I am new to Python. I would like to perform hierarchical clustering on N by P dataset that contains some missing values. I am planning to use scipy.cluster.hierarchy.linkage function that takes distance matrix in condensed form. Does Python have a method to compute distance matrix for missing value contained data? (In R dist function automatically takes care of missing values... but scipy.spatial.distance.pdist seems not handling missing values!)
I could not find a method to compute distance matrix for data with missing values. So here is my naive solution using Euclidean distance.
import numpy as np
def getMissDist(x,y):
return np.nanmean( (x - y)**2 )
def getMissDistMat(dat):
Npat = dat.shape[0]
dist = np.ndarray(shape=(Npat,Npat))
dist.fill(0)
for ix in range(0,Npat):
x = dat[ix,]
if ix >0:
for iy in range(0,ix):
y = dat[iy,]
dist[ix,iy] = getMissDist(x,y)
dist[iy,ix] = dist[ix,iy]
return dist
Then assume that dat is N (= number of cases) by P (=number of features) data matrix with missing values then one can perform hierarchical clustering on this dat as:
distMat = getMissDistMat(dat)
condensDist = dist.squareform(distMat)
link = hier.linkage(condensDist, method='average')
Related
I'm calculating with Python. Let's say i have this kind of DataFrame where it consists of long lat of some points
import pandas as pd
dfa=pd.DataFrame(([1,2],[1,3],[1,1],[1,4]), columns=['y','x'])
before, i used distance matrix from scipy.spatial and create another DataFrame with this code. but it seems that it can't precisely calculate the distance between points (with long lat)
from scipy.spatial import distance_matrix
pd.DataFrame(distance_matrix(dfa.values, dfa.values), index=dfa.index, columns=dfa.index)
Do you think it's possible to change the calculation with geodesic? here what i've tried.
from geopy.distance import geodesic
pd.DataFrame(geodesic(dfa.values[0], dfa.values[0]).kilometers, index=dfa.index, columns=dfa.index)
# i don't know how to change [0] adjusted to column and index
any suggestion?
Given a list or list-like object locations, you can do
distances = pd.DataFrame([[geodesic(a,b) for a in locations]
for b in locations])
This will be redundant, though, since it will calculate distance for both a,b and b,a, even though they should be the same. Depending on the cost of geodesic, you may find the some of the following alternatives faster:
distances = pd.DataFrame([[geodesic(a,b) if a > b else 0
for a in locations]
for b in locations])
distances = distances.add(distances.T)
size = len(locations)
distances = pd.DataFrame(columns = range(size), index = range(size))
def get_distance(i,j):
if distances.loc[j,i]:
return distances.loc[j,i]
if i == j:
return 0
return geodesic(locations[i], locations[j])
for i in range(size):
for j in range(size):
distances.loc[i,j] = get_distance(i,j)
You can also store the data as a dictionary with the keys being output from itertools.combinations. There's also this article on creating a symmetric matrix class.
I'm working on an assignment where I am tasked to implement PCA in Python for an online course. Unfortunately, when I try to run a comparison (provided by the course) between my implementation and SKLearn's, my results appear to differ too greatly.
After many hours of review, I am still unsure where it is going wrong. If someone could take a look and determine what step I have coded or interpreted incorrectly, I would greatly appreciate it.
def normalize(X):
"""
Normalize the given dataset X to have zero mean.
Args:
X: ndarray, dataset of shape (N,D)
Returns:
(Xbar, mean): tuple of ndarray, Xbar is the normalized dataset
with mean 0; mean is the sample mean of the dataset.
Note:
You will encounter dimensions where the standard deviation is zero.
For those ones, the process of normalization results in normalized data with NaN entries.
We can handle this by setting the std = 1 for those dimensions when doing normalization.
"""
# YOUR CODE HERE
### Uncomment and modify the code below
mu = np.mean(X, axis = 0) # Setting axis = 0 will compute means column-wise. Setting it to 1 will compute the mean across rows.
std = np.std(X, axis = 0) # Computing the std dev column wise using axis = 0.
std_filled = std.copy()
std_filled[std == 0] = 1
# Compute the normalized data as Xbar
Xbar = (X - mu)/std_filled
return Xbar, mu, # std_filled
def eig(S):
"""
Compute the eigenvalues and corresponding unit eigenvectors for the covariance matrix S.
