generic-library/opencv
1
0
Fork 0
Code Issues Pull Requests Releases Wiki Activity

Doxygen tutorials: python basic

Browse Source
This commit is contained in:
Maksim Shabunin
2014-11-28 17:18:32 +03:00
parent 36a04ef8de
commit 875f922332
80 changed files with 9240 additions and 2 deletions
Download Patch File Download Diff File Expand all files Collapse all files

10
doc/py_tutorials/py_ml/py_kmeans/py_kmeans_index.markdown Normal file
View File

@@ -0,0 +1,10 @@
K-Means Clustering {#tutorial_py_kmeans_index}
==================
- @subpage tutorial_py_kmeans_understanding
Read to get an intuitive understanding of K-Means Clustering
- @subpage tutorial_py_kmeans_opencv
Now let's try K-Means functions in OpenCV

201
doc/py_tutorials/py_ml/py_kmeans/py_kmeans_opencv/py_kmeans_opencv.markdown Normal file
View File

@@ -0,0 +1,201 @@
K-Means Clustering in OpenCV {#tutorial_py_kmeans_opencv}
============================
Goal
----
- Learn to use **cv2.kmeans()** function in OpenCV for data clustering
Understanding Parameters
------------------------
### Input parameters
-# **samples** : It should be of **np.float32** data type, and each feature should be put in a
single column.
2. **nclusters(K)** : Number of clusters required at end
3.
**criteria** : It is the iteration termination criteria. When this criteria is satisfied, algorithm iteration stops. Actually, it should be a tuple of 3 parameters. They are \`( type, max_iter, epsilon )\`:
-
3.a - type of termination criteria : It has 3 flags as below:
**cv2.TERM_CRITERIA_EPS** - stop the algorithm iteration if specified accuracy,
*epsilon*, is reached. **cv2.TERM_CRITERIA_MAX_ITER** - stop the algorithm
after the specified number of iterations, *max_iter*. **cv2.TERM_CRITERIA_EPS +
cv2.TERM_CRITERIA_MAX_ITER** - stop the iteration when any of the above
condition is met.
- 3.b - max_iter - An integer specifying maximum number of iterations.
- 3.c - epsilon - Required accuracy
-# **attempts** : Flag to specify the number of times the algorithm is executed using different
initial labellings. The algorithm returns the labels that yield the best compactness. This
compactness is returned as output.
5. **flags** : This flag is used to specify how initial centers are taken. Normally two flags are
used for this : **cv2.KMEANS_PP_CENTERS** and **cv2.KMEANS_RANDOM_CENTERS**.
### Output parameters
-# **compactness** : It is the sum of squared distance from each point to their corresponding
centers.
2. **labels** : This is the label array (same as 'code' in previous article) where each element
marked '0', '1'.....
3. **centers** : This is array of centers of clusters.
Now we will see how to apply K-Means algorithm with three examples.
-# Data with Only One Feature
-----------------------------
Consider, you have a set of data with only one feature, ie one-dimensional. For eg, we can take our
t-shirt problem where you use only height of people to decide the size of t-shirt.
So we start by creating data and plot it in Matplotlib
@code{.py}
import numpy as np
import cv2
from matplotlib import pyplot as plt
x = np.random.randint(25,100,25)
y = np.random.randint(175,255,25)
z = np.hstack((x,y))
z = z.reshape((50,1))
z = np.float32(z)
plt.hist(z,256,[0,256]),plt.show()
@endcode
So we have 'z' which is an array of size 50, and values ranging from 0 to 255. I have reshaped 'z'
to a column vector. It will be more useful when more than one features are present. Then I made data
of np.float32 type.
We get following image :
![image](images/oc_1d_testdata.png)
Now we apply the KMeans function. Before that we need to specify the criteria. My criteria is such
that, whenever 10 iterations of algorithm is ran, or an accuracy of epsilon = 1.0 is reached, stop
the algorithm and return the answer.
@code{.py}
# Define criteria = ( type, max_iter = 10 , epsilon = 1.0 )
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 10, 1.0)
