I want to find the association between variables and cramer V works like a treat for matrices of sizes greater than 2X2. However, for matrices with low frequencies, it does not work well. For the following contingency matrix, i get the result as 0.5. How can I account for the same?
1 2
a 2 0
b 0 2
Here is my code:
def cramers_stat(confusion_matrix):
chi2 = ss.chi2_contingency(confusion_matrix)[0]
n = confusion_matrix.sum().sum()
return np.sqrt(chi2 / (n*(min(confusion_matrix.shape)-1)))
result=cramers_stat(confusion_matrix)
print(result)
confusion_matrix is my input, in this case the matrix i mentioned above. I understand for good results, i need a matrix frequency above 5, but for perfect association as the case above I expected the result to be 1.
When you compute the Cramér coefficient, you must compute chi2 without continuity correction. For a 2x2 matrix, chi2_contingency uses continuity correction by default. So you must tell chi2_contingency to not use continuity correction by giving the argument correction=False:
chi2 = ss.chi2_contingency(confusion_matrix, correction=False)[0]
Related
I was going through the book called Hands-On Machine Learning with Scikit-Learn, Keras and Tensorflow and the author was explaining how the pseudo-inverse (Moore-Penrose inverse) of a matrix is calculated in the context of Linear Regression. I'm quoting verbatim here:
The pseudoinverse itself is computed using a standard matrix
factorization technique called Singular Value Decomposition (SVD) that
can decompose the training set matrix X into the matrix
multiplication of three matrices U Σ VT (see numpy.linalg.svd()). The
pseudoinverse is calculated as X+ = V * Σ+ * UT. To compute the matrix
Σ+, the algorithm takes Σ and sets to zero all values smaller than a
tiny threshold value, then it replaces all nonzero values with their
inverse, and finally it transposes the resulting matrix. This approach
is more efficient than computing the Normal equation.
I've got an understanding of how the pseudo-inverse and SVD are related from this post. But I'm not able to grasp the rationale behind setting all values less than the threshold to zero. The inverse of a diagonal matrix is obtained by taking the reciprocals of the diagonal elements. Then small values would be converted to large values in the inverse matrix, right? Then why are we removing the large values?
I went and looked into the numpy code, and it looks like follows, just for reference:
#array_function_dispatch(_pinv_dispatcher)
def pinv(a, rcond=1e-15, hermitian=False):
a, wrap = _makearray(a)
rcond = asarray(rcond)
if _is_empty_2d(a):
m, n = a.shape[-2:]
res = empty(a.shape[:-2] + (n, m), dtype=a.dtype)
return wrap(res)
a = a.conjugate()
u, s, vt = svd(a, full_matrices=False, hermitian=hermitian)
# discard small singular values
cutoff = rcond[..., newaxis] * amax(s, axis=-1, keepdims=True)
large = s > cutoff
s = divide(1, s, where=large, out=s)
s[~large] = 0
res = matmul(transpose(vt), multiply(s[..., newaxis], transpose(u)))
return wrap(res)
It's almost certainly an adjustment for numerical error. To see why this might be necessary, look what happens when you take the svd of a rank-one 2x2 matrix. We can create a rank-one matrix by taking the outer product of a vector like so:
>>> a = numpy.arange(2) + 1
>>> A = a[:, None] * a[None, :]
>>> A
array([[1, 2],
[2, 4]])
Although this is a 2x2 matrix, it only has one linearly independent column, and so its rank is one instead of two. So we should expect that when we pass it to svd, one of the singular values will be zero. But look what happens:
>>> U, s, V = numpy.linalg.svd(A)
>>> s
array([5.00000000e+00, 1.98602732e-16])
What we actually get is a singular value that is not quite zero. This result is inevitable in many cases given that we are working with finite-precision floating point numbers. So although the problem you have identified is a real one, we will not be able to tell in practice the difference between a matrix that really has a very small singular value and a matrix that ought to have a zero singular value but doesn't. Setting small values to zero is the safest practical way to handle that problem.
I've got a 2x2 matrix defined by the variables J00, J01, J10, J11 coming in from other inputs. Since the matrix is small, I was able to compute the spectral norm by first computing the trace and determinant
J_T = tf.reduce_sum([J00, J11])
J_ad = tf.reduce_prod([J00, J11])
J_cb = tf.reduce_prod([J01, J10])
J_det = tf.reduce_sum([J_ad, -J_cb])
and then solving the quadratic
L1 = J_T/2.0 + tf.sqrt(J_T**2/4.0 - J_det)
L2 = J_T/2.0 - tf.sqrt(J_T**2/4.0 - J_det)
spectral_norm = tf.maximum(L1, L2)
This works, but it looks rather ugly and it isn't generalizable to larger matrices. Is there cleaner way (maybe a method call that I'm missing) to compute spectral_norm?
The spectral norm of a matrix J equals the largest singular value of the matrix.
Therefore you can use tf.svd() to perform the singular value decomposition, and take the largest singular value:
spectral_norm = tf.svd(J,compute_uv=False)[...,0]
where J is your matrix.