Args:
S: ndarray, covariance matrix
Returns:
(eigvals, eigvecs): ndarray, the eigenvalues and eigenvectors
Note:
the eigenvals and eigenvecs should be sorted in descending
order of the eigen values
"""
# YOUR CODE HERE
# Uncomment and modify the code below
# Compute the eigenvalues and eigenvectors
# You can use library routines in `np.linalg.*` https://numpy.org/doc/stable/reference/routines.linalg.html for this
eigvals, eigvecs = np.linalg.eig(S)
# The eigenvalues and eigenvectors need to be sorted in descending order according to the eigenvalues
# We will use `np.argsort` (https://docs.scipy.org/doc/numpy/reference/generated/numpy.argsort.html) to find a permutation of the indices
# of eigvals that will sort eigvals in ascending order and then find the descending order via [::-1], which reverse the indices
sort_indices = np.argsort(eigvals)[::-1]
# Notice that we are sorting the columns (not rows) of eigvecs since the columns represent the eigenvectors.
return eigvals[sort_indices], eigvecs[:, sort_indices]
def projection_matrix(B):
"""Compute the projection matrix onto the space spanned by the columns of `B`
Args:
B: ndarray of dimension (D, M), the basis for the subspace
Returns:
P: the projection matrix
"""
# YOUR CODE HERE
P = B # (np.linalg.inv(B.T # B)) # B.T
return P
def select_components(eig_vals, eig_vecs, num_components):
"""
Selects the n components desired for projecting the data upon.
Args:
eig_vals: The eigenvalues sorted in descending order of magnitude.
eig_vecs: The eigenvectors sorted in order relative to that of the eigenvalues.
num_components: the number of principal components to use.
Returns:
The number of desired components to keep for projection of the data upon.
"""
principal_vals, principal_components = eig_vals[:num_components], eig_vecs[:, range(num_components)]
return principal_vals, principal_components
def PCA(X, num_components):
"""
Projects normalized data onto the 'n' desired principal components.
Args:
X: ndarray of size (N, D), where D is the dimension of the data,
and N is the number of datapoints
num_components: the number of principal components to use.
Returns:
the reconstructed data, the sample mean of the X, principal values
and principal components
"""
# Normalize to have mean 0 and variance 1.
Z, mean_vec = normalize(X)
# Calculate the covariance matrix
S = np.cov(Z, rowvar=False, bias=True) # Set rowvar = False to treat columns as variables. Set bias = True to ensure normalization is done with N and not N-1
# Calculate the (unit) eigenvectors and eigenvalues of S. Sort them in descending order of importance relative to the magnitude of the eigenvalues.
eig_vals, eig_vecs = eig(S)
# Keep only the n largest Principle Components of the sorted unit eigenvectors.
principal_vals, principal_components = select_components(eig_vals, eig_vecs, num_components)
# Compute the projection matrix using only the n largest Principle Components of the sorted unit eigenvectors, where n = num_components.
#P = projection_matrix(eig_vecs[:, :num_components])
P = projection_matrix(principal_components)
# Reconstruct the data by using the projection matrix to project the data onto the principal component vectors we've kept
X_reconst = (P # X.T).T
return X_reconst, mean_vec, principal_vals, principal_components
And here is the test case I'm supposed to pass:
random = np.random.RandomState(0)
X = random.randn(10, 5)
from sklearn.decomposition import PCA as SKPCA
for num_component in range(1, 4):
# We can compute a standard solution given by scikit-learn's implementation of PCA
pca = SKPCA(n_components=num_component, svd_solver="full")
sklearn_reconst = pca.inverse_transform(pca.fit_transform(X))
reconst, _, _, _ = PCA(X, num_component)
# The difference in the result should be very small (<10^-20)
print(
"difference in reconstruction for num_components = {}: {}".format(
num_component, np.square(reconst - sklearn_reconst).sum()
)
)
np.testing.assert_allclose(reconst, sklearn_reconst)
As far as I can tell, there are a few things wrong with your code.
Your projection matrix is wrong.