# Set flags (Just to avoid line break in the code)
flags = cv2.KMEANS_RANDOM_CENTERS
# Apply KMeans
compactness,labels,centers = cv2.kmeans(z,2,None,criteria,10,flags)
@endcode
This gives us the compactness, labels and centers. In this case, I got centers as 60 and 207. Labels
will have the same size as that of test data where each data will be labelled as '0','1','2' etc.
depending on their centroids. Now we split the data to different clusters depending on their labels.
@code{.py}
A = z[labels==0]
B = z[labels==1]
@endcode
Now we plot A in Red color and B in Blue color and their centroids in Yellow color.
@code{.py}
# Now plot 'A' in red, 'B' in blue, 'centers' in yellow
plt.hist(A,256,[0,256],color = 'r')
plt.hist(B,256,[0,256],color = 'b')
plt.hist(centers,32,[0,256],color = 'y')
plt.show()
@endcode
Below is the output we got:
![image](images/oc_1d_clustered.png)
-# Data with Multiple Features
------------------------------
In previous example, we took only height for t-shirt problem. Here, we will take both height and
weight, ie two features.
Remember, in previous case, we made our data to a single column vector. Each feature is arranged in
a column, while each row corresponds to an input test sample.
For example, in this case, we set a test data of size 50x2, which are heights and weights of 50
people. First column corresponds to height of all the 50 people and second column corresponds to
their weights. First row contains two elements where first one is the height of first person and
second one his weight. Similarly remaining rows corresponds to heights and weights of other people.
Check image below:
![image](images/oc_feature_representation.jpg)
Now I am directly moving to the code:
@code{.py}
import numpy as np
import cv2
from matplotlib import pyplot as plt
X = np.random.randint(25,50,(25,2))
Y = np.random.randint(60,85,(25,2))
Z = np.vstack((X,Y))
# convert to np.float32
Z = np.float32(Z)
# define criteria and apply kmeans()
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 10, 1.0)
ret,label,center=cv2.kmeans(Z,2,None,criteria,10,cv2.KMEANS_RANDOM_CENTERS)
# Now separate the data, Note the flatten()
A = Z[label.ravel()==0]
B = Z[label.ravel()==1]
# Plot the data
plt.scatter(A[:,0],A[:,1])
plt.scatter(B[:,0],B[:,1],c = 'r')
plt.scatter(center[:,0],center[:,1],s = 80,c = 'y', marker = 's')
plt.xlabel('Height'),plt.ylabel('Weight')
plt.show()
@endcode
Below is the output we get:
![image](images/oc_2d_clustered.jpg)
-# Color Quantization
---------------------
Color Quantization is the process of reducing number of colors in an image. One reason to do so is
to reduce the memory. Sometimes, some devices may have limitation such that it can produce only
limited number of colors. In those cases also, color quantization is performed. Here we use k-means
clustering for color quantization.
There is nothing new to be explained here. There are 3 features, say, R,G,B. So we need to reshape
the image to an array of Mx3 size (M is number of pixels in image). And after the clustering, we
apply centroid values (it is also R,G,B) to all pixels, such that resulting image will have
specified number of colors. And again we need to reshape it back to the shape of original image.
Below is the code:
@code{.py}
import numpy as np
import cv2
img = cv2.imread('home.jpg')
Z = img.reshape((-1,3))
# convert to np.float32
Z = np.float32(Z)
# define criteria, number of clusters(K) and apply kmeans()
criteria = (cv2.TERM_CRITERIA_EPS + cv2.TERM_CRITERIA_MAX_ITER, 10, 1.0)
K = 8
ret,label,center=cv2.kmeans(Z,K,None,criteria,10,cv2.KMEANS_RANDOM_CENTERS)
# Now convert back into uint8, and make original image
center = np.uint8(center)
res = center[label.flatten()]
res2 = res.reshape((img.shape))
cv2.imshow('res2',res2)
cv2.waitKey(0)
cv2.destroyAllWindows()
@endcode
See the result below for K=8:
![image](images/oc_color_quantization.jpg)
Additional Resources
--------------------
Exercises
---------

