Using matplotlib(or if there exists anything else), i want to populate a scatterplot image by using a grey scale image as its distribution. I have found many resource to create heat maps from images but not the other way around.
The input image will be like this one.
I think I understand what you're going for, but I'm not certain. I also don't really understand what this would be used for so I'm extra uncertain about this answer, but here goes:
So by loading the image we can evaluate each pixel position and its intensity. We can use that intensity as a "fitness" value and probabilistically add it to our plot so that we can get some of that "density" of points that you want to see. I picked a really simple equation as a decider (I just cubed the value), but feel free to replace that with whatever you want.
import cv2
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
import random
# select func
def selection(value):
return value**3 >= random.randint(0, 255**3);
# populate the sample
def populate(img):
# get res
h, w = img.shape;
# go through and populate
sx = [];
sy = [];
for y in range(0, h):
for x in range(0, w):
val = img[y, x];
# use intensity to decide if it gets in
# replace with what you want this function to look like
if selection(val):
sx.append(x);
sy.append(h - y); # opencv is top-left origin
return sx, sy;
# I'm using opencv to pull the image into code, use whatever you like
# matplotlib can also do something similar, but I'm not familiar with its format
img = cv2.imread("circ.png");
img = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY);
# lets take a sample
sx, sy = populate(img);
# find the bigger square size
h, w = img.shape;
side = None;
if h > w:
side = h;
else:
side = w;
# make a square graph
fig, ax = plt.subplots();
ax.scatter(sx, sy, s = 4);
ax.set_xlim((0, side));
ax.set_ylim((0, side));
x0,x1 = ax.get_xlim();
y0,y1 = ax.get_ylim();
ax.set_aspect(abs(x1-x0)/abs(y1-y0));
fig.savefig("out.png", dpi=600);
plt.show();
Feel free to replace opencv with whatever image library you're comfortable with. I'm pretty sure matplotlib can open images as well, but openCV is what I'm most familiar with so I used that.
As far as I can tell, you're trying to generate random coordinates that follow a distribution described by a grayscale image: the brighter each point, the more likely that point's coordinates will be generated. Your problem can thus be solved by a rejection sampler, as follows.
Assume you know the width and height of the image in pixels, call them w and h.
Generate two random numbers: one in the interval [0, w), and [0, h). These are the x and y coordinates, respectively.
Get the pixel at the given coordinates x and y in the image. This can be done using interpolation, but describing interpolation techniques is beyond the scope of this answer. For this reason, we will use only the nearest pixel ("nearest neighbor") in the image: take the pixel at coordinate floor(x) and floor(y) (and step 1 devolves to generating random integers). Convert the pixel somehow to a number p in the interval [0, 1]; in this answer we will assume black is 0 and white is 1, to simplify matters.
With probability p, return the point (x, y). Otherwise, go to step 1.
Roughly speaking, the time complexity of this algorithm depends on the numbers of "bright points" the input image has, compared to the number of "dark points". In general, the "brighter" the image, the higher the acceptance rate (and the faster the algorithm runs).
Related
I have made a workflow code to detect the edges of a flame in an image. I could get the edge line. It consists of many pixel points stored in an array (data in my code). Now based on the data, I would like to calculate the length of the edge. The idea is to calculate the distance between every point in data and sum them all to get the length. I really stuck in making that. Please help me, many thanks.
Here is a processed image:
Here is the original image that converted to the processed image, I put in the code is to compare the result:
import cv2
import matplotlib.pyplot as plt
if __name__ == '__main__':
path = '1897_1.jpg' #processed image
pic = cv2.imread(path)
original = cv2.imread('1897_2.jpg') #original image
img2 = cv2.flip(original, 1)
b,g,r = cv2.split(pic)
img4 = cv2.flip(b, 1)
h,w = img4.shape
data = []
th_val = 20
for i in range(h):
for j in range(w):
val = img4[i, j]
if (val >= th_val):
data.append(j)
break
b1 = range(len(data))
b2 = len(data)
result = [b2]
print (b2)
plt.figure(figsize = (10, 8))
plt.subplot(121)
plt.imshow(img4)
plt.plot(data, b1)
plt.axis('off');
plt.subplot(122)
plt.plot(data, b1)
plt.imshow(img2)
plt.axis('off')
I came up with a very simple solution, it is far from optimal, but it works for this example, and it is a good starting point. Unfortunately, this solution is not optimal for the blue chanell, where the curve is not smooth, but it works for green and red chanells.
