I would like to change the size of a point amongst scatter points in BQPLOT. What is the best way to do it without affecting the size of the rest of the points?
Thanks
I think I have solved it, I am leaving this here in case it can help somebody.
I go about it using a size array:
size_r = np.array([0.1,0.5,0.5, 0.2, 0.5, 0.5, 0.5, 0.8,0.5])
scat = bplt.scatter(x,y,size = size_r)
The issue is that the smallest marker (of size 0.1 in this case) will not appear. I go about it by adding a dummy scatter point at the beginning of my data, and I give it the smallest number of the size array (it can be 0 for example).
Related
given an array of points my program should in theory, Find the two furthest points from each other. Then calculate the angle that those two points make with the x axis. Then in rotate all the points in the array around the averaged center of all the points by that angle. For some reason my translation function to rotate all the points around the center is not working it is giving me unexpected values. I am fairly sure the math I am using to do this is accurate since I tested the formula I am using using wolfram alpha and plotted the points on desmos. I am not sure what's wrong with my code because it keeps giving me unexpected output. Any help would greatly be appreciated.
This is the code to translate the array:
def translation(array,centerArray):
array1=array
maxDistance=0
point1=[]
point2=[]
global angle
for i in range(len(array1)):
for idx in range(len(array1)):
if(maxDistance<math.sqrt(((array1[i][0]-array1[idx][0])**2)+((array1[i][1]-array1[idx][1])**2)+((array1[i][2]-array1[idx][2])**2))):
maxDistance=math.sqrt(((array1[i][0]-array1[idx][0])**2)+((array1[i][1]-array1[idx][1])**2)+((array1[i][2]-array1[idx][2])**2))
point1 = array1[i]
point2 = array1[idx]
angle=math.atan2(point1[1]-point2[1],point1[0]-point2[0]) #gets the angle between two furthest points and xaxis
for i in range(len(array1)): #this is the problem here
array1[i][0]=((array[i][0]-centerArray[0])*math.cos(angle)-(array[i][1]-centerArray[1])*math.sin(angle))+centerArray[0] #rotate x cordiate around center of all points
array1[i][1]=((array[i][1]-centerArray[1])*math.cos(angle)+(array[i][0]-centerArray[0])*math.sin(angle))+centerArray[1] #rotate y cordiate around center of all points
return array1
This is the code I am using to test it. tortose is what I set turtle graphics name as
tortose.color("violet")
testarray=[[200,400,9],[200,-100,9]] #array of 2 3d points but don't worry about z axis it will not be used for in function translation
print("testsarray",testarray)
for i in range(len(testarray)): #graph points in testarray
tortose.setposition(testarray[i][0],testarray[i][1])
tortose.dot()
testcenter=findCenter(testarray) # array of 1 point in the center of all the points format center=[x,y,z] but again don't worry about z
print("center",testcenter)
translatedTest=translation(testarray,testcenter) # array of points after they have been translated same format and size of testarray
print("translatedarray",translatedTest) #should give the output [[-50,150,9]] as first point but instead give output of [-50,-99.999999997,9] not sure why
tortose.color("green")
for i in range(len(testarray)): #graphs rotated points
tortose.setposition(translatedTest[i][0],translatedTest[i][1])
tortose.dot()
print(angle*180/3.14) #checks to make sure angle is 90 degrees because it should be in this case this is working fine
tortose.color("red")
tortose.setposition(testcenter[0],testcenter[1])
tortose.dot()
find center code finds the center of all points in array don't worry about z axis since it is not used in translation:
def findCenter(array):
sumX = 0
sumY = 0
sumZ = 0
for i in range(len(array)):
sumX += array[i][0]
sumY += array[i][1]
sumZ += array[i][2]
centerX= sumX/len(array)
centerY= sumY/len(array)
centerZ= sumZ/len(array)
#print(centerX)
#print(centerY)
#print(centerZ)
centerArray=[centerX,centerY,centerZ]
return centerArray
import math
import turtle
tortose = turtle.Turtle()
tortose.penup()
my expected output should be a point at (-50,150) but it is giving me a point at (-50,-99.99999999999997)
This is a common mistake when doing in-place rotations:
array1[i][0]= ...
array1[i][1]= ... array[i][0] ...
First you update array1[i][0]. Then you update array1[i][1], but you use the new value when you should use the old value. Instead, temporarily store the old value:
x = array1[i][0]
array1[i][0]=((array[i][0]-centerArray[0])*math.cos(angle)-(array[i][1]-centerArray[1])*math.sin(angle))+centerArray[0] #rotate x cordiate around center of all points
array1[i][1]=((array[i][1]-centerArray[1])*math.cos(angle)+(x-centerArray[0])*math.sin(angle))+centerArray[1] #rotate y cordiate around center of all points
c = canvas.Canvas('data.pdf', pagesize= [width*inch,height*inch])
c.drawImage('dataptah',x,y, width,height)
c.save()
I can't height center the picture,
so I need know x and y unit,
or put something.
First of all, you can use units for the x, y, width, and height values in the drawImage call, just as you did for the pagesize. Thus, if you know the aspect ratio of your image, you can calculate these values for the exact centered position.