85
doc/py_tutorials/py_ml/py_kmeans/py_kmeans_understanding/py_kmeans_understanding.markdown Normal file
View File

@@ -0,0 +1,85 @@
Understanding K-Means Clustering {#tutorial_py_kmeans_understanding}
================================
Goal
----
In this chapter, we will understand the concepts of K-Means Clustering, how it works etc.
Theory
------
We will deal this with an example which is commonly used.
### T-shirt size problem
Consider a company, which is going to release a new model of T-shirt to market. Obviously they will
have to manufacture models in different sizes to satisfy people of all sizes. So the company make a
data of people's height and weight, and plot them on to a graph, as below:
![image](images/tshirt.jpg)
Company can't create t-shirts with all the sizes. Instead, they divide people to Small, Medium and
Large, and manufacture only these 3 models which will fit into all the people. This grouping of
people into three groups can be done by k-means clustering, and algorithm provides us best 3 sizes,
which will satisfy all the people. And if it doesn't, company can divide people to more groups, may
be five, and so on. Check image below :
![image](images/tshirt_grouped.jpg)
### How does it work ?
This algorithm is an iterative process. We will explain it step-by-step with the help of images.
Consider a set of data as below ( You can consider it as t-shirt problem). We need to cluster this
data into two groups.
![image](images/testdata.jpg)
**Step : 1** - Algorithm randomly chooses two centroids, \f$C1\f$ and \f$C2\f$ (sometimes, any two data are
taken as the centroids).
**Step : 2** - It calculates the distance from each point to both centroids. If a test data is more
closer to \f$C1\f$, then that data is labelled with '0'. If it is closer to \f$C2\f$, then labelled as '1'
(If more centroids are there, labelled as '2','3' etc).
In our case, we will color all '0' labelled with red, and '1' labelled with blue. So we get
following image after above operations.
![image](images/initial_labelling.jpg)
**Step : 3** - Next we calculate the average of all blue points and red points separately and that
will be our new centroids. That is \f$C1\f$ and \f$C2\f$ shift to newly calculated centroids. (Remember, the
images shown are not true values and not to true scale, it is just for demonstration only).
And again, perform step 2 with new centroids and label data to '0' and '1'.
So we get result as below :
![image](images/update_centroid.jpg)
Now **Step - 2** and **Step - 3** are iterated until both centroids are converged to fixed points.
*(Or it may be stopped depending on the criteria we provide, like maximum number of iterations, or a
specific accuracy is reached etc.)* **These points are such that sum of distances between test data
and their corresponding centroids are minimum**. Or simply, sum of distances between
\f$C1 \leftrightarrow Red_Points\f$ and \f$C2 \leftrightarrow Blue_Points\f$ is minimum.
\f[minimize \;\bigg[J = \sum_{All\: Red_Points}distance(C1,Red_Point) + \sum_{All\: Blue_Points}distance(C2,Blue_Point)\bigg]\f]
Final result almost looks like below :
![image](images/final_clusters.jpg)
So this is just an intuitive understanding of K-Means Clustering. For more details and mathematical
explanation, please read any standard machine learning textbooks or check links in additional
resources. It is just a top layer of K-Means clustering. There are a lot of modifications to this
algorithm like, how to choose the initial centroids, how to speed up the iteration process etc.
Additional Resources
--------------------
-# [Machine Learning Course](https://www.coursera.org/course/ml), Video lectures by Prof. Andrew Ng
(Some of the images are taken from this)
Exercises
---------

10
doc/py_tutorials/py_ml/py_knn/py_knn_index.markdown Normal file
View File

@@ -0,0 +1,10 @@
K-Nearest Neighbour {#tutorial_py_knn_index}
===================
- @subpage tutorial_py_knn_understanding
Get a basic understanding of what kNN is
- @subpage tutorial_py_knn_opencv
Now let's use kNN in OpenCV for digit recognition OCR

121
doc/py_tutorials/py_ml/py_knn/py_knn_opencv/py_knn_opencv.markdown Normal file
View File

