Different papers/libraries seem to have a different way of computing the chi squared distance, for instance in OpenCV it's expressed in one way while in this paper it's expressed in a different manner.
My first question is, what's the difference between the two formulas, i.e. why in one formula we divide by the value of one bin while in the other we divide by the sum of the two bins?
Secondly, should the histograms be normalized, if so why? The chi-squared statistic doesn't require that but the general consensus is to normalize the histogram before using a chi-squared distance.
The documentation is wrong. The implementation is correct inside OpenCV. Take a look at this bug post.
Also, normalising a histogram does not really change its pattern or "shape". Only the scale is brought down. So as long as you're working independent of scale, which you probably are if you're looking at how much one histogram "resembles" another, normalising should only make calculations faster (hopefully).
Related
I am trying to understand the math behind the Decision tree(Regression). I came across 2 article and both of them explain differently on how the split is done in regression tree. Can anyone point out which one is correct or both are similar just the method is different ?
https://www.saedsayad.com/decision_tree_reg.htm
https://www.python-course.eu/Regression_Trees.php
Thanks,
Both are correct. Method 1 uses standard deviation for spliiting the nodes and method 2 uses variance. Both s.d and variance are used since the target value is continuous.
Variance is one of the most commonly used splitting criteria for
regression trees.
Variance
The variance is the average of the squared differences from the mean. To figure out the variance, first calculate the difference between each point and the mean; then, square and average the results.
Standard Deviation
Standard deviation is a statistic that looks at how far from the mean a group of numbers is, by using the square root of the variance. The calculation of variance uses squares because it weights outliers more heavily than data very near the mean. This calculation also prevents differences above the mean from canceling out those below, which can sometimes result in a variance of zero.
I have a bunch of gray-scale images decomposed into superpixels. Each superpixel in these images have a label in the rage of [0-1]. You can see one sample of images below.
Here is the challenge: I want the spatially (locally) neighboring superpixels to have consistent labels (close in value).
I'm kind of interested in smoothing local labels but do not want to apply Gaussian smoothing functions or whatever, as some colleagues suggested. I have also heard about Conditional Random Field (CRF). Is it helpful?
Any suggestion would be welcome.
I'm kind of interested in smoothing local labels but do not want to apply Gaussian smoothing functions or whatever, as some colleagues suggested.
And why is that? Why do you not consider helpful advice of your colleagues, which are actually right. Applying smoothing function is the most reasonable way to go.
I have also heard about Conditional Random Field (CRF). Is it helpful?
This also suggests, that you should rather go with collegues advice, as CRF has nothing to do with your problem. CRF is a classifier, sequence classifier to be exact, requiring labeled examples to learn from and has nothing to do with the setting presented.
What are typical approaches?
The exact thing proposed by your collegues, you should define a smoothing function and apply it to your function values (I will not use a term "labels" as it is missleading, you do have values in [0,1], continuous values, "label" denotes categorical variable in machine learning) and its neighbourhood.
Another approach would be to define some optimization problem, where your current assignment of values is one goal, and the second one is "closeness", for example:
Let us assume that you have points with values {(x_i, y_i)}_{i=1}^N and that n(x) returns indices of neighbouring points of x.
Consequently you are trying to find {a_i}_{i=1}^N such that they minimize
SUM_{i=1}^N (y_i - a_i)^2 + C * SUM_{i=1}^N SUM_{j \in n(x_i)} (a_i - a_j)^2
------------------------- - --------------------------------------------
closeness to current constant to closeness to neighbouring values
values weight each part
You can solve the above optimization problem using many techniques, for example through scipy.optimize.minimize module.
I am not sure that your request makes any sense.
Having close label values for nearby superpixels is trivial: take some smooth function of (X, Y), such as constant or affine, taking values in the range [0,1], and assign the function value to the superpixel centered at (X, Y).
You could also take the distance function from any point in the plane.
But this is of no use as it is unrelated to the image content.
I am using Linear regression to predict data. But, I am getting totally contrasting results when I Normalize (Vs) Standardize variables.
