marcoE's profile - activity

I asked this question : Question about second momentos, which @Tetragramm helped me a lot .

For me it's not clear how can I get the Second Moments of Area, I'm not sure what cv:moments returns, what "moments" .

#!/usr/bin/python
import cv2
import numpy as np

# mesh image
imMesh = cv2.imread('new_refined2__DXF.png', -1)
# layer image
imLayer1 = cv2.imread('Layer1.png', -1)
"""
binary conversions
"""
imMeshGray = cv2.cvtColor(imMesh, cv2.COLOR_BGR2GRAY)
ret, imMeshBw = cv2.threshold(imMeshGray, 250, 255, cv2.THRESH_BINARY)

imLayer1Gray = cv2.cvtColor(imLayer1, cv2.COLOR_BGR2GRAY)
(thresh, imLayerBw) = cv2.threshold(imLayer1Gray, 128, 255,
                                cv2.THRESH_BINARY | cv2.THRESH_OTSU)

cv2.imwrite('bwImlayer.jpg', imLayerBw)
connectivity = 4
output = cv2.connectedComponentsWithStats(imMeshBw, connectivity, cv2.CV_32S)
num_labels = output[0]
labels = output[1]
stats = output[2]
centroids = output[3]
file = open('moments_.txt', 'w+')
# if I want for each label
for label in np.unique(labels):
   if label == 0:
       continue
   mask = np.zeros(imMeshGray.shape, dtype="uint8")
   mask[labels == label] = 255
   maskedLayer = cv2.bitwise_and(mask, imLayerBw)

   M = cv2.moments(maskedLayer)
   m2 = cv2.moments(mask)
   Vr = float(M["m00"] / m2["m00"])

   if M["m00"] != 0:
       cX = int(M["m10"] / M["m00"])
       cY = int(M["m01"] / M["m00"])
   else:
        cX, cY = 0, 0

    file.write('{} {} {} {} {} \n'.format(Vr, M["m10"], M["m00"], cX, cY))
    cv2.putText(imMesh, "*",
                (cX, cY), cv2.FONT_HERSHEY_SIMPLEX, 0.5, (0, 0, 255), 1)
file.close()
cv2.imwrite('tstCountoursConnected_3.jpg', imMesh)

The outputs: BwImLayer

OutputCountoursConnected

Output Text

Inputs :

source Imesh

Layer image

I think this peace of code:

cX = int(M["m10"] / M["m00"])

cY = int(M["m01"] / M["m00"])

calculates the centroids, because they are related to the masked layer, so the "centers" are not the center of the "triangles". I'm not sure !

I want to calculate eigenvalues and eigenvector of the second moments of area from the center of triangle. With this result I'll have the magnitude and the direction of the "piece of the layer" in each triangle which has something.

I think with this piece of code :

Vr = float(M["m00"] / m2["m00"])

I'm computing the ration between the "white" and "black". I've followed the @Tetragramm tip

Ok, I think it is closer. This is my updated code: http://pastebin.com/u4WkuH40 and the outputs https://dl.dropboxusercontent.com/u/7... https://dl.dropboxusercontent.com/u/7... https://dl.dropboxusercontent.com/u/7... Is this right? Another Question, i want to calculate the % of white for each triangle ? The mask "cuts" the triangle contour.

a litle update without success : http://pastebin.com/xkrMatcv image : https://dl.dropboxusercontent.com/u/7...

imLayerBw is what your are saying :) https://dl.dropboxusercontent.com/u/7... Your code update doesn't have the contour code. Without the code to take the countour how can we get the moments ?

Ok, I did an update http://pastebin.com/kaC8aVQu and the output is https://dl.dropboxusercontent.com/u/7... I think, something is missing, anyway ! But is close

@Tetragramm Thank you for your support. I think there is a litle thing to solve. This is the updated code, and I think that's what you've recommended http://pastebin.com/NpjDz7Yw Is not possible do the maskedLayer = cv2.bitwise_and(mask, imLayer1) , because they don't have the same dimensions, So I assumed that was maskedLayer = cv2.bitwise_and(mask, imLayerBw) , but it does not work either. I think this is the "litle thing" to solve. it is almost done !

@Tetragramm I didn't get it :) What bitwise stuff ? I know that i need some oeprations, but with so many comments and misunderstandings (of me ) I don't know what they are :) Anyway, I added features to my code, that I think with your help will be helpful. Code: http://pastebin.com/6TgZKXmD output: https://dl.dropboxusercontent.com/u/7...

hi ! my current code : http://pastebin.com/PgucizcR This are my images https://dl.dropboxusercontent.com/u/7... https://dl.dropboxusercontent.com/u/7... But I have a problem. My "triangles" onde the layer are not "good" Because there black and white parts of the triangle, it depends on the geometry of the layer. Should I do other bitwise operation or use this image https://dl.dropboxusercontent.com/u/7...

I think that I get it : http://pastebin.com/m6FJnQpi I have the handicap with this code. I'll only see the label 14, because it's the last of the set. But I can do for all of the components. What's the next step ?