data contains width coordinates of the first red pixel overcoming threshold. So, all first pixels are separated by 1 pixel step on vertical axes and data[i+1] - data[i] on horizontal axes. These two values can be considered as two cathetus of the squeare triangle, and the hypothenuse is the distance we want to calculate. So, here is the solution:
length = 0
for i in range(0,len(data)-1):
cathetus = data[i+1]-data[i]
hypothenuse = (cathetus**2 + 1**2)**1/2
length += hypothenuse
print(length)
Update
I have came up with two solutions: a hardcoded one and one released in the form of the function. Let us start with the first one: mean is a rather good approximator for the signal + noise. In the situation, when you do not have very strong noise or missing data, you may use this approach. In the example below we select points with x in [1,2,3] then we calculate mean y for these points and assign mean to coordinate x=2. Next we select points x in [2,3,4] and so on. As a result, we obtain mean_data list with y coordinates and mean_x with x coordinates. We can calculate length with the approach described above. You may also increase the power of smoothing by averaging over 4 and more points from data.
mean_data = []
mean_x = range(1,len(data)-1)
for i in range(0,len(data)-2):
mean_d = (data[i] + data[i+1] + data[i+2])/3
mean_data.append(mean_d)
Another approach is to use smoothing tools from scipy package. One of them is described below. When calculating the length you will have to adjust to new x axes xnew.
from scipy.interpolate import spline
import numpy as np
#transform to np.arrays initial data
b1_ = np.array(b1)
data_ = np.array(data)
# create new x with more data points
xnew = np.linspace(b1_.min(),b1_.max(),50) #50 is a number of points in between
smoothed_data = spline(b1_,data_,xnew)
I am looking to generate a linear gradient for a polycollection in matplotlib. I have the following code that gives me a single block color
v = []
for k in range(0, len(xs) - 1):
x = [xs[k], xs[k+1], xs[k+1], xs[k]]
y = [ys[k], ys[k+1], ys[k+1], ys[k]]
z = [zs[k], zs[k+1], h, h]
v.append(zip(x, y, z))
poly3dCollection = Poly3DCollection(v, facecolor='yellow', edgecolors='none')
I have played around with colormaps and interpolation but I can't quite seem to get this to work in the way I want. Specifically I need a linear gradient based on a number of greyscale values I supply (so no default jet color scheme). The gradient can be set once i.e. top x% of each polygon black then various shades of gray for same x% rather than this needing to be dynamically calculated for each polygon.
I did get the effect I wanted by saving to SVG and manually editing the image, but I need to do this all in code.
It strikes me this should be very straight forward so im not sure what I am missing
I have a historical time sequence of seafloor images scanned from film that need registration.
from pylab import *
import cv2
import urllib
urllib.urlretrieve('http://geoport.whoi.edu/images/frame014.png','frame014.png');
urllib.urlretrieve('http://geoport.whoi.edu/images/frame015.png','frame015.png');
gray1=cv2.imread('frame014.png',0)
gray2=cv2.imread('frame015.png',0)
figure(figsize=(14,6))
subplot(121);imshow(gray1,cmap=cm.gray);
subplot(122);imshow(gray2,cmap=cm.gray);
I want to use the black region on the left of each image to do the registration, since that region was inside the camera and should be fixed in time. So I just need to compute the affine transformation between the black regions.