Normalization = x -xmin/ xmax – xmin
Zero Score Standardization = x - xmean/ xstd
a) Also, when to Normalize (Vs) Standardize ?
b) How Normalization affects Linear Regression?
c) Is it okay if I don't normalize all the attributes/lables in the linear regression?
Thanks,
Santosh
Note that the results might not necessarily be so different. You might simply need different hyperparameters for the two options to give similar results.
The ideal thing is to test what works best for your problem. If you can't afford this for some reason, most algorithms will probably benefit from standardization more so than from normalization.
See here for some examples of when one should be preferred over the other:
For example, in clustering analyses, standardization may be especially crucial in order to compare similarities between features based on certain distance measures. Another prominent example is the Principal Component Analysis, where we usually prefer standardization over Min-Max scaling, since we are interested in the components that maximize the variance (depending on the question and if the PCA computes the components via the correlation matrix instead of the covariance matrix; but more about PCA in my previous article).
However, this doesn’t mean that Min-Max scaling is not useful at all! A popular application is image processing, where pixel intensities have to be normalized to fit within a certain range (i.e., 0 to 255 for the RGB color range). Also, typical neural network algorithm require data that on a 0-1 scale.
One disadvantage of normalization over standardization is that it loses some information in the data, especially about outliers.
Also on the linked page, there is this picture:
As you can see, scaling clusters all the data very close together, which may not be what you want. It might cause algorithms such as gradient descent to take longer to converge to the same solution they would on a standardized data set, or it might even make it impossible.
"Normalizing variables" doesn't really make sense. The correct terminology is "normalizing / scaling the features". If you're going to normalize or scale one feature, you should do the same for the rest.
That makes sense because normalization and standardization do different things.
Normalization transforms your data into a range between 0 and 1
Standardization transforms your data such that the resulting distribution has a mean of 0 and a standard deviation of 1
Normalization/standardization are designed to achieve a similar goal, which is to create features that have similar ranges to each other. We want that so we can be sure we are capturing the true information in a feature, and that we dont over weigh a particular feature just because its values are much larger than other features.
If all of your features are within a similar range of each other then theres no real need to standardize/normalize. If, however, some features naturally take on values that are much larger/smaller than others then normalization/standardization is called for
If you're going to be normalizing at least one variable/feature, I would do the same thing to all of the others as well
First question is why we need Normalisation/Standardisation?
=> We take a example of dataset where we have salary variable and age variable.
Age can take range from 0 to 90 where salary can be from 25thousand to 2.5lakh.
We compare difference for 2 person then age difference will be in range of below 100 where salary difference will in range of thousands.
So if we don't want one variable to dominate other then we use either Normalisation or Standardization. Now both age and salary will be in same scale
but when we use standardiztion or normalisation, we lose original values and it is transformed to some values. So loss of interpretation but extremely important when we want to draw inference from our data.
Normalization rescales the values into a range of [0,1]. also called min-max scaled.
Standardization rescales data to have a mean (μ) of 0 and standard deviation (σ) of 1.So it gives a normal graph.
Example below:
Another example:
In above image, you can see that our actual data(in green) is spread b/w 1 to 6, standardised data(in red) is spread around -1 to 3 whereas normalised data(in blue) is spread around 0 to 1.
Normally many algorithm required you to first standardise/normalise data before passing as parameter. Like in PCA, where we do dimension reduction by plotting our 3D data into 1D(say).Here we required standardisation.
But in Image processing, it is required to normalise pixels before processing.
But during normalisation, we lose outliers(extreme datapoints-either too low or too high) which is slight disadvantage.
So it depends on our preference what we chose but standardisation is most recommended as it gives a normal curve.
None of the mentioned transformations shall matter for linear regression as these are all affine transformations.
Found coefficients would change but explained variance will ultimately remain the same. So, from linear regression perspective, Outliers remain as outliers (leverage points).
