Sliding window using as_strided function in numpy? - image-processing

As I get to implement a sliding window using python to detect objects in still images, I get to know the nice function:
numpy.lib.stride_tricks.as_strided
So I tried to achieve a general rule to avoid mistakes I may fail in while changing the size of the sliding windows I need. Finally I got this representation:
all_windows = as_strided(x,((x.shape[0] - xsize)/xstep ,(x.shape[1] - ysize)/ystep ,xsize,ysize), (x.strides[0]*xstep,x.strides[1]*ystep,x.strides[0],x.strides[1])
which results in a 4 dim matrix. The first two represents the number of windows on the x and y axis of the image. and the others represent the size of the window (xsize,ysize)
and the step represents the displacement from between two consecutive windows.
This representation works fine if I choose a squared sliding windows. but still I have a problem in getting this to work for windows of e.x. (128,64), where I get usually unrelated data to the image.
What is wrong my code. Any ideas? and if there is a better way to get a sliding windows nice and neat in python for image processing?
Thanks

There is an issue in your code. Actually this code work good for 2D and no reason to use multi dimensional version (Using strides for an efficient moving average filter). Below is a fixed version:
A = np.arange(100).reshape((10, 10))
print A
all_windows = as_strided(A, ((A.shape[0] - xsize + 1) / xstep, (A.shape[1] - ysize + 1) / ystep, xsize, ysize),
(A.strides[0] * xstep, A.strides[1] * ystep, A.strides[0], A.strides[1]))
print all_windows

Check out the answers to this question: Using strides for an efficient moving average filter. Basically strides are not a great option, although they work.

For posteriority:
This is implemented in scikit-learn in the function sklearn.feature_extraction.image.extract_patches.

I had a similar use-case where I needed to create sliding windows over a batch of multi-channel images and ended up coming up with the below function. I've written a more in-depth blog post covering this in regards to manually creating a Convolution layer. This function implements the sliding windows and also includes dilating or adding padding to the input array.
The function takes as input:
input - Size of (Batch, Channel, Height, Width)
output_size - Depends on usage, comments below.
kernel_size - size of the sliding window you wish to create (square)
padding - amount of 0-padding added to the outside of the (H,W) dimensions
stride - stride the sliding window should take over the inputs
dilate - amount to spread the cells of the input. This adds 0-filled rows/cols between elements
Typically, when performing forward convolution, you do not need to perform dilation so your output size can be found be using the following formula (replace x with input dimension):
(x - kernel_size + 2 * padding) // stride + 1
When performing the backwards pass of convolution with this function, use a stride of 1 and set your output_size to the size of your forward pass's x-input
Sample code with an example of using this function can be found at this link.
def getWindows(input, output_size, kernel_size, padding=0, stride=1, dilate=0):
working_input = input
working_pad = padding
# dilate the input if necessary
if dilate != 0:
working_input = np.insert(working_input, range(1, input.shape[2]), 0, axis=2)
working_input = np.insert(working_input, range(1, input.shape[3]), 0, axis=3)
# pad the input if necessary
if working_pad != 0:
working_input = np.pad(working_input, pad_width=((0,), (0,), (working_pad,), (working_pad,)), mode='constant', constant_values=(0.,))
in_b, in_c, out_h, out_w = output_size
out_b, out_c, _, _ = input.shape
batch_str, channel_str, kern_h_str, kern_w_str = working_input.strides
return np.lib.stride_tricks.as_strided(
working_input,
(out_b, out_c, out_h, out_w, kernel_size, kernel_size),
(batch_str, channel_str, stride * kern_h_str, stride * kern_w_str, kern_h_str, kern_w_str)
)