I determined these regions by thresholding and finding the largest contour:
def find_biggest_contour(gray,threshold=40):
# threshold a grayscale image
ret,thresh = cv2.threshold(gray,threshold,255,1)
# find the contours
contours,h = cv2.findContours(thresh,mode=cv2.RETR_LIST,method=cv2.CHAIN_APPROX_NONE)
# measure the perimeter
perim = [cv2.arcLength(cnt,True) for cnt in contours]
# find contour with largest perimeter
i=perim.index(max(perim))
return contours[i]
c1=find_biggest_contour(gray1)
c2=find_biggest_contour(gray2)
x1=c1[:,0,0];y1=c1[:,0,1]
x2=c2[:,0,0];y2=c2[:,0,1]
figure(figsize=(8,8))
imshow(gray1,cmap=cm.gray, alpha=0.5);plot(x1,y1,'b-')
imshow(gray2,cmap=cm.gray, alpha=0.5);plot(x2,y2,'g-')
axis([0,1500,1000,0]);
The blue is the longest contour from the 1st frame, the green is the longest contour from the 2nd frame.
What is the best way to determine the rotation and offset between the blue and green contours?
I only want to use the right side of the contours in some region surrounding the step, something like the region between the arrows.
Of course, if there is a better way to register these images, I'd love to hear it. I already tried a standard feature matching approach on the raw images, and it didn't work well enough.
Following Shambool's suggested approach, here's what I've come up with. I used a Ramer-Douglas-Peucker algorithm to simplify the contour in the region of interest and identified the two turning points. I was going to use the two turning points to get my three unknowns (xoffset, yoffset and angle of rotation), but the 2nd turning point is a bit too far toward the right because RDP simplified away the smoother curve in this region. So instead I used the angle of the line segment leading up to the 1st turning point. Differencing this angle between image1 and image2 gives me the rotation angle. I'm still not completely happy with this solution. It worked well enough for these two images, but I'm not sure it will work well on the entire image sequence. We'll see.
It would really be better to fit the contour to the known shape of the black border.
# select region of interest from largest contour
ind1=where((x1>190.) & (y1>200.) & (y1<900.))[0]
ind2=where((x2>190.) & (y2>200.) & (y2<900.))[0]
figure(figsize=(10,10))
imshow(gray1,cmap=cm.gray, alpha=0.5);plot(x1[ind1],y1[ind1],'b-')
imshow(gray2,cmap=cm.gray, alpha=0.5);plot(x2[ind2],y2[ind2],'g-')
axis([0,1500,1000,0])
def angle(x1,y1):
# Returns angle of each segment along an (x,y) track
return array([math.atan2(y,x) for (y,x) in zip(diff(y1),diff(x1))])
def simplify(x,y, tolerance=40, min_angle = 60.*pi/180.):
"""
Use the Ramer-Douglas-Peucker algorithm to simplify the path
http://en.wikipedia.org/wiki/Ramer-Douglas-Peucker_algorithm
Python implementation: https://github.com/sebleier/RDP/
"""
from RDP import rdp
points=vstack((x,y)).T
simplified = array(rdp(points.tolist(), tolerance))
sx, sy = simplified.T
theta=abs(diff(angle(sx,sy)))
# Select the index of the points with the greatest theta
# Large theta is associated with greatest change in direction.
idx = where(theta>min_angle)[0]+1
return sx,sy,idx
sx1,sy1,i1 = simplify(x1[ind1],y1[ind1])
sx2,sy2,i2 = simplify(x2[ind2],y2[ind2])
fig = plt.figure(figsize=(10,6))
ax =fig.add_subplot(111)
ax.plot(x1, y1, 'b-', x2, y2, 'g-',label='original path')
ax.plot(sx1, sy1, 'ko-', sx2, sy2, 'ko-',lw=2, label='simplified path')
ax.plot(sx1[i1], sy1[i1], 'ro', sx2[i2], sy2[i2], 'ro',
markersize = 10, label='turning points')
ax.invert_yaxis()
plt.legend(loc='best')
# determine x,y offset between 1st turning points, and
# angle from difference in slopes of line segments approaching 1st turning point
xoff = sx2[i2[0]] - sx1[i1[0]]
yoff = sy2[i2[0]] - sy1[i1[0]]
iseg1 = [i1[0]-1, i1[0]]
iseg2 = [i2[0]-1, i2[0]]
ang1 = angle(sx1[iseg1], sy1[iseg1])
ang2 = angle(sx2[iseg2], sy2[iseg2])
ang = -(ang2[0] - ang1[0])
print xoff, yoff, ang*180.*pi
-28 14 5.07775871644
# 2x3 affine matrix M
M=array([cos(ang),sin(ang),xoff,-sin(ang),cos(ang),yoff]).reshape(2,3)
print M
[[ 9.99959685e-01 8.97932821e-03 -2.80000000e+01]
[ -8.97932821e-03 9.99959685e-01 1.40000000e+01]]
# warp 2nd image into coordinate frame of 1st
Minv = cv2.invertAffineTransform(M)
gray2b = cv2.warpAffine(gray2,Minv,shape(gray2.T))
figure(figsize=(10,10))
imshow(gray1,cmap=cm.gray, alpha=0.5);plot(x1[ind1],y1[ind1],'b-')
imshow(gray2b,cmap=cm.gray, alpha=0.5);
axis([0,1500,1000,0]);
title('image1 and transformed image2 overlain with 50% transparency');
Good question.