And these transformations also will not change the distribution. Shape of the distribution remains the same.
lot of people use Normalisation and Standardisation interchangeably. The purpose remains the same is to bring features into the same scale. The approach is to subtract each value from min value or mean and divide by max value minus min value or SD respectively. The difference you can observe that when using min value u will get all value + ve and mean value u will get bot + ve and -ve values. This is also one of the factors to decide which approach to use.
Problem:
To cluster the similar colour pixels in CIE LAB using K means.
I want to use CIE 94 for distance between 2 pixels
Formula of CIE94
What i read was Kmeans work in "Euclidean space" where the positional cordinates are minimised by cost function which is (sum of squared difference)
The reason of not Using kmeans in space other than euclidean is
"""algorithm is often presented as assigning objects to the nearest cluster by distance. The standard algorithm aims at minimizing the within-cluster sum of squares (WCSS) objective, and thus assigns by "least sum of squares", which is exactly equivalent to assigning by the smallest Euclidean distance. Using a different distance function other than (squared) Euclidean distance may stop the algorithm from converging""(source wiki)
So how to use distance CIE 94 in LAB SPACE for similar colour clustering ?
So how to approach the problem ? What should be the minimisation function here ? HOW to map euclidean space to lab space if for the k mean euclidean formula to work ? Any other approach here ?
The reason that CIE LAB is often used for clustering is because it reduces the color to 2 dimensions (as opposed to RGB with 3 color channels). You can easily think of the color for each pixel in a Cartesian coordinate system, instead of points (x,y) you have points (a,b) From here you simply perform a 2d kmeans.
Exactly how you implement kmeans is up to you. The nice thing about reducing colors to a 2d space is we can imagine the data on a grid, and now we can use any regular distance measure we want. Mahalonobis, euclidean, 1 norm, city block, etc. The possibilities are really endless here.
You don't have to use CIELAB, you can just as easily use YCbCr, YUV, or any other colorspace that represents color in 2 dimensions. IF you wanted to try a 3d kmeans you could use rgb, hsv, etc. One problem with higher dimensionality is sparsity of clusters (large variance) and most importantly, increased computation time.
Just for fun I've included two images clustered using kmeans, one in LAB and one in YCbCr, you can see the clustering is nearly identical (except that the labels are different), just proving that the exact color space is irrelevant, the main point is to match the dimensionality of your kmeans with that of your data
EDIT
You made some good points in your comments. I was merely demonstrating that by abstracting the problem you can imagine many variations for the same basic clustering algorithm. But you are right, there are advantages to using CIELAB
Back to the distance measure. Kmeans has two steps, assignment, and update (it is very similar to the Expectation Maximization algorithm). This distance is used in assignment step of k-means. Here is some psuedo code
for each pixel 1 to rows*cols
for each cluster 1 to k
dist[k] = calculate_distance(pixel, mu[k])
pixel_id = index k of minimum dist
you would create a function calculate_distance that uses the delta_e calculation from cielab94. This formula uses all 3 channels to calculate distance. Hopefully this answers your questions
NOTE
My examples only use the 2 color channels, ignoring the luminance channel. I used this technique since often the goal is group colors despite lighting disparities(such as shadows). The delta_E measure is not lighting invariant. This may or may not be a concern for your application, but it is something to keep in mind.
results using square euclidean distance
results using cityblock distance
There are k-means variations for other distance functions.
In particular k-medoids (PAM) works with arbitrary distance functions.
I have implemented k-means clustering for determining the clusters in 300 objects. Each of my object
has about 30 dimensions. The distance is calculated using the Euclidean metric.
I need to know
How would I determine if my algorithms works correctly? I can't have a graph which will
give some idea about the correctness of my algorithm.
Is Euclidean distance the correct method for calculating distances? What if I have 100 dimensions
instead of 30 ?
The two questions in the OP are separate topics (i.e., no overlap in the answers), so I'll try to answer them one at a time staring with item 1 on the list.
How would I determine if my [clustering] algorithms works correctly?
k-means, like other unsupervised ML techniques, lacks a good selection of diagnostic tests to answer questions like "are the cluster assignments returned by k-means more meaningful for k=3 or k=5?"