@@ -0,0 +1,121 @@
OCR of Hand-written Data using kNN {#tutorial_py_knn_opencv}
==================================
Goal
----
In this chapter
- We will use our knowledge on kNN to build a basic OCR application.
- We will try with Digits and Alphabets data available that comes with OpenCV.
OCR of Hand-written Digits
--------------------------
Our goal is to build an application which can read the handwritten digits. For this we need some
train_data and test_data. OpenCV comes with an image digits.png (in the folder
opencv/samples/python2/data/) which has 5000 handwritten digits (500 for each digit). Each digit is
a 20x20 image. So our first step is to split this image into 5000 different digits. For each digit,
we flatten it into a single row with 400 pixels. That is our feature set, ie intensity values of all
pixels. It is the simplest feature set we can create. We use first 250 samples of each digit as
train_data, and next 250 samples as test_data. So let's prepare them first.
@code{.py}
import numpy as np
import cv2
from matplotlib import pyplot as plt
img = cv2.imread('digits.png')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
# Now we split the image to 5000 cells, each 20x20 size
cells = [np.hsplit(row,100) for row in np.vsplit(gray,50)]
# Make it into a Numpy array. It size will be (50,100,20,20)
x = np.array(cells)
# Now we prepare train_data and test_data.
train = x[:,:50].reshape(-1,400).astype(np.float32) # Size = (2500,400)
test = x[:,50:100].reshape(-1,400).astype(np.float32) # Size = (2500,400)
# Create labels for train and test data
k = np.arange(10)
train_labels = np.repeat(k,250)[:,np.newaxis]
test_labels = train_labels.copy()
# Initiate kNN, train the data, then test it with test data for k=1
knn = cv2.KNearest()
knn.train(train,train_labels)
ret,result,neighbours,dist = knn.find_nearest(test,k=5)
# Now we check the accuracy of classification
# For that, compare the result with test_labels and check which are wrong
matches = result==test_labels
correct = np.count_nonzero(matches)
accuracy = correct*100.0/result.size
print accuracy
@endcode
So our basic OCR app is ready. This particular example gave me an accuracy of 91%. One option
improve accuracy is to add more data for training, especially the wrong ones. So instead of finding
this training data everytime I start application, I better save it, so that next time, I directly
read this data from a file and start classification. You can do it with the help of some Numpy
functions like np.savetxt, np.savez, np.load etc. Please check their docs for more details.
@code{.py}
# save the data
np.savez('knn_data.npz',train=train, train_labels=train_labels)
# Now load the data
with np.load('knn_data.npz') as data:
print data.files
train = data['train']
train_labels = data['train_labels']
@endcode
In my system, it takes around 4.4 MB of memory. Since we are using intensity values (uint8 data) as
features, it would be better to convert the data to np.uint8 first and then save it. It takes only
1.1 MB in this case. Then while loading, you can convert back into float32.
OCR of English Alphabets
------------------------
Next we will do the same for English alphabets, but there is a slight change in data and feature
set. Here, instead of images, OpenCV comes with a data file, letter-recognition.data in
opencv/samples/cpp/ folder. If you open it, you will see 20000 lines which may, on first sight, look
like garbage. Actually, in each row, first column is an alphabet which is our label. Next 16 numbers
following it are its different features. These features are obtained from [UCI Machine Learning
Repository](http://archive.ics.uci.edu/ml/). You can find the details of these features in [this
page](http://archive.ics.uci.edu/ml/datasets/Letter+Recognition).
There are 20000 samples available, so we take first 10000 data as training samples and remaining
10000 as test samples. We should change the alphabets to ascii characters because we can't work with
alphabets directly.
@code{.py}
import cv2
import numpy as np
import matplotlib.pyplot as plt
# Load the data, converters convert the letter to a number
data= np.loadtxt('letter-recognition.data', dtype= 'float32', delimiter = ',',
converters= {0: lambda ch: ord(ch)-ord('A')})
# split the data to two, 10000 each for train and test
train, test = np.vsplit(data,2)
# split trainData and testData to features and responses
responses, trainData = np.hsplit(train,[1])
labels, testData = np.hsplit(test,[1])
# Initiate the kNN, classify, measure accuracy.
knn = cv2.KNearest()
knn.train(trainData, responses)
ret, result, neighbours, dist = knn.find_nearest(testData, k=5)
correct = np.count_nonzero(result == labels)
accuracy = correct*100.0/10000
print accuracy
@endcode
It gives me an accuracy of 93.22%. Again, if you want to increase accuracy, you can iteratively add
error data in each level.
Additional Resources
--------------------
Exercises
---------

153
doc/py_tutorials/py_ml/py_knn/py_knn_understanding/py_knn_understanding.markdown Normal file
View File