This is my example code : http://pastebin.com/WvPwebS8 How can I get just one triangle ? I'm tryng selec only the labels which are one and then write a image without success

@Tetragramm thx and I'll try. I think my problem is with "masks" and bitwise operations, to get the images that I want. But I'll do the "tests" anyway.

Completly stucked . :\

@Tetragramm I understood what you said, but i'm not able to see the triangle or how to extract it. I'll look into the docs.

@Tetragramm I've tried another approach, without success. :\ http://pastebin.com/kU5ZaSaj https://dl.dropboxusercontent.com/u/7...

No, I don't know how many triangle do I have. However, I know that are lot of labels = 0, and labels =1 ; so I can assume that are triangles with the same label. I did the tresh = 250 .

However there are more triangles than labels. For istance with label 0, it has 7884676 triangles. With the label 1, 19673, etc.

I think that finally I get the correct connectComponents value, 13053 . I still don't know how to do what I need to do :)

The highest number that I get of connect components labels is : 1740 and countours is : 172187 with this config : ret, imMesh_bw11 = cv2.threshold(imGrayMsh1, 128, 255, cv2.THRESH_BINARY) and I get this image: https://dl.dropboxusercontent.com/u/7...

@Tetragramm when I use _, contours, hier_ = cv2.findContours(imMesh_bw1,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE) cv2.drawContours(colorMask,np.array(contours),-1,(0,0,255),6) it returns 125091 contours (probably the total number of triangles), and if I draw it, it returns this image : https://dl.dropboxusercontent.com/u/7... . Could this be an alternative? i'm not handling properly connectedComponents, I think.

@Tetragramm something is wrong. I look into the labels matrix. It's size is the same of the original image. However, they are 3079 different connectedComponents . If select according to the label value, I select a lot of triangles not only one.

It seems odd to me. I have more than 3080 components. There are far more triangles.

Ok, I think that I'm understanding what you are saying. This is my current code: http://pastebin.com/HfuGUD2N I have 3080 connectComponents. But I don't know how to deal with the information

when I appy the ret, markers = cv2.connectedComponents(imMeshly_bw), with imMeshLayer = cv2.imread('testLayer6_.png', -1) imGrayMshLayer = cv2.cvtColor(imMeshLayer, cv2.COLOR_BGR2GRAY) (thresh, imMeshly_bw) = cv2.threshold(imGrayMshLayer, 128, 255, cv2.THRESH_BINARY | cv2.THRESH_OTSU) I got : https://dl.dropboxusercontent.com/u/7...

@Tetragramm you've said " Make a copy of your layer image and use bitwise_and with the single triangle to erase everything but what is inside that triangle." How can I select just one triangle for each time. I think that I'm understanding the process. Having n-masks, one for each triangle. Then for each mesh-triangle I can calc moments, centroids, etc . However. I don't get the "process" to reach it.

@Tetragramm the connectComponents function is not available in python :\ Any alternative, or I should just implement in C++ ? It won't be easy to me, I'm used to use opencv python

@Tetragramm is it something like this that is need to calc the moments per triangle? https://dl.dropboxusercontent.com/u/7...

another alternative approach. If I have the mesh and the "layer" where the mesh is. Is it easier to isolate? Mesh image: https://dl.dropboxusercontent.com/u/7... Layer image: https://dl.dropboxusercontent.com/u/7... code: http://pastebin.com/Q3sN5akv

@Tetragramm It's not easy. Should I keep all the mesh contours in white? Or some in white and others in black? Could I use other color to all mesh, and then calculate moments?

link text I've cut a litle piece. I want to calculate the volumes and the second moments of intertia of each triangle. I've put some numbers just to have an example. So, the number 1 has 0% of Volume, it's empty, the seconds moments are null. The number 2 has 100% of volume, second moments would have big values. Number 3 would have 60% of volume more or less, and the second moments would be intermediate. Number 4 it's a big hole, I don't want to calc anything.

I think it's clearer what I'd like to calculate.

@sturkmen , what is big ? :)

Hello! Finally I was able to properly fit he mesh with the layer. For this image (7158x16392 8MB)

I need to calculate the percentage of blacks and browns per triangle, and also the seconds moments of area per triangle. I'm facing some problems. The mesh layer should always be the master. For instance, If I have a triangle with all black, it should return 100% empty and without second moments.

So, I'm facing some problems to find out the proper bit-wise function, which allows me to make the mesh being the "master", and then I don't know how to iterate over a such huge amount of contours and make the calcs. I think the second moments which are native supported by opencv are the seconds moments of inertia, not the second moments of area..

@theodore another question, that could be significant. Having the stl coordinates in numpy array, is it possible draw the mesh in opencv instead of using matplotlib to render it? I think it could reduce the errors between scale resolution

@theodore I agree with you. I was thinking if it would be better create at source image (the one which I use to generate the mesh) 3 or 4 points as guides, and then plot them on the mesh with the transformation and them draw them on the layer.

@theodore did you understand what my problem is? I think in a simple chat i could explain easily