Related

Placing a shape inside another shape using opencv

I have two images and I need to place the second image inside the first image. The second image can be resized, rotated or skewed such that it covers a larger area of the other images as possible. As an example, in the figure shown below, the green circle need to be placed inside the blue shape:
Here the green circle is transformed such that it covers a larger area. Another example is shown below:
Note that there may be some multiple results. However, any similar result is acceptable as shown in the above example.
How do I solve this problem?
Thanks in advance!
I tested the idea I mentioned earlier in the comments and the output is almost good. It may be better but it takes time. The final code was too much and it depends on one of my old personal projects, so I will not share. But I will explain step by step how I wrote such an algorithm. Note that I have tested the algorithm many times. Not yet 100% accurate.
for N times do this:
1. Copy from shape
2. Transform it randomly
3. Put the shape on the background
4-1. It is not acceptable if the shape exceeds the background. Go to
the first step.
4.2. Otherwise we will continue to step 5.
5. We calculate the length, width and number of shape pixels.
6. We keep a list of the best candidates and compare these three
parameters (W, H, Pixels) with the members of the list. If we
find a better item, we will save it.
I set the value of N to 5,000. The larger the number, the slower the algorithm runs, but the better the result.
You can use anything for Transform. Mirror, Rotate, Shear, Scale, Resize, etc. But I used warpPerspective for this one.
im1 = cv2.imread(sys.path[0]+'/Back.png')
im2 = cv2.imread(sys.path[0]+'/Shape.png')
bH, bW = im1.shape[:2]
sH, sW = im2.shape[:2]
# TopLeft, TopRight, BottomRight, BottomLeft of the shape
_inp = np.float32([[0, 0], [sW, 0], [sW, sH], [0, sH]])
cx = random.randint(5, sW-5)
ch = random.randint(5, sH-5)
o = 0
# Random transformed output
_out = np.float32([
[random.randint(-o, cx-1), random.randint(1-o, ch-1)],
[random.randint(cx+1, sW+o), random.randint(1-o, ch-1)],
[random.randint(cx+1, sW+o), random.randint(ch+1, sH+o)],
[random.randint(-o, cx-1), random.randint(ch+1, sH+o)]
])
# Transformed output
M = cv2.getPerspectiveTransform(_inp, _out)
t = cv2.warpPerspective(shape, M, (bH, bW))
You can use countNonZero to find the number of pixels and findContours and boundingRect to find the shape size.
def getSize(msk):
cnts, _ = cv2.findContours(msk, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)
cnts.sort(key=lambda p: max(cv2.boundingRect(p)[2],cv2.boundingRect(p)[3]), reverse=True)
w,h=0,0
if(len(cnts)>0):
_, _, w, h = cv2.boundingRect(cnts[0])
pix = cv2.countNonZero(msk)
return pix, w, h
To find overlaping of back and shape you can do something like this:
make a mask from back and shape and use bitwise methods; Change this section according to the software you wrote. This is just an example :)
mskMix = cv2.bitwise_and(mskBack, mskShape)
mskMix = cv2.bitwise_xor(mskMix, mskShape)
isCandidate = not np.any(mskMix == 255)
For example this is not a candidate answer; This is because if you look closely at the image on the right, you will notice that the shape has exceeded the background.
I just tested the circle with 4 different backgrounds; And the results:
After 4879 Iterations:
After 1587 Iterations:
After 4621 Iterations:
After 4574 Iterations:
A few additional points. If you use a method like medianBlur to cover the noise in the Background mask and Shape mask, you may find a better solution.
I suggest you read about Evolutionary Computation, Metaheuristic and Soft Computing algorithms for better understanding of this algorithm :)

How to calculate the output size of a convoluitonal layer in YOLO?

This is the architecture of YOLO. I am trying to calculate the output size of each layer myself, but I can't get the size as described in the paper.
For example, in the first Conv Layer, the input size is 448x448 but it uses a 7x7 filter with stride 2, but according to this equation W2=(W1−F+2P)/S+1 = (448 - 7 + 0)/2 + 1, I can't get an integer result, so the filter size seems to be unsuitable to the input size.
So anyone can explain this problem? Did I miss something or misunderstand the YOLO architecture?
As Hawx Won said, the input image has been added extra 3 paddings, and here is how it works from the source code.
For convolution layers, if pad is enabled, The padding value of each layer will be calculated by:
# In parser.c
if(pad) padding = size/2;
# In convolutional_layer.c
l.pad = padding;
Where size is the shape of the filter.
So, for the first layer: padding = size/2 = 7/2=3
Then the output of first convolutional layer should be:
output_w = (input_w+2*pad-size)/stride+1 = (448+6-7)/2+1 = 224
output_h = (input_h+2*pad-size)/stride+1 = (448+6-7)/2+1 = 224
Well, I spent some time learning the source code, and learned about that the input image has added extra 3 paddings on top,down,left and right side of the image, so the image size becomes (448+2x3)=454, the out put size of valid padding should be calculated in this way:
Output_size=ceil((W-F+1)/S)=(454-7+1)/2=224, therefore, output size should be 224x224x64
I hope this could be helpful

How to apply a Hanning window without changing calibration scale of an image in Digital Micrograph?