@@ -0,0 +1,153 @@
Understanding k-Nearest Neighbour {#tutorial_py_knn_understanding}
=================================
Goal
----
In this chapter, we will understand the concepts of k-Nearest Neighbour (kNN) algorithm.
Theory
------
kNN is one of the simplest of classification algorithms available for supervised learning. The idea
is to search for closest match of the test data in feature space. We will look into it with below
image.
![image](images/knn_theory.png)
In the image, there are two families, Blue Squares and Red Triangles. We call each family as
**Class**. Their houses are shown in their town map which we call feature space. *(You can consider
a feature space as a space where all datas are projected. For example, consider a 2D coordinate
space. Each data has two features, x and y coordinates. You can represent this data in your 2D
coordinate space, right? Now imagine if there are three features, you need 3D space. Now consider N
features, where you need N-dimensional space, right? This N-dimensional space is its feature space.
In our image, you can consider it as a 2D case with two features)*.
Now a new member comes into the town and creates a new home, which is shown as green circle. He
should be added to one of these Blue/Red families. We call that process, **Classification**. What we
do? Since we are dealing with kNN, let us apply this algorithm.
One method is to check who is his nearest neighbour. From the image, it is clear it is the Red
Triangle family. So he is also added into Red Triangle. This method is called simply **Nearest
Neighbour**, because classification depends only on the nearest neighbour.
But there is a problem with that. Red Triangle may be the nearest. But what if there are lot of Blue
Squares near to him? Then Blue Squares have more strength in that locality than Red Triangle. So
just checking nearest one is not sufficient. Instead we check some k nearest families. Then whoever
is majority in them, the new guy belongs to that family. In our image, let's take k=3, ie 3 nearest
families. He has two Red and one Blue (there are two Blues equidistant, but since k=3, we take only
one of them), so again he should be added to Red family. But what if we take k=7? Then he has 5 Blue
families and 2 Red families. Great!! Now he should be added to Blue family. So it all changes with
value of k. More funny thing is, what if k = 4? He has 2 Red and 2 Blue neighbours. It is a tie !!!
So better take k as an odd number. So this method is called **k-Nearest Neighbour** since
classification depends on k nearest neighbours.
Again, in kNN, it is true we are considering k neighbours, but we are giving equal importance to
all, right? Is it justice? For example, take the case of k=4. We told it is a tie. But see, the 2
Red families are more closer to him than the other 2 Blue families. So he is more eligible to be
added to Red. So how do we mathematically explain that? We give some weights to each family
depending on their distance to the new-comer. For those who are near to him get higher weights while
those are far away get lower weights. Then we add total weights of each family separately. Whoever
gets highest total weights, new-comer goes to that family. This is called **modified kNN**.
So what are some important things you see here?
- You need to have information about all the houses in town, right? Because, we have to check
the distance from new-comer to all the existing houses to find the nearest neighbour. If there
are plenty of houses and families, it takes lots of memory, and more time for calculation
also.
- There is almost zero time for any kind of training or preparation.
Now let's see it in OpenCV.
kNN in OpenCV
-------------
We will do a simple example here, with two families (classes), just like above. Then in the next
chapter, we will do an even better example.
So here, we label the Red family as **Class-0** (so denoted by 0) and Blue family as **Class-1**
(denoted by 1). We create 25 families or 25 training data, and label them either Class-0 or Class-1.
We do all these with the help of Random Number Generator in Numpy.
Then we plot it with the help of Matplotlib. Red families are shown as Red Triangles and Blue
families are shown as Blue Squares.
@code{.py}
import cv2
import numpy as np
import matplotlib.pyplot as plt
# Feature set containing (x,y) values of 25 known/training data
trainData = np.random.randint(0,100,(25,2)).astype(np.float32)
# Labels each one either Red or Blue with numbers 0 and 1
responses = np.random.randint(0,2,(25,1)).astype(np.float32)
# Take Red families and plot them
red = trainData[responses.ravel()==0]
plt.scatter(red[:,0],red[:,1],80,'r','^')
# Take Blue families and plot them
blue = trainData[responses.ravel()==1]
plt.scatter(blue[:,0],blue[:,1],80,'b','s')
plt.show()
@endcode
You will get something similar to our first image. Since you are using random number generator, you
will be getting different data each time you run the code.
Next initiate the kNN algorithm and pass the trainData and responses to train the kNN (It constructs
a search tree).
Then we will bring one new-comer and classify him to a family with the help of kNN in OpenCV. Before
going to kNN, we need to know something on our test data (data of new comers). Our data should be a
floating point array with size \f$number \; of \; testdata \times number \; of \; features\f$. Then we
find the nearest neighbours of new-comer. We can specify how many neighbours we want. It returns:
-# The label given to new-comer depending upon the kNN theory we saw earlier. If you want Nearest
Neighbour algorithm, just specify k=1 where k is the number of neighbours.
2. The labels of k-Nearest Neighbours.
3. Corresponding distances from new-comer to each nearest neighbour.
So let's see how it works. New comer is marked in green color.
@code{.py}
newcomer = np.random.randint(0,100,(1,2)).astype(np.float32)
plt.scatter(newcomer[:,0],newcomer[:,1],80,'g','o')
knn = cv2.KNearest()
knn.train(trainData,responses)
ret, results, neighbours ,dist = knn.find_nearest(newcomer, 3)
print "result: ", results,"\n"
print "neighbours: ", neighbours,"\n"
print "distance: ", dist
plt.show()
@endcode
I got the result as follows:
@code{.py}
result: [[ 1.]]
neighbours: [[ 1. 1. 1.]]
distance: [[ 53. 58. 61.]]
@endcode
It says our new-comer got 3 neighbours, all from Blue family. Therefore, he is labelled as Blue
family. It is obvious from plot below:
![image](images/knn_simple.png)
If you have large number of data, you can just pass it as array. Corresponding results are also
obtained as arrays.
@code{.py}
# 10 new comers
newcomers = np.random.randint(0,100,(10,2)).astype(np.float32)
ret, results,neighbours,dist = knn.find_nearest(newcomer, 3)
# The results also will contain 10 labels.
@endcode
Additional Resources
--------------------
-# [NPTEL notes on Pattern Recognition, Chapter
11](http://www.nptel.iitm.ac.in/courses/106108057/12)
Exercises
---------

131
doc/py_tutorials/py_ml/py_svm/py_svm_basics/py_svm_basics.markdown Normal file
View File