One approach is to represent contours as 2d point clouds and then do registration.
More simple and clear code in Matlab that can give you affine transform.
And more complex C++ code(using VXL lib) with python and matlab wrapper included.
Or you can use some modificated ICP(iterative closest point) algorithm that is robust to noise and can handle affine transform.
Also your contours seems to be not very accurate so it can be a problem.
Another approach is to use some kind of registration that use pixel values.
Matlab code (I think it's using some kind of minimizer+ crosscorrelation metric)
Also maybe there is some kind of optical flow registration(or some other kind) that is used in medical imaging.
Also you can use point features as SIFT(SURF).
You can try it quick in FIJI(ImageJ)
also this link.
Open 2 images
Plugins->feature extraction-> sift (or other)
Set expected transformation to affine
Look at estimated transformation model [3,3] homography matrix in ImageJ log.
If it works good then you can implement it in python using OpenCV or maybe using Jython with ImageJ.
And it will be better if you post original images and describe all conditions (it seems that image is changing between frames)
You can represent these contours with their respective ellipses. These ellipses are centered on the centroid of the contour and they are oriented towards the main density axis. You can compare the centroids and the orientation angle.
1) Fill the contours => drawContours with thickness=CV_FILLED
2) Find moments => cvMoments()
3) And use them.
Centroid: { x, y } = {M10/M00, M01/M00 }
Orientation (theta):
EDIT: I customized the sample code from legacy (enteringblobdetection.cpp) for your case.
/* Image moments */
double M00,X,Y,XX,YY,XY;
CvMoments m;
CvRect r = ((CvContour*)cnt)->rect;
CvMat mat;
cvMoments( cvGetSubRect(pImgFG,&mat,r), &m, 0 );
M00 = cvGetSpatialMoment( &m, 0, 0 );
X = cvGetSpatialMoment( &m, 1, 0 )/M00;
Y = cvGetSpatialMoment( &m, 0, 1 )/M00;
XX = (cvGetSpatialMoment( &m, 2, 0 )/M00) - X*X;
YY = (cvGetSpatialMoment( &m, 0, 2 )/M00) - Y*Y;
XY = (cvGetSpatialMoment( &m, 1, 1 )/M00) - X*Y;
/* Contour description */
CvPoint myCentroid(r.x+(float)X,r.y+(float)Y);
double myTheta = atan( 2*XY/(XX-YY) );
Also, check this with OpenCV 2.0 examples.
If you don't want to find the homography between the two images and want to find the affine transformation you have three unknowns, rotation angle (R), and the displacement in x and y (X,Y). Therefore minimum of two points (with two known values for each) are needed to find the unknowns. Two points should be matched between the two images or two lines, each has two known values, the intercept and slope. If you go with the point matching approach, the further the points are from each other the more robust is the found transform to noise (this is very simple if you remember error propagation rules).