I want to apply a Hanning window to an image before I process 'FFT'. I found a script written by Ruben Bjorge:
number size, sizeX, sizeY, top, left, bottom, right, ii, posX, posY
image front, hannX, hannY, hann, avg, hannout
front := GetFrontImage();
GetSize(front, sizeX, sizeY);
GetSelection(front, top, left, bottom, right);
// Create Hanning window.
ii = 1;
hannX := CreateFloatImage("", (right-left), (bottom-top));
hannX = 0;
hannX[0, 0, 1, (right-left)] = 1 - cos( 2 * Pi() * icol / (right-left));
while( ii < (bottom-top) )
{
hannX[ii, 0, 2*ii, (right-left)] = hannX[0, 0, ii, (right-left)];
ii = ii * 2;
}
ii = 1;
hannY := CreateFloatImage("", (right-left), (bottom-top));
hannY = 0;
hannY[0, 0, (bottom-top), 1] = 1 - cos( 2 * Pi() * irow / (bottom-top));
while( ii < (right-left) )
{
hannY[0, ii, (bottom-top), 2*ii] = hannY[0, 0, (bottom-top), ii];
ii = ii * 2;
}
hann = hannX * hannY;
// Subtract average from image.
avg = front - Average(front);
// Multiply with Hanning window.
hannout = avg[top, left, bottom, right] * hann;
// Do fast Fourier transform and display image.
fft = RealFFT(hannout);
By using this script, the calibration scale of the FFT is changed to 1. But it should be 0.11948, as all shown in the pictures below.
My question are:
Is there a way to apply hanning window without changing the calibration scale of image?
Or how to calculate the scale of the FFT image in terms of the scale of original image?
Since the rest part of my script needs the correct scale of the fft image, I'll appreciate that if anyone could answer this long question. Thanks.
The key to preserving calibration info in DM image expressions is to work with image references and in-place operations on clones of image objects as much as possible. Your example script becomes much simpler and more efficient when you transform it to use such techniques, as follows:
Image frontImage := GetFrontImage();
// Step 1 - extract and get info about the front image selection
Image frontSelection := frontImage[];
Number selW = ImageGetDimensionSize(frontSelection, 0);
Number selH = ImageGetDimensionSize(frontSelection, 1);
// Step 2 - subtract average value from selection and apply Hanning window
Image filteredSelection := ImageClone(frontSelection);
filteredSelection -= Average(frontSelection);
filteredSelection *= (1 - cos(2 * Pi() * icol / selW));
filteredSelection *= (1 - cos(2 * Pi() * irow / selH));
String selectionName = ImageGetName(frontImage) + " filtered selection";
ImageSetName(filteredSelection, selectionName);
// Step 3 - take FFT of filtered selection and display result
Image filteredFFT := RealFFT(filteredSelection);
ShowImage(filteredFFT);
The main differences here are in the two sections marked Step 1 and Step 2.
In Step 1, this script directly accesses the front image selection by using the selection operator, '[ ]'. This operator preserves the calibration (and tag) info of the original image.
In Step 2, the ImageClone function makes a complete copy of the selection's image object, including its calibration (and tag) data. The next three lines do the mathematical processing in place, directly on the cloned selection. Note, in particular, that the Hanning window factors are applied very simply with a single image expression that is automatically applied to all the pixels of the result image. There is no need for the while loops used in the example script. In addition to adding code complexity, these are significantly slower than the implied looping done by the one-line image expressions. In fact, the Hanning window can be applied in a single line that includes both the x- and y-dependent factors, as follows:
filteredSelection *= (1-cos(2*Pi() * icol/selW)) * (1-cos(2*Pi() * irow/selH));
Your main question is about the propagation of calibration info when performing mathematical operations on DM image objects. The short answer is that such info, as well as all the tag data attached to an image, is not transferred to the result image when one uses a simple assignment with an equal sign. Each such image expression effectively allocates a new real image (with no calibration, tags, or name) and only its pixel values are transferred to the result image (which is also an uncalibrated, unnamed real image, by default).
There are several ways to make this script do what you want much more simply and efficiently (I will post a second answer to show this), but the minimal change to get the desired result is to replace the last line with the following three lines:
ImageCopyCalibrationFrom(hannout, front);
Image fft := RealFFT(hannout);
ShowImage(fft);
The first line transfers the calibration from the original front image to the filtered one from which you actually want a calibrated Fourier transform. The second line uses the ':=' operator to have the image variable 'fft' point directly to the output of the RealFFT function, thereby preserving its calibration info. In other words, this is an assignment by reference, not an image expression, and it bypasses the allocation of default (uncalibrated) real images for the intermediate and final result. The third line actually shows the result (which seems to be missing from your sample code).