@@ -0,0 +1,131 @@
Understanding SVM {#tutorial_py_svm_basics}
=================
Goal
----
In this chapter
- We will see an intuitive understanding of SVM
Theory
------
### Linearly Separable Data
Consider the image below which has two types of data, red and blue. In kNN, for a test data, we used
to measure its distance to all the training samples and take the one with minimum distance. It takes
plenty of time to measure all the distances and plenty of memory to store all the training-samples.
But considering the data given in image, should we need that much?
![image](images/svm_basics1.png)
Consider another idea. We find a line, \f$f(x)=ax_1+bx_2+c\f$ which divides both the data to two
regions. When we get a new test_data \f$X\f$, just substitute it in \f$f(x)\f$. If \f$f(X) > 0\f$, it belongs
to blue group, else it belongs to red group. We can call this line as **Decision Boundary**. It is
very simple and memory-efficient. Such data which can be divided into two with a straight line (or
hyperplanes in higher dimensions) is called **Linear Separable**.
So in above image, you can see plenty of such lines are possible. Which one we will take? Very
intuitively we can say that the line should be passing as far as possible from all the points. Why?
Because there can be noise in the incoming data. This data should not affect the classification
accuracy. So taking a farthest line will provide more immunity against noise. So what SVM does is to
find a straight line (or hyperplane) with largest minimum distance to the training samples. See the
bold line in below image passing through the center.
![image](images/svm_basics2.png)
So to find this Decision Boundary, you need training data. Do you need all? NO. Just the ones which
are close to the opposite group are sufficient. In our image, they are the one blue filled circle
and two red filled squares. We can call them **Support Vectors** and the lines passing through them
are called **Support Planes**. They are adequate for finding our decision boundary. We need not
worry about all the data. It helps in data reduction.
What happened is, first two hyperplanes are found which best represents the data. For eg, blue data
is represented by \f$w^Tx+b_0 > 1\f$ while red data is represented by \f$w^Tx+b_0 < -1\f$ where \f$w\f$ is
**weight vector** ( \f$w=[w_1, w_2,..., w_n]\f$) and \f$x\f$ is the feature vector
(\f$x = [x_1,x_2,..., x_n]\f$). \f$b_0\f$ is the **bias**. Weight vector decides the orientation of decision
boundary while bias point decides its location. Now decision boundary is defined to be midway
between these hyperplanes, so expressed as \f$w^Tx+b_0 = 0\f$. The minimum distance from support vector
to the decision boundary is given by, \f$distance_{support \, vectors}=\frac{1}{||w||}\f$. Margin is
twice this distance, and we need to maximize this margin. i.e. we need to minimize a new function
\f$L(w, b_0)\f$ with some constraints which can expressed below:
\f[\min_{w, b_0} L(w, b_0) = \frac{1}{2}||w||^2 \; \text{subject to} \; t_i(w^Tx+b_0) \geq 1 \; \forall i\f]
where \f$t_i\f$ is the label of each class, \f$t_i \in [-1,1]\f$.
### Non-Linearly Separable Data
Consider some data which can't be divided into two with a straight line. For example, consider an
one-dimensional data where 'X' is at -3 & +3 and 'O' is at -1 & +1. Clearly it is not linearly
separable. But there are methods to solve these kinds of problems. If we can map this data set with
a function, \f$f(x) = x^2\f$, we get 'X' at 9 and 'O' at 1 which are linear separable.
Otherwise we can convert this one-dimensional to two-dimensional data. We can use \f$f(x)=(x,x^2)\f$