In the two point matching method:
find two points (A and B) in the first image I1 and their corresponding points (A',B') in the second image I2
find the middle point between A and B: C, and the middle point between A' and B': C'
the difference C and C' (C-C') gives the translation between the images (X and Y)
using the dot product of C-A and C'-A' you can find the rotation angle (R)
To detect robust points, I would find the the points along the side of counter you have found with highest absolute value of the second derivative (Hessian) and then try to match them. Since you mentioned this is a video footage you can easily make the assumption the transformation between each two frames is small to reject the outliers.
I have a bunch of images like this one:
The corresponding data is not available. I need to automatically retrieve about 100 points (regularly x-spaced) on the blue curve. All curves are very similar, so I need at least 1 pixel precision, but sub-pixel would be preferred. The good news is all curves start from 0,0 and end at 1,1, so we may forget about the grid.
Any hint on Python libs that could help or any other approach ? Thanks !
I saved your image to a file 14154233_input.png. Then this program
import pylab as plt
import numpy as np
# Read image from disk and filter all grayscale
im = plt.imread("14154233_input.png")[:,:,:3]
im -= im.mean(axis=2).reshape(im.shape[0], im.shape[1], 1).repeat(3,axis=2)
im_maxnorm = im.max(axis=2)
# Find y-position of remaining line
ypos = np.ones((im.shape[1])) * np.nan
for i in range(im_maxnorm.shape[1]):
if im_maxnorm[:,i].max()<0.01:
continue
ypos[i] = np.argmax(im_maxnorm[:,i])
# Pick only values that are set
ys = 1-ypos[np.isfinite(ypos)]
# Normalize to 0,1
ys -= ys.min()
ys /= ys.max()
# Create x values
xs = np.linspace(0,1,ys.shape[0])
# Create plot of both
# read and filtered image and
# data extracted
plt.figure(figsize=(4,8))
plt.subplot(211)
plt.imshow(im_maxnorm)
plt.subplot(212, aspect="equal")
plt.plot(xs,ys)
plt.show()
Produces this plot:
You can then do with xs and ys whatever you want. Maybe you should put this code in a function that returns xs and ys or so.
One could improve the precision by fitting gaussians on each column or so. If you really need it, tell me.
First, read the image via
from scipy.misc import imread
im = imread("thefile.png")
This gives a 3D numpy array with the third dimension being the color channels (RGB+alpha). The curve is in the blue channel, but the grid is there also. But in the red channel, you have the grid and not the curve. So we use
a = im[:,:,2] - im[:,:,0]
Now, we want the position of the maximum along each column. With one pixel precision, it is given by
y0 = np.argmax(a, axis=0)
The result of this is zero when there is no blue curve in the column , i.e. outside the frame. On can get the limits of the frame by
xmin, xmax = np.where(y0>0)[0][[0,-1]
With this, you may be able to rescale x axis.
Then, you want subpixel resolution. Let us focus on a single column
f=a[:,x]
We use a single iteration of the Newton method to refine the position of an extrema
y1 = y0 - f'[y]/f''[y]
Note that we cannot iterate further because of the discreet sampling. Nontheless, we want a good approximation of the derivatives, so we will use a 5-point scheme for both.
coefprime = np.array([1,-8, 0, 8, -1], float)
coefsec = np.array([-1, 16, -30, 16, -1], float)
y1 = y0 - np.dot(f[y0-2:y0+3], coefprime)/np.dot(f[y0-2:y0+3], coefsec)
P.S. : Thorsten Kranz was faster than me (at least here), but my answer has the subpixel precision and my way of extracting the blue curve is probably more understandable.
How can I plot an 2D array as an image with Matplotlib having the y scale relative to the power of two of the y value?
For instance the first row of my array will have a height in the image of 1, the second row will have a height of 4, etc. (units are irrelevant)
It's not simple to explain with words so look at this image please (that's the kind of result I want):
alt text http://support.sas.com/rnd/app/da/new/802ce/iml/chap1/images/wavex1k.gif
As you can see the first row is 2 times smaller that the upper one, and so on.
For those interested in why I am trying to do this:
I have a pretty big array (10, 700000) of floats, representing the discrete wavelet transform coefficients of a sound file. I am trying to plot the scalogram using those coefficients.