Automatic approach for removing colord object shadow on white background?

I am working on some leaf images using OpenCV (Java). The leaves are captured on a white paper and some has shadows like this one:
Of course, it's somehow the extreme case (there are milder shadows).
Now, I want to threshold the leaf and also remove the shadow (while reserving the leaf's details).
My current flow is this:
1) Converting to HSV and extracting the Saturation channel:
Imgproc.cvtColor(colorMat, colorMat, Imgproc.COLOR_RGB2HSV);
ArrayList<Mat> channels = new ArrayList<Mat>();
Core.split(colorMat, channels);
satImg = channels.get(1);
2) De-noising (median) and applying adaptiveThreshold:
Imgproc.medianBlur(satImg , satImg , 11);
Imgproc.adaptiveThreshold(satImg , satImg , 255, Imgproc.ADAPTIVE_THRESH_MEAN_C, Imgproc.THRESH_BINARY, 401, -10);
And the result is this:
It looks OK, but the shadow is causing some anomalies along the left boundary. Also, I have this feeling that I am not using the white background to my benefit.
Now, I have 2 questions:
1) How can I improve the result and get rid of the shadow?
2) Can I get good results without working on saturation channel?. The reason I ask is that on most of my images, working on L channel (from HLS) gives way better results (apart from the shadow, of course).
Update: Using the Hue channel makes threshdolding better, but makes the shadow situation worse:
Update2: In some cases, the assumption that the shadow is darker than the leaf doesn't always hold. So, working on intensities won't help. I'm looking more toward a color channels approach.
I don't use opencv, instead I was trying to use matlab image processing toolbox to extract the leaf. Hopefully opencv has all the processing functions for you. Please see my result below. I did all the operations in your original image channel 3 and channel 1.
First I used your channel 3, threshold it with 100 (left top). Then I remove the regions on the border and regions with the pixel size smaller than 100, filling in the hole in the leaf, the result is shown in right top.
Next I used your channel 1, did the same thing as I did in channel 3, the result is shown in left bottom. Then I found out the connected regions (there are only two as you can see in the left bottom figure), remove the one with smaller area (shown in right bottom).
Suppose the right top image is I1, and the right bottom image is I, the leaf is extracted by implement ~I && I1. The leaf is:
Hope it helps. Thanks
I tried two different things:
1. other thresholding on the saturation channel
2. try to find two contours: shadow and leaf
I use c++ so your code snippets will look a little different.
trying otsu-thresholding instead of adaptive thresholding:
cv::threshold(hsv_imgs,mask,0,255,CV_THRESH_BINARY|CV_THRESH_OTSU);
leading to following images (just OTSU thresholding on saturation channel):
the other thing is computing gradient information (i used sobel, see oppenCV documentation), thresholding that and after an opening-operator I used findContours giving something like this, not useable yet (gradient contour approach):
I'm trying to do the same thing with photos of butterflies, but with more uneven and unpredictable backgrounds such as this. Once you've identified a good portion of the background (e.g. via thresholding, or as we do, flood filling from random points), what works well is to use the GrabCut algorithm to get all those bits you might miss on the initial pass. In python, assuming you still want to identify an initial area of background by thresholding on the saturation channel, try something like