Notes:
I use compute_uv=False since we are interested only in singular values, not singular vectors.
J does not need to be square.
This solution works also for the case where J has any number of batch dimensions (as long as the two last dimensions are the matrix dimensions).
The elipsis ... operation works as in NumPy.
I take the 0 index because we are interested only in the largest singular value.
I am trying to understand this optimized code to find cosine similarity between users matrix.
def fast_similarity(ratings,epsilon=1e-9):
# epsilon -> small number for handling dived-by-zero errors
sim = ratings.T.dot(ratings) + epsilon
norms = np.array([np.sqrt(np.diagonal(sim))])
return (sim / norms / norms.T)
If ratings =
items
u [
s [1,2,3]
e [4,5,6]
r [7,8,9]
s ]
nomrs will be equal to = [1^2 + 5^2 + 9^2]
but why we are writing sim/norms/norms.T to calculate cosine similarity?
Any help is appreciated.
Going through the code we have that:
And this means that, one the diagonal of the sim matrix we have the result of the multiplication of each column.
You can give it a try if you want using a simple matrix:
And you can easily check that this gram matrix (that's how this matrix product is named) has this property.
Now the code defines norms that is nothing but an array taking the diagonal of our gram matrix and apply a sqrt on each element of it.
This will give us an array containing the norm value for each column:
So basically the norms vector contains the norm value of each column of the result matrix.
Once we have all those data we can evaluate the cosine similarity between those users, so we know that cosine similarity is evaluated like:
Note that :
So we have that our similarity is going to be:
So we just have to substitute the terms with our code variable to get:
And this explain why you have this line of code:
return sim / norms / norms.T
EDIT:
Since it seems that I was not clear, every time I am talking about matrix multiplication in this answer I am reffering to the DOT PRODUCT of two matrices.
This actually means that when it's written A*B we actually develop and
solve as A.T * B
I would like to compute a distance matrix using the Jaccard distance. And do so as fast as possible. I used to use scikit-learn's pairwise_distances function. But scikit-learn doesn't plan to support GPU, and there's even a known bug that makes the function slower when run in parallel.
My only constraint is that the resulting distance matrix can then be fed to scikit-learn's DBSCAN clustering algorithm. I was thinking about implementing the computation with tensorflow but couldn't find a nice and simple way to do it.
PS: I have reasons to precompute the distance matrix instead of letting DBSCAN do it as needed.
Hej I was facing the same problem.
Given the idea that the jaccard similarity is the ratio of true postives (tp) to the sum of true positives, false negatives (fn) and false positives (fp), I came up with this solution:
def jaccard_distance(self):
tp = tf.reduce_sum(tf.mul(self.target, self.prediction), 1)
fn = tf.reduce_sum(tf.mul(self.target, 1-self.prediction), 1)
fp = tf.reduce_sum(tf.mul(1-self.target, self.prediction), 1)
return 1 - (tp / (tp + fn + fp))
Hope this helps!
I am not a tensorflow expert, but here is the solution I got. As far as I know, the only ways in tensorflow to do a computation on all-pairs of a list is to do a matrix multiplication or use the broadcasting rules, this solution uses both at some point.
So let's assume we have an input boolean matrix of n_samples rows, one per set, and n_features columns, one per possible element. A value True in the i-th row, j-th column means the i-th set contains the element j. Just like scikit-learn's pairwise_distances expect. We can then proceed as follow.
Cast the matrix to numbers, getting 1 for True and 0 for False.
Multiply the matrix by its own transpose. This produce a matrix where each element M[i][j] contains size of the intersection between the i-th and j-th sets.
Compute a cardv vector that contains the cardinality of all the sets by summing the input matrix by rows.
Make a row and a column vector from cardv.
Compute 1 - M / (cardvrow + cardvcol - M). The broadcasting rules will do all the work when adding a row and a column vector.
This algorithm as a whole seems a bit hack-ish, but it works and produce results within a reasonable margin from the result computed by scikit-learn's pairwise_distances function. A better algorithm should probably make a single pass on every pair of input vectors and compute only half of the matrix as it is symmetric. Any improvement is welcome.
setsin = tf.placeholder(tf.bool, shape=(N, M))
sets = tf.cast(setsin, tf.float16)
mat = tf.matmul(sets, sets, transpose_b=True, name="Main_matmul")
#mat = tf.cast(mat, tf.float32, name="Upgrade_mat")
#sets = tf.cast(sets, tf.float32, name="Upgrade_sets")
cardinal = tf.reduce_sum(sets, 1, name="Richelieu")
cardinalrow = tf.expand_dims(cardinal, 0)
cardinalcol = tf.expand_dims(cardinal, 1)
mat = 1 - mat / (cardinalrow + cardinalcol - mat)
I used float16 type as it seems much faster than float32. Casting to float32 might only be useful if the cardinals are large enough to make them inaccurate or if more precision is needed when performing the division. But even when the casts are needed, it seems to be still relevant to do the matrix multiplication as float16.