If the eigenvectors of your covariance matrix is B with dimension D x M where M is the number of components you select and D is the dimension of the original data, then the projection matrix is just B # B.T.
In standard implementation of PCA, we typically do not scale the data by the inverse of the standard deviation. You seem to be trying to do an approximation of a whitened PCA (ZCA), but even then it looks wrong.
As a quick test, you can compute the normalized data without dividing by the standard deviation, and when you compute the covariance matrix, set bias=False.
You should also subtract the mean from the data before multiplying it by the projection operator, and adding it back after that, i.e.,
X_reconst = (P # (X - mean_vec).T).T + mean_vec.
PCA essentially is just a change of basis, followed by discarding coordinates corresponding to directions with low variance. The eigenvectors of the covariance matrix corresponds to the new orthogonal basis, and the eigenvalues tells you the variance of the data along the direction of the corresponding eigenvectors. P = B # B.T is just the change of basis followed to the new basis (and discarding some coordinates), B, followed by a change back to the original basis.
Edit
I'm curious to know which online course teaches people to implement PCA this way.
I'm trying to follow an example on k-Nearest Neighbors and I'm not sure about the numpy command syntax. I'm supposed to be doing a matrix-wise distance calculation and the code given is
def classify(inputVector, trainingData,labels,k):
dataSetSize=trainingData.shape[0]
diffMat=tile(inputVector,(dataSetSize,1))-trainingData
sqDiffMat = diffMat**2
sqDistances = sqDiffMat.sum(axis=1)
distances = sqDistances**0.5
sortedDistIndicies = distances.argsort()
for i in range(k):
voteIlabel = labels[sortedDistIndicies[i]]
classCount[voteIlabel] = classCount.get(voteIlabel,0) + 1
sortedClassCount = sorted(classCount.iteritems(), key=operator.itemgetter(1), reverse=True)
return sortedClassCount[0][0]
def createDataSet():
group = array([[1.0,1.1],[1.0,1.0],[0,0],[0,0.1]])
labels = ['A','A','B','B']
return group, labels
my question is how does sqDistances**0.5 amount to the distance equation ((A[0]-B[0])+(A[1]-B[1]))^1/2? I don't follow how tile influences it specifically how the matrix is made from (datasetsize,1)-training data.
I hope the following will explain the working.
Numpy tile : https://docs.scipy.org/doc/numpy/reference/generated/numpy.tile.html
Using this function, you are creating matrix from input vector same to the shape of training data. From this matrix you are subtracting training data which will give you some part from what you mentioned say test[0]-train[0] i.e. element wise difference.
Then you squared each obtained element by using diffMat**2 and then taken sum along axis = 1 (https://docs.scipy.org/doc/numpy/reference/generated/numpy.sum.html). This resulted in equations like (test[0] - train[0])^2 + (test[1] - train[1])^2.
Next by taking sqDistances**0.5 , it will give Euclidean distance.
To calculate Euclidean distance, this might be helpful
https://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.euclidean.html#scipy.spatial.distance.euclidean
I have the following setup for kmeans clustering algorithm that I am implementing for a project:
import numpy as np
import scipy
import sys
import random
import matplotlib.pyplot as plt
import operator
class KMeansClass:
#takes in an npArray like object
def __init__(self,dataset,k):
self.dataset=np.array(dataset)
#initialize mins to maximum possible value
self.min_x = sys.maxint
self.min_y = sys.maxint
#initialize maxs to minimum possible value
self.max_x = -(sys.maxint)-1
self.max_y = -(sys.maxint)-1
self.k = k
#a is the coefficient matrix that is continually updated as the centroids of the clusters change respectively.