import cv2
import numpy as np
img = cv2.imread("leaf.jpg")
sat = cv2.cvtColor(img, cv2.COLOR_BGR2HSV)[:,:,1]
sat = cv2.medianBlur(sat, 11)
thresh = cv2.adaptiveThreshold(sat , 255, cv2.ADAPTIVE_THRESH_MEAN_C, cv2.THRESH_BINARY, 401, 10);
cv2.imwrite("thresh.jpg", thresh)
h, w = img.shape[:2]
bgdModel = np.zeros((1,65),np.float64)
fgdModel = np.zeros((1,65),np.float64)
grabcut_mask = thresh/255*3 #background should be 0, probable foreground = 3
cv2.grabCut(img, grabcut_mask,(0,0,w,h),bgdModel,fgdModel,5,cv2.GC_INIT_WITH_MASK)
grabcut_mask = np.where((grabcut_mask ==2)|(grabcut_mask ==0),0,1).astype('uint8')
cv2.imwrite("GrabCut1.jpg", img*grabcut_mask[...,None])
This actually gets rid of the shadows for you in this case, because the edge of the shadow actually has high saturation levels, so is included in the grab cut deletion. (I would post images, but don't have enough reputation)
Usually, however, you can't trust shadows to be included in the background detection. In this case you probably want to compare areas in the image with colour of the now-known background using the chromacity distortion measure proposed by Horprasert et. al. (1999) in "A Statistical Approach for Real-time Robust Background Subtraction and Shadow Detection". This measure takes account of the fact that for desaturated colours, hue is not a relevant measure.
Note that the pdf of the preprint you find online has a mistake (no + signs) in equation 6. You can use the version re-quoted in Rodriguez-Gomez et al (2012), equations 1 & 2. Or you can use my python code below:
def brightness_distortion(I, mu, sigma):
return np.sum(I*mu/sigma**2, axis=-1) / np.sum((mu/sigma)**2, axis=-1)
def chromacity_distortion(I, mu, sigma):
alpha = brightness_distortion(I, mu, sigma)[...,None]
return np.sqrt(np.sum(((I - alpha * mu)/sigma)**2, axis=-1))
You can feed the known background mean & stdev as the last two parameters of the chromacity_distortion function, and the RGB pixel image as the first parameter, which should show you that the shadow is basically the same chromacity as the background, and very different from the leaf. In the code below, I've then thresholded on chromacity, and done another grabcut pass. This works to remove the shadow even if the first grabcut pass doesn't (e.g. if you originally thresholded on hue)
mean, stdev = cv2.meanStdDev(img, mask = 255-thresh)
mean = mean.ravel() #bizarrely, meanStdDev returns an array of size [3,1], not [3], so flatten it
stdev = stdev.ravel()
chrom = chromacity_distortion(img, mean, stdev)
chrom255 = cv2.normalize(chrom, alpha=0, beta=255, norm_type=cv2.NORM_MINMAX).astype(np.uint8)[:,:,None]
cv2.imwrite("ChromacityDistortionFromBackground.jpg", chrom255)
thresh2 = cv2.adaptiveThreshold(chrom255 , 255, cv2.ADAPTIVE_THRESH_MEAN_C, cv2.THRESH_BINARY, 401, 10);
cv2.imwrite("thresh2.jpg", thresh2)
grabcut_mask[...] = 3
grabcut_mask[thresh==0] = 0 #where thresh == 0, definitely background, set to 0
grabcut_mask[np.logical_and(thresh == 255, thresh2 == 0)] = 2 #could try setting this to 2 or 0
cv2.grabCut(img, grabcut_mask,(0,0,w,h),bgdModel,fgdModel,5,cv2.GC_INIT_WITH_MASK)
grabcut_mask = np.where((grabcut_mask ==2)|(grabcut_mask ==0),0,1).astype('uint8')
cv2.imwrite("final_leaf.jpg", grabcut_mask[...,None]*img)
I'm afraid with the parameters I tried, this still removes the stalk, though. I think that's because GrabCut thinks that it looks a similar colour to the shadows. Let me know if you find a way to keep it.