My code:
from numpy import *
def pca(orig_data):
data = array(orig_data)
data = (data - data.mean(axis=0)) / data.std(axis=0)
u, s, v = linalg.svd(data)
print s #should be s**2 instead!
print v
def load_iris(path):
lines = []
with open(path) as input_file:
lines = input_file.readlines()
data = []
for line in lines:
cur_line = line.rstrip().split(',')
cur_line = cur_line[:-1]
cur_line = [float(elem) for elem in cur_line]
data.append(array(cur_line))
return array(data)
if __name__ == '__main__':
data = load_iris('iris.data')
pca(data)
The iris dataset: http://archive.ics.uci.edu/ml/machine-learning-databases/iris/iris.data
Output:
[ 20.89551896 11.75513248 4.7013819 1.75816839]
[[ 0.52237162 -0.26335492 0.58125401 0.56561105]
[-0.37231836 -0.92555649 -0.02109478 -0.06541577]
[ 0.72101681 -0.24203288 -0.14089226 -0.6338014 ]
[ 0.26199559 -0.12413481 -0.80115427 0.52354627]]
Desired Output:
Eigenvalues - [2.9108 0.9212 0.1474 0.0206]
Principal Components - Same as I got but transposed so okay I guess
Also, what's with the output of the linalg.eig function? According to the PCA description on wikipedia, I'm supposed to this:
cov_mat = cov(orig_data)
val, vec = linalg.eig(cov_mat)
print val
But it doesn't really match the output in the tutorials I found online. Plus, if I have 4 dimensions, I thought I should have 4 eigenvalues and not 150 like the eig gives me. Am I doing something wrong?
edit: I've noticed that the values differ by 150, which is the number of elements in the dataset. Also, the eigenvalues are supposed to add to be equal to the number of dimensions, in this case, 4. What I don't understand is why this difference is happening. If I simply divided the eigenvalues by len(data) I could get the result I want, but I don't understand why. Either way the proportion of the eigenvalues isn't altered, but they are important to me so I'd like to understand what's going on.
You decomposed the wrong matrix.
Principal Component Analysis requires manipulating the eigenvectors/eigenvalues
of the covariance matrix, not the data itself. The covariance matrix, created from an m x n data matrix, will be an m x m matrix with ones along the main diagonal.
You can indeed use the cov function, but you need further manipulation of your data. It's probably a little easier to use a similar function, corrcoef:
import numpy as NP
import numpy.linalg as LA
# a simulated data set with 8 data points, each point having five features
data = NP.random.randint(0, 10, 40).reshape(8, 5)
# usually a good idea to mean center your data first:
data -= NP.mean(data, axis=0)
# calculate the covariance matrix
C = NP.corrcoef(data, rowvar=0)
# returns an m x m matrix, or here a 5 x 5 matrix)
# now get the eigenvalues/eigenvectors of C:
eval, evec = LA.eig(C)
To get the eigenvectors/eigenvalues, I did not decompose the covariance matrix using SVD,
though, you certainly can. My preference is to calculate them using eig in NumPy's (or SciPy's)
LA module--it is a little easier to work with than svd, the return values are the eigenvectors
and eigenvalues themselves, and nothing else. By contrast, as you know, svd doesn't return these these directly.
Granted the SVD function will decompose any matrix, not just square ones (to which the eig function is limited); however when doing PCA, you'll always have a square matrix to decompose,
regardless of the form that your data is in. This is obvious because the matrix you
are decomposing in PCA is a covariance matrix, which by definition is always square
(i.e., the columns are the individual data points of the original matrix, likewise
for the rows, and each cell is the covariance of those two points, as evidenced
by the ones down the main diagonal--a given data point has perfect covariance with itself).
The left singular values returned by SVD(A) are the eigenvectors of AA^T.
The covariance matrix of a dataset A is : 1/(N-1) * AA^T
Now, when you do PCA by using the SVD, you have to divide each entry in your A matrix by (N-1) so you get the eigenvalues of the covariance with the correct scale.
In your case, N=150 and you haven't done this division, hence the discrepancy.
This is explained in detail here
(Can you ask one question, please? Or at least list your questions separately. Your post reads like a stream of consciousness because you are not asking one single question.)
You probably used cov incorrectly by not transposing the matrix first. If cov_mat is 4-by-4, then eig will produce four eigenvalues and four eigenvectors.
Note how SVD and PCA, while related, are not exactly the same. Let X be a 4-by-150 matrix of observations where each 4-element column is a single observation. Then, the following are equivalent:
a. the left singular vectors of X,
b. the principal components of X,
c. the eigenvectors of X X^T.
Also, the eigenvalues of X X^T are equal to the square of the singular values of X. To see all this, let X have the SVD X = QSV^T, where S is a diagonal matrix of singular values. Then consider the eigendecomposition D = Q^T X X^T Q, where D is a diagonal matrix of eigenvalues. Replace X with its SVD, and see what happens.
Question already adressed: Principal component analysis in Python