I could copy the array x times until I get the desired image row size but the memory cannot hold so much information...
Have you tried to transform the axis? For example:
ax = subplot(111)
ax.yaxis.set_ticks([0, 2, 4, 8])
imshow(data)
This means there must be gaps in the data for the non-existent coordinates, unless there is a way to provide a transform function instead of just lists (never tried).
Edit:
I admit it was just a lead, not a complete solution. Here is what I meant in more details.
Let's assume you have your data in an array, a. You can use a transform like this one:
class arr(object):
#staticmethod
def mylog2(x):
lx = 0
while x > 1:
x >>= 1
lx += 1
return lx
def __init__(self, array):
self.array = array
def __getitem__(self, index):
return self.array[arr.mylog2(index+1)]
def __len__(self):
return 1 << len(self.array)
Basically it will transform the first coordinate of an array or list with the mylog2 function (that you can transform as you wish - it's home-made as a simplification of log2). The advantage is, you can re-use that for another transform should you need it, and you can easily control it too.
Then map your array to this one, which doesn't make a copy but a local reference in the instance:
b = arr(a)
Now you can display it, for example:
ax = subplot(111)
ax.yaxis.set_ticks([16, 8, 4, 2, 1, 0])
axis([-0.5, 4.5, 31.5, 0.5])
imshow(b, interpolation="nearest")
Here is a sample (with an array containing random values):
alt text http://img691.imageshack.us/img691/8883/clipboard01f.png
The best way I've found to make a scalogram using matplotlib is to use imshow, similar to the implementation of specgram. Using rectangles is slow, because you're having to make a separate glyph for each value. Similarly, you don't want to have to bake things into a uniform NumPy array, because you'll probably run out of memory fast, since your highest level is going to be about as long as half your signal.
Here's an example using SciPy and PyWavelets:
from pylab import *
import pywt
import scipy.io.wavfile as wavfile
# Find the highest power of two less than or equal to the input.
def lepow2(x):
return 2 ** floor(log2(x))
# Make a scalogram given an MRA tree.
def scalogram(data):
bottom = 0
vmin = min(map(lambda x: min(abs(x)), data))
vmax = max(map(lambda x: max(abs(x)), data))
gca().set_autoscale_on(False)
for row in range(0, len(data)):
scale = 2.0 ** (row - len(data))
imshow(
array([abs(data[row])]),
interpolation = 'nearest',
vmin = vmin,
vmax = vmax,
extent = [0, 1, bottom, bottom + scale])
bottom += scale
# Load the signal, take the first channel, limit length to a power of 2 for simplicity.
rate, signal = wavfile.read('kitten.wav')
signal = signal[0:lepow2(len(signal)),0]
tree = pywt.wavedec(signal, 'db5')
# Plotting.
gray()
scalogram(tree)
show()
You may also want to scale values adaptively per-level.
This works pretty well for me. The only problem I have is that matplotlib creates a hairline-thin space between levels. I'm still looking for a way to fix this.
P.S. - Even though this question is pretty old now, I figured I'd respond here, because this page came up on Google when I was looking for a method of creating scalograms using MPL.
You can look at matplotlib.image.NonUniformImage. But that only assists with having nonuniform axis - I don't think you're going to be able to plot adaptively like you want to (I think each point in the image is always going to have the same area - so you are going to have to have the wider rows multiple times). Is there any reason you need to plot the full array? Obviously the full detail isn't going to show up in any plot - so I would suggest heavily downsampling the original matrix so you can copy rows as required to get the image without running out of memory.
If you want both to be able to zoom and save memory, you could do the drawing "by hand". Matplotlib allows you to draw rectangles (they would be your "rectangular pixels"):
from matplotlib import patches
axes = subplot(111)
axes.add_patch(patches.Rectangle((0.2, 0.2), 0.5, 0.5))
Note that the extents of the axes are not set by add_patch(), but you can set them yourself to the values you want (axes.set_xlim,…).
PS: I looks to me like thrope's response (matplotlib.image.NonUniformImage) can actually do what you want, in a simpler way that the "manual" method described here!