How to define the markers for Watershed in OpenCV?

I'm writing for Android with OpenCV. I'm segmenting an image similar to below using marker-controlled watershed, without the user manually marking the image. I'm planning to use the regional maxima as markers.
minMaxLoc() would give me the value, but how can I restrict it to the blobs which is what I'm interested in? Can I utilize the results from findContours() or cvBlob blobs to restrict the ROI and apply maxima to each blob?
First of all: the function minMaxLoc finds only the global minimum and global maximum for a given input, so it is mostly useless for determining regional minima and/or regional maxima. But your idea is right, extracting markers based on regional minima/maxima for performing a Watershed Transform based on markers is totally fine. Let me try to clarify what is the Watershed Transform and how you should correctly use the implementation present in OpenCV.
Some decent amount of papers that deal with watershed describe it similarly to what follows (I might miss some detail, if you are unsure: ask). Consider the surface of some region you know, it contains valleys and peaks (among other details that are irrelevant for us here). Suppose below this surface all you have is water, colored water. Now, make holes in each valley of your surface and then the water starts to fill all the area. At some point, differently colored waters will meet, and when this happen, you construct a dam such that they don't touch each other. In the end you have a collection of dams, which is the watershed separating all the different colored water.
Now, if you make too many holes in that surface, you end up with too many regions: over-segmentation. If you make too few you get an under-segmentation. So, virtually any paper that suggests using watershed actually presents techniques to avoid these problems for the application the paper is dealing with.
I wrote all this (which is possibly too naïve for anyone that knows what the Watershed Transform is) because it reflects directly on how you should use watershed implementations (which the current accepted answer is doing in a completely wrong manner). Let us start on the OpenCV example now, using the Python bindings.
The image presented in the question is composed of many objects that are mostly too close and in some instances overlapping. The usefulness of watershed here is to separate correctly these objects, not to group them into a single component. So you need at least one marker for each object and good markers for the background. As an example, first binarize the input image by Otsu and perform a morphological opening for removing small objects. The result of this step is shown below in the left image. Now with the binary image consider applying the distance transform to it, result at right.
With the distance transform result, we can consider some threshold such that we consider only the regions most distant to the background (left image below). Doing this, we can obtain a marker for each object by labeling the different regions after the earlier threshold. Now, we can also consider the border of a dilated version of the left image above to compose our marker. The complete marker is shown below at right (some markers are too dark to be seen, but each white region in the left image is represented at the right image).
This marker we have here makes a lot of sense. Each colored water == one marker will start to fill the region, and the watershed transformation will construct dams to impede that the different "colors" merge. If we do the transform, we get the image at left. Considering only the dams by composing them with the original image, we get the result at right.
import sys
import cv2
import numpy
from scipy.ndimage import label
def segment_on_dt(a, img):
border = cv2.dilate(img, None, iterations=5)
border = border - cv2.erode(border, None)
dt = cv2.distanceTransform(img, 2, 3)
dt = ((dt - dt.min()) / (dt.max() - dt.min()) * 255).astype(numpy.uint8)
_, dt = cv2.threshold(dt, 180, 255, cv2.THRESH_BINARY)
lbl, ncc = label(dt)
lbl = lbl * (255 / (ncc + 1))
# Completing the markers now.
lbl[border == 255] = 255
lbl = lbl.astype(numpy.int32)
cv2.watershed(a, lbl)
lbl[lbl == -1] = 0
lbl = lbl.astype(numpy.uint8)
return 255 - lbl
img = cv2.imread(sys.argv[1])
# Pre-processing.
img_gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
_, img_bin = cv2.threshold(img_gray, 0, 255,
cv2.THRESH_OTSU)
img_bin = cv2.morphologyEx(img_bin, cv2.MORPH_OPEN,
numpy.ones((3, 3), dtype=int))
result = segment_on_dt(img, img_bin)
cv2.imwrite(sys.argv[2], result)
result[result != 255] = 0
result = cv2.dilate(result, None)
img[result == 255] = (0, 0, 255)
cv2.imwrite(sys.argv[3], img)
I would like to explain a simple code on how to use watershed here. I am using OpenCV-Python, but i hope you won't have any difficulty to understand.
In this code, I will be using watershed as a tool for foreground-background extraction. (This example is the python counterpart of the C++ code in OpenCV cookbook). This is a simple case to understand watershed. Apart from that, you can use watershed to count the number of objects in this image. That will be a slightly advanced version of this code.
