.) Just compute the angles from your vec.1 and vec.2 to the basis vectors:
e1 <- c(1, 0)
e2 <- c(0, 1)
as you have done.