The reference documentation mentions two other parameters, that could be helpful:
preserveAspectRation=True keeps the aspect ratio of the image even if the box specified with x, y, width, height has a different aspect ratio.
anchor='c' specifies the anchor position of the image, center in this case.
Thus, if you add these two parameters and center the box on the page, then your image should appear centered as well. Here is an example:
c.drawImage('dataptah',
width/4*inch, height/4*inch,
width/2*inch, height/2*inch,
preserveAspectRatio=True, anchor='c'
)
I have a weird issue in my bar graph realized using d3.js: the 1 px padding between each rectangle appears irregular. I gather either or both the width or x position are the culprit but i don't understand what i'm doing wrong: the width is a fraction of the svg area and the X position is obtained via a D3 scale.
I've put a demo here: http://jsfiddle.net/pixeline/j679N/4/
The code ( a scale) controling the x position:
var xScale = d3.time.scale().domain([minDate, maxDate]).rangeRound([padding, w - padding]);
The code controlling the width:
var barWidth = Math.floor((w/dataset.length))-barPadding;
Thank you for your insight.
It's irregular because you are rounding your output range (rangeRound). In some cases, the distance between two bars is 3 pixels and sometimes only 2. This is because the actual x position is a fractional value and ends up being rounded one way in some cases and the other way on other cases.
You can mitigate the effect but changing rangeRound to range, but that won't eliminate it entirely as you'll still get fractional pixel values for positions. The best thing to do is probably to simply increase the padding so that the differences aren't as obvious.
I'm trying to fill in a structured grid with an analytical field, but despite reading the vtk docs, I haven't found out how to actually set scalar values at the grid points or the set the spacing/origin info of the grid. Starting from the code below, how do I
associate spatial information with the grid (ie cell 0,0,0 is at coordinates 0,0,0, the spacing is dx in every direction)
associate scalar values with each grid point. To start, I just need one, but eventually I'd like to store 3 pieces of data at each point (not a vector, 3 distinct scalars).
grid = vtk.vtkStructuredGrid()
numPoints = int((maxGrid - minGrid)/dx)
grid.SetDimensions(numPoints, numPoints, numPoints)
In VTK there are 3 types of "structured" grids, vtkImageData (vtkUniformGrid derives from this), vtkRectilinearGrid, and vtkStructuredGrid. They are all structured in the sense that the topology is set. vtkImageData has constant spacing between points and is axis aligned, vtkRectilinearGrid is axis aligned but can vary the spacing in each axis direction, and vtkStructuredGrid has arbitrarily located points (cells may not be valid though).
For what you want to do you should do:
from vtk import *
dx = 2.0
grid = vtkImageData()
grid.SetOrigin(0, 0, 0) # default values
grid.SetSpacing(dx, dx, dx)
grid.SetDimensions(5, 8, 10) # number of points in each direction
# print grid.GetNumberOfPoints()
# print grid.GetNumberOfCells()
array = vtkDoubleArray()
array.SetNumberOfComponents(1) # this is 3 for a vector
array.SetNumberOfTuples(grid.GetNumberOfPoints())
for i in range(grid.GetNumberOfPoints()):
array.SetValue(i, 1)
grid.GetPointData().AddArray(array)
# print grid.GetPointData().GetNumberOfArrays()
array.SetName("unit array")
Given a "shape" drawn by the user, I would like to "normalize" it so they all have similar size and orientation. What we have is a set of points. I can approximate the size using bounding box or circle, but the orientation is a bit more tricky.
The right way to do it, I think, is to calculate the majoraxis of its bounding ellipse. To do that you need to calculate the eigenvector of the covariance matrix. Doing so likely will be way too complicated for my need, since I am looking for some good-enough estimate. Picking min, max, and 20 random points could be some starter. Is there an easy way to approximate this?
Edit:
I found Power method to iteratively approximate eigenvector. Wikipedia article.
So far I am liking David's answer.
You'd be calculating the eigenvectors of a 2x2 matrix, which can be done with a few simple formulas, so it's not that complicated. In pseudocode:
// sums are over all points
b = -(sum(x * x) - sum(y * y)) / (2 * sum(x * y))
evec1_x = b + sqrt(b ** 2 + 1)
evec1_y = 1
evec2_x = b - sqrt(b ** 2 + 1)
evec2_y = 1
You could even do this by summing over only some of the points to get an estimate, if you expect that your chosen subset of points would be representative of the full set.
Edit: I think x and y must be translated to zero-mean, i.e. subtract mean from all x, y first (eed3si9n).
Here's a thought... What if you performed a linear regression on the points and used the slope of the resulting line? If not all of the points, at least a sample of them.
The r^2 value would also give you information about the general shape. The closer to 0, the more circular/uniform the shape is (circle/square). The closer to 1, the more stretched out the shape is (oval/rectangle).
The ultimate solution to this problem is running PCA
I wish I could find a nice little implementation for you to refer to...
Here you go! (assuming x is a nx2 vector)
def majAxis(x):
e,v = np.linalg.eig(np.cov(x.T)); return v[:,np.argmax(e)]