function to map this data. Then 'X' becomes (-3,9) and (3,9) while 'O' becomes (-1,1) and (1,1).
This is also linear separable. In short, chance is more for a non-linear separable data in
lower-dimensional space to become linear separable in higher-dimensional space.
In general, it is possible to map points in a d-dimensional space to some D-dimensional space
\f$(D>d)\f$ to check the possibility of linear separability. There is an idea which helps to compute the
dot product in the high-dimensional (kernel) space by performing computations in the low-dimensional
input (feature) space. We can illustrate with following example.
Consider two points in two-dimensional space, \f$p=(p_1,p_2)\f$ and \f$q=(q_1,q_2)\f$. Let \f$\phi\f$ be a
mapping function which maps a two-dimensional point to three-dimensional space as follows:
\f[\phi (p) = (p_{1}^2,p_{2}^2,\sqrt{2} p_1 p_2)
\phi (q) = (q_{1}^2,q_{2}^2,\sqrt{2} q_1 q_2)\f]
Let us define a kernel function \f$K(p,q)\f$ which does a dot product between two points, shown below:
\f[K(p,q) = \phi(p).\phi(q) &= \phi(p)^T \phi(q) \\
&= (p_{1}^2,p_{2}^2,\sqrt{2} p_1 p_2).(q_{1}^2,q_{2}^2,\sqrt{2} q_1 q_2) \\
&= p_1 q_1 + p_2 q_2 + 2 p_1 q_1 p_2 q_2 \\
&= (p_1 q_1 + p_2 q_2)^2 \\
\phi(p).\phi(q) &= (p.q)^2\f]
It means, a dot product in three-dimensional space can be achieved using squared dot product in
two-dimensional space. This can be applied to higher dimensional space. So we can calculate higher
dimensional features from lower dimensions itself. Once we map them, we get a higher dimensional
space.
In addition to all these concepts, there comes the problem of misclassification. So just finding
decision boundary with maximum margin is not sufficient. We need to consider the problem of
misclassification errors also. Sometimes, it may be possible to find a decision boundary with less
margin, but with reduced misclassification. Anyway we need to modify our model such that it should
find decision boundary with maximum margin, but with less misclassification. The minimization
criteria is modified as:
\f[min \; ||w||^2 + C(distance \; of \; misclassified \; samples \; to \; their \; correct \; regions)\f]
Below image shows this concept. For each sample of the training data a new parameter \f$\xi_i\f$ is
defined. It is the distance from its corresponding training sample to their correct decision region.
For those who are not misclassified, they fall on their corresponding support planes, so their
distance is zero.
![image](images/svm_basics3.png)
So the new optimization problem is :
\f[\min_{w, b_{0}} L(w,b_0) = ||w||^{2} + C \sum_{i} {\xi_{i}} \text{ subject to } y_{i}(w^{T} x_{i} + b_{0}) \geq 1 - \xi_{i} \text{ and } \xi_{i} \geq 0 \text{ } \forall i\f]
How should the parameter C be chosen? It is obvious that the answer to this question depends on how
the training data is distributed. Although there is no general answer, it is useful to take into
account these rules:
- Large values of C give solutions with less misclassification errors but a smaller margin.
Consider that in this case it is expensive to make misclassification errors. Since the aim of
the optimization is to minimize the argument, few misclassifications errors are allowed.
- Small values of C give solutions with bigger margin and more classification errors. In this
case the minimization does not consider that much the term of the sum so it focuses more on
finding a hyperplane with big margin.
Additional Resources
--------------------
-# [NPTEL notes on Statistical Pattern Recognition, Chapters
25-29](http://www.nptel.iitm.ac.in/courses/106108057/26).
Exercises
---------