1 - First we load our image, convert it to grayscale, and threshold it with a suitable value. I took Otsu's binarization, so it would find the best threshold value.
import cv2
import numpy as np
img = cv2.imread('sofwatershed.jpg')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
ret,thresh = cv2.threshold(gray,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
Below is the result I got:
( even that result is good, because great contrast between foreground and background images)
2 - Now we have to create the marker. Marker is the image with same size as that of original image which is 32SC1 (32 bit signed single channel).
Now there will be some regions in the original image where you are simply sure, that part belong to foreground. Mark such region with 255 in marker image. Now the region where you are sure to be the background are marked with 128. The region you are not sure are marked with 0. That is we are going to do next.
A - Foreground region:- We have already got a threshold image where pills are white color. We erode them a little, so that we are sure remaining region belongs to foreground.
fg = cv2.erode(thresh,None,iterations = 2)
fg :
B - Background region :- Here we dilate the thresholded image so that background region is reduced. But we are sure remaining black region is 100% background. We set it to 128.
bgt = cv2.dilate(thresh,None,iterations = 3)
ret,bg = cv2.threshold(bgt,1,128,1)
Now we get bg as follows :
C - Now we add both fg and bg :
marker = cv2.add(fg,bg)
Below is what we get :
Now we can clearly understand from above image, that white region is 100% foreground, gray region is 100% background, and black region we are not sure.
Then we convert it into 32SC1 :
marker32 = np.int32(marker)
3 - Finally we apply watershed and convert result back into uint8 image:
cv2.watershed(img,marker32)
m = cv2.convertScaleAbs(marker32)
m :
4 - We threshold it properly to get the mask and perform bitwise_and with the input image:
ret,thresh = cv2.threshold(m,0,255,cv2.THRESH_BINARY+cv2.THRESH_OTSU)
res = cv2.bitwise_and(img,img,mask = thresh)
res :
Hope it helps!!!
ARK
Foreword
I'm chiming in mostly because I found both the watershed tutorial in the OpenCV documentation (and C++ example) as well as mmgp's answer above to be quite confusing. I revisited a watershed approach multiple times to ultimately give up out of frustration. I finally realized I needed to at least give this approach a try and see it in action. This is what I've come up with after sorting out all of the tutorials I've come across.
Aside from being a computer vision novice, most of my trouble probably had to do with my requirement to use the OpenCVSharp library rather than Python. C# doesn't have baked-in high-power array operators like those found in NumPy (though I realize this has been ported via IronPython), so I struggled quite a bit in both understanding and implementing these operations in C#. Also, for the record, I really despise the nuances of, and inconsistencies in most of these function calls. OpenCVSharp is one of the most fragile libraries I've ever worked with. But hey, it's a port, so what was I expecting? Best of all, though -- it's free.
Without further ado, let's talk about my OpenCVSharp implementation of the watershed, and hopefully clarify some of the stickier points of watershed implementation in general.
Application
First of all, make sure watershed is what you want and understand its use. I am using stained cell plates, like this one:
It took me a good while to figure out I couldn't just make one watershed call to differentiate every cell in the field. On the contrary, I first had to isolate a portion of the field, then call watershed on that small portion. I isolated my region of interest (ROI) via a number of filters, which I will explain briefly here:
Start with source image (left, cropped for demonstration purposes)
Isolate the red channel (left middle)
Apply adaptive threshold (right middle)
Find contours then eliminate those with small areas (right)
Once we have cleaned the contours resulting from the above thresholding operations, it is time to find candidates for watershed. In my case, I simply iterated through all contours greater than a certain area.
Code
Say we've isolated this contour from the above field as our ROI:
Let's take a look at how we'll code up a watershed.
We'll start with a blank mat and draw only the contour defining our ROI:
var isolatedContour = new Mat(source.Size(), MatType.CV_8UC1, new Scalar(0, 0, 0));
Cv2.DrawContours(isolatedContour, new List<List<Point>> { contour }, -1, new Scalar(255, 255, 255), -1);
In order for the watershed call to work, it will need a couple of "hints" about the ROI. If you're a complete beginner like me, I recommend checking out the CMM watershed page for a quick primer. Suffice to say we're going to create hints about the ROI on the left by creating the shape on the right:
To create the white part (or "background") of this "hint" shape, we'll just Dilate the isolated shape like so:
var kernel = Cv2.GetStructuringElement(MorphShapes.Ellipse, new Size(2, 2));
var background = new Mat();
Cv2.Dilate(isolatedContour, background, kernel, iterations: 8);