# It is an mxk matrix where each row corresponds to a training_instance and each column corresponds to a centroid of a cluster
#Values are either 0 or 1. A value for a particular training_instance (data_point) is 1 only for that centroid to which the training_instance
# has the least distance else the value is 0.
self.a = np.zeros(shape=[self.dataset.shape[0],self.k])
self.distanceMatrix = np.empty(shape =[self.dataset.shape[0],self.k])
#initialize mu to zeros of the requisite shape array for now. Change this after implementing max and min methods.
self.mu = np.empty(shape=[k,2])
self.findMinMaxdataPoints()
self.initializeCentroids()
self.createDistanceMatrix()
self.scatterPlotOfInitializedPoints()
#pointa and pointb are npArray like vecors.
def euclideanDistance(self,pointa,pointb):
return np.sqrt(np.sum((pointa - pointb)**2))
""" Problem Initialization And Visualization Helper methods"""
##############################################################################
##param: dataset : list of tuples [(x1,y1),(x2,y2),...(xm,ym)]
def findMinMaxdataPoints(self):
for item in self.dataset:
self.min_x = min(self.min_x,item[0])
self.min_y = min(self.min_y,item[1])
self.max_x = max(self.max_x,item[0])
self.max_y = max(self.max_y,item[1])
def initializeCentroids(self):
for i in range(self.k):
#each value of mu is a tuple with a random number between (min_x - max_x) and (min_y - max_y)
self.mu[i] = (random.randint(self.min_x,self.max_x),random.randint(self.min_y,self.max_y))
self.sortCentroids()
print self.mu
def sortCentroids(self):
#the following 3 lines of code are to ensure that the mu values are always sorted in ascending order first with respect to the
#x values and then with respect to the y values.
half_sorted = sorted(self.mu,key=operator.itemgetter(1)) #sort wrt y values
full_sorted = sorted(half_sorted,key=operator.itemgetter(0)) #sort the y-sorted array wrt x-values
self.mu = np.array(full_sorted)
def scatterPlotOfInitializedPoints(self):
plt.scatter([item[0] for item in self.dataset],[item[1] for item in self.dataset],color='b')
plt.scatter([item[0] for item in self.mu],[item[1] for item in self.mu],color='r')
plt.show()
###############################################################################
#minimizing euclidean distance is the same as minimizing the square of the euclidean distance.
def calcSquareEuclideanDistanceBetweenTwoPoints(point_a,point_b):
return np.sum((pointa-pointb)**2)
def createDistanceMatrix(self):
for i in range(self.dataset.shape[0]):
for j in range(self.k):
self.distanceMatrix[i,j] = calcSquareEuclideanDistanceBetweenTwoPoints(self.dataset[i],self.mu[j])
def createCoefficientMatrix(self):
for i in range(self.dataset.shape[0]):
self.a[i,self.distanceMatrix[i].argmin()] = 1
#update functions for CoefficientMatrix and Centroid values:
def updateCoefficientMatrix(self):
for i in range(self.dataset.shape[0]):
self.a[i,self.distanceMatrix[i].argmin()]= 1
def updateCentroids(self):
for j in range(self.k):
non_zero_indices = np.nonzero(self.a[:,j])
avg = 0
for i in range(len(non_zero_indices[0])):
avg+=self.a[non_zero_indices[0][i],j]
self.mu[j] = avg/len(non_zero_indices[0])
############################################################
def lossFunction(self):
loss=0;
for j in range(self.k):
#vectorized this implementation.
loss+=np.sum(np.dot(self.a[:,j],self.distanceMatrix[:,j]))
return loss
Here my question pertains to the lossFunction and how to use this with the scipy.optimize package. I would like to minimize the loss function iteratively by performing the following steps:
Repeat until convergence:
a> Optimize 'a' by keeping mu constant ( I have an
updateCoefficientMatrix method for updating 'a' matrix which is an
mXk matrix where we have m training instances and k clusters.)
b> Optimize 'mu' by keeping 'a' constant (I have an updateCentroids
method to do this. where mu is a mXk matrix wherein m is number of
training instances and k is the number of clusters and the number of
centroids)
But I am very new to using scipy.optimize package so I am writing to ask for help as to how to invoke the scipy.optimize to achieve my optimization goal as stated above?
Basically I have 2 mxk matrices and I would like to minimize a lossFunction() by first optimizing one mxk matrix keeping the other constant and in the succeeding step optimize the second matrix keeping the first constant. This can be considered a special case of the expectation maximization problem but unfortunately I haven't quite gotten what the documentation is trying to say so far hence thought I'd turn to SO for help.
Thanks in advance!
And this is part of a class assignment so please do not post code! Any guidance or explanation would be highly appreciated.
Use scipy.optimize