10
doc/py_tutorials/py_ml/py_svm/py_svm_index.markdown Normal file
View File

@@ -0,0 +1,10 @@
Support Vector Machines (SVM) {#tutorial_py_svm_index}
=============================
- @subpage tutorial_py_svm_basics
Get a basic understanding of what SVM is
- @subpage tutorial_py_svm_opencv
Let's use SVM functionalities in OpenCV

138
doc/py_tutorials/py_ml/py_svm/py_svm_opencv/py_svm_opencv.markdown Normal file
View File

@@ -0,0 +1,138 @@
OCR of Hand-written Data using SVM {#tutorial_py_svm_opencv}
==================================
Goal
----
In this chapter
- We will revisit the hand-written data OCR, but, with SVM instead of kNN.
OCR of Hand-written Digits
--------------------------
In kNN, we directly used pixel intensity as the feature vector. This time we will use [Histogram of
Oriented Gradients](http://en.wikipedia.org/wiki/Histogram_of_oriented_gradients) (HOG) as feature
vectors.
Here, before finding the HOG, we deskew the image using its second order moments. So we first define
a function **deskew()** which takes a digit image and deskew it. Below is the deskew() function:
@code{.py}
def deskew(img):
m = cv2.moments(img)
if abs(m['mu02']) < 1e-2:
return img.copy()
skew = m['mu11']/m['mu02']
M = np.float32([[1, skew, -0.5*SZ*skew], [0, 1, 0]])
img = cv2.warpAffine(img,M,(SZ, SZ),flags=affine_flags)
return img
@endcode
Below image shows above deskew function applied to an image of zero. Left image is the original
image and right image is the deskewed image.
![image](images/deskew.jpg)
Next we have to find the HOG Descriptor of each cell. For that, we find Sobel derivatives of each
cell in X and Y direction. Then find their magnitude and direction of gradient at each pixel. This
gradient is quantized to 16 integer values. Divide this image to four sub-squares. For each
sub-square, calculate the histogram of direction (16 bins) weighted with their magnitude. So each
sub-square gives you a vector containing 16 values. Four such vectors (of four sub-squares) together
gives us a feature vector containing 64 values. This is the feature vector we use to train our data.
@code{.py}
def hog(img):
gx = cv2.Sobel(img, cv2.CV_32F, 1, 0)
gy = cv2.Sobel(img, cv2.CV_32F, 0, 1)
mag, ang = cv2.cartToPolar(gx, gy)
# quantizing binvalues in (0...16)
bins = np.int32(bin_n*ang/(2*np.pi))
# Divide to 4 sub-squares
bin_cells = bins[:10,:10], bins[10:,:10], bins[:10,10:], bins[10:,10:]
mag_cells = mag[:10,:10], mag[10:,:10], mag[:10,10:], mag[10:,10:]
hists = [np.bincount(b.ravel(), m.ravel(), bin_n) for b, m in zip(bin_cells, mag_cells)]
hist = np.hstack(hists)
return hist
@endcode
Finally, as in the previous case, we start by splitting our big dataset into individual cells. For
every digit, 250 cells are reserved for training data and remaining 250 data is reserved for
testing. Full code is given below:
@code{.py}
import cv2
import numpy as np
SZ=20
bin_n = 16 # Number of bins
svm_params = dict( kernel_type = cv2.SVM_LINEAR,
svm_type = cv2.SVM_C_SVC,
C=2.67, gamma=5.383 )
affine_flags = cv2.WARP_INVERSE_MAP|cv2.INTER_LINEAR
def deskew(img):
m = cv2.moments(img)
if abs(m['mu02']) < 1e-2:
return img.copy()
skew = m['mu11']/m['mu02']
M = np.float32([[1, skew, -0.5*SZ*skew], [0, 1, 0]])
img = cv2.warpAffine(img,M,(SZ, SZ),flags=affine_flags)
return img
def hog(img):
gx = cv2.Sobel(img, cv2.CV_32F, 1, 0)
gy = cv2.Sobel(img, cv2.CV_32F, 0, 1)
mag, ang = cv2.cartToPolar(gx, gy)
bins = np.int32(bin_n*ang/(2*np.pi)) # quantizing binvalues in (0...16)
bin_cells = bins[:10,:10], bins[10:,:10], bins[:10,10:], bins[10:,10:]
mag_cells = mag[:10,:10], mag[10:,:10], mag[:10,10:], mag[10:,10:]
hists = [np.bincount(b.ravel(), m.ravel(), bin_n) for b, m in zip(bin_cells, mag_cells)]
hist = np.hstack(hists) # hist is a 64 bit vector
return hist
img = cv2.imread('digits.png',0)
cells = [np.hsplit(row,100) for row in np.vsplit(img,50)]
# First half is trainData, remaining is testData
train_cells = [ i[:50] for i in cells ]
test_cells = [ i[50:] for i in cells]
###### Now training ########################
deskewed = [map(deskew,row) for row in train_cells]
hogdata = [map(hog,row) for row in deskewed]
trainData = np.float32(hogdata).reshape(-1,64)
responses = np.float32(np.repeat(np.arange(10),250)[:,np.newaxis])
svm = cv2.SVM()
svm.train(trainData,responses, params=svm_params)
svm.save('svm_data.dat')
###### Now testing ########################
deskewed = [map(deskew,row) for row in test_cells]
hogdata = [map(hog,row) for row in deskewed]
testData = np.float32(hogdata).reshape(-1,bin_n*4)
result = svm.predict_all(testData)
####### Check Accuracy ########################
mask = result==responses
correct = np.count_nonzero(mask)
print correct*100.0/result.size
@endcode
This particular technique gave me nearly 94% accuracy. You can try different values for various
parameters of SVM to check if higher accuracy is possible. Or you can read technical papers on this
area and try to implement them.
Additional Resources
--------------------
-# [Histograms of Oriented Gradients Video](www.youtube.com/watch?v=0Zib1YEE4LU‎)
Exercises
---------
-# OpenCV samples contain digits.py which applies a slight improvement of the above method to get
improved result. It also contains the reference. Check it and understand it.

16
doc/py_tutorials/py_ml/py_table_of_contents_ml/py_table_of_contents_ml.markdown Normal file
View File

@@ -0,0 +1,16 @@
Machine Learning {#tutorial_py_table_of_contents_ml}
================
- @subpage tutorial_py_knn_index
Learn to use kNN for classification
Plus learn about handwritten digit recognition using kNN
- @subpage tutorial_py_svm_index
Understand concepts of SVM
- @subpage tutorial_py_kmeans_index
Learn to use K-Means Clustering to group data to a number of clusters.
Plus learn to do color quantization using K-Means Clustering