To create the black part in the middle (or "foreground"), we'll use a distance transform followed by threshold, which takes us from the shape on the left to the shape on the right:
This takes a few steps, and you may need to play around with the lower bound of your threshold to get results that work for you:
var foreground = new Mat(source.Size(), MatType.CV_8UC1);
Cv2.DistanceTransform(isolatedContour, foreground, DistanceTypes.L2, DistanceMaskSize.Mask5);
Cv2.Normalize(foreground, foreground, 0, 1, NormTypes.MinMax); //Remember to normalize!
foreground.ConvertTo(foreground, MatType.CV_8UC1, 255, 0);
Cv2.Threshold(foreground, foreground, 150, 255, ThresholdTypes.Binary);
Then we'll subtract these two mats to get the final result of our "hint" shape:
var unknown = new Mat(); //this variable is also named "border" in some examples
Cv2.Subtract(background, foreground, unknown);
Again, if we Cv2.ImShow unknown, it would look like this:
Nice! This was easy for me to wrap my head around. The next part, however, got me quite puzzled. Let's look at turning our "hint" into something the Watershed function can use. For this we need to use ConnectedComponents, which is basically a big matrix of pixels grouped by the virtue of their index. For example, if we had a mat with the letters "HI", ConnectedComponents might return this matrix:
0 0 0 0 0 0 0 0 0
0 1 0 1 0 2 2 2 0
0 1 0 1 0 0 2 0 0
0 1 1 1 0 0 2 0 0
0 1 0 1 0 0 2 0 0
0 1 0 1 0 2 2 2 0
0 0 0 0 0 0 0 0 0
So, 0 is the background, 1 is the letter "H", and 2 is the letter "I". (If you get to this point and want to visualize your matrix, I recommend checking out this instructive answer.) Now, here's how we'll utilize ConnectedComponents to create the markers (or labels) for watershed:
var labels = new Mat(); //also called "markers" in some examples
Cv2.ConnectedComponents(foreground, labels);
labels = labels + 1;
//this is a much more verbose port of numpy's: labels[unknown==255] = 0
for (int x = 0; x < labels.Width; x++)
{
for (int y = 0; y < labels.Height; y++)
{
//You may be able to just send "int" in rather than "char" here:
var labelPixel = (int)labels.At<char>(y, x); //note: x and y are inexplicably
var borderPixel = (int)unknown.At<char>(y, x); //and infuriatingly reversed
if (borderPixel == 255)
labels.Set(y, x, 0);
}
}
Note that the Watershed function requires the border area to be marked by 0. So, we've set any border pixels to 0 in the label/marker array.
At this point, we should be all set to call Watershed. However, in my particular application, it is useful just to visualize a small portion of the entire source image during this call. This may be optional for you, but I first just mask off a small bit of the source by dilating it:
var mask = new Mat();
Cv2.Dilate(isolatedContour, mask, new Mat(), iterations: 20);
var sourceCrop = new Mat(source.Size(), source.Type(), new Scalar(0, 0, 0));
source.CopyTo(sourceCrop, mask);
And then make the magic call:
Cv2.Watershed(sourceCrop, labels);
Results
The above Watershed call will modify labels in place. You'll have to go back to remembering about the matrix resulting from ConnectedComponents. The difference here is, if watershed found any dams between watersheds, they will be marked as "-1" in that matrix. Like the ConnectedComponents result, different watersheds will be marked in a similar fashion of incrementing numbers. For my purposes, I wanted to store these into separate contours, so I created this loop to split them up:
var watershedContours = new List<Tuple<int, List<Point>>>();
for (int x = 0; x < labels.Width; x++)
{
for (int y = 0; y < labels.Height; y++)
{
var labelPixel = labels.At<Int32>(y, x); //note: x, y switched
var connected = watershedContours.Where(t => t.Item1 == labelPixel).FirstOrDefault();
if (connected == null)
{
connected = new Tuple<int, List<Point>>(labelPixel, new List<Point>());
watershedContours.Add(connected);
}
connected.Item2.Add(new Point(x, y));
if (labelPixel == -1)
sourceCrop.Set(y, x, new Vec3b(0, 255, 255));
}
}
Then, I wanted to print these contours with random colors, so I created the following mat:
var watershed = new Mat(source.Size(), MatType.CV_8UC3, new Scalar(0, 0, 0));
foreach (var component in watershedContours)
{
if (component.Item2.Count < (labels.Width * labels.Height) / 4 && component.Item1 >= 0)
{
var color = GetRandomColor();
foreach (var point in component.Item2)
watershed.Set(point.Y, point.X, color);
}
}
Which yields the following when shown:
If we draw on the source image the dams that were marked by a -1 earlier, we get this:
Edits:
I forgot to note: make sure you're cleaning up your mats after you're done with them. They WILL stay in memory and OpenCVSharp may present with some unintelligible error message. I should really be using using above, but mat.Release() is an option as well.
Also, mmgp's answer above includes this line: dt = ((dt - dt.min()) / (dt.max() - dt.min()) * 255).astype(numpy.uint8), which is a histogram stretching step applied to the results of the distance transform. I omitted this step for a number of reasons (mostly because I didn't think the histograms I saw were too narrow to begin with), but your mileage may vary.

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