Still, there is one widely accepted test that yields intuitive results and that is straightforward to apply. This diagnostic metric is just this ratio:
inter-centroidal separation / intra-cluster variance
As the value of this ratio increase, the quality of your clustering result increases.
This is intuitive. The first of these metrics is just how far apart is each cluster from the others (measured according to the cluster centers)?
But inter-centroidal separation alone doesn't tell the whole story, because two clustering algorithms could return results having the same inter-centroidal separation though one is clearly better, because the clusters are "tighter" (i.e., smaller radii); in other words, the cluster edges have more separation. The second metric--intra-cluster variance--accounts for this. This is just the mean variance, calculated per cluster.
In sum, the ratio of inter-centroidal separation to intra-cluster variance is a quick, consistent, and reliable technique for comparing results from different clustering algorithms, or to compare the results from the same algorithm run under different variable parameters--e.g., number of iterations, choice of distance metric, number of centroids (value of k).
The desired result is tight (small) clusters, each one far away from the others.
The calculation is simple:
For inter-centroidal separation:
calculate the pair-wise distance between cluster centers; then
calculate the median of those distances.
For intra-cluster variance:
for each cluster, calculate the distance of every data point in a given cluster from
its cluster center; next
(for each cluster) calculate the variance of the sequence of distances from the step above; then
average these variance values.
That's my answer to the first question. Here's the second question:
Is Euclidean distance the correct method for calculating distances? What if I have 100 dimensions instead of 30 ?
First, the easy question--is Euclidean distance a valid metric as dimensions/features increase?
Euclidean distance is perfectly scalable--works for two dimensions or two thousand. For any pair of data points:
subtract their feature vectors element-wise,
square each item in that result vector,
sum that result,
take the square root of that scalar.
Nowhere in this sequence of calculations is scale implicated.
But whether Euclidean distance is the appropriate similarity metric for your problem, depends on your data. For instance, is it purely numeric (continuous)? Or does it have discrete (categorical) variables as well (e.g., gender? M/F) If one of your dimensions is "current location" and of the 200 users, 100 have the value "San Francisco" and the other 100 have "Boston", you can't really say that, on average, your users are from somewhere in Kansas, but that's sort of what Euclidean distance would do.
In any event, since we don't know anything about it, i'll just give you a simple flow diagram so that you can apply it to your data and identify an appropriate similarity metric.
To identify an appropriate similarity metric given your data:
Euclidean distance is good when dimensions are comparable and on the same scale. If one dimension represents length and another - weight of item - euclidean should be replaced with weighted.
Make it in 2d and show the picture - this is good option to see visually if it works.
Or you may use some sanity check - like to find cluster centers and see that all items in the cluster aren't too away of it.
Can't you just try sum |xi - yi| instead if (xi - yi)^2
in your code, and see if it makes much difference ?
I can't have a graph which will give some idea about the correctness of my algorithm.
A couple of possibilities:
look at some points midway between 2 clusters in detail
vary k a bit, see what happens (what is your k ?)
use
PCA
to map 30d down to 2d; see the plots under
calculating-the-percentage-of-variance-measure-for-k-means,
also SO questions/tagged/pca
By the way, scipy.spatial.cKDTree
can easily give you say 3 nearest neighbors of each point,
in p=2 (Euclidean) or p=1 (Manhattan, L1), to look at.
It's fast up to ~ 20d, and with early cutoff works even in 128d.
Added: I like Cosine distance in high dimensions; see euclidean-distance-is-usually-not-good-for-sparse-data for why.
Euclidean distance is the intuitive and "normal" distance between continuous variable. It can be inappropriate if too noisy or if data has a non-gaussian distribution.
You might want to try the Manhattan distance (or cityblock) which is robust to that (bear in mind that robustness always comes at a cost : a bit of the information is lost, in this case).
There are many further distance metrics for specific problems (for example Bray-Curtis distance for count data). You might want to try some of the distances implemented in pdist from python module scipy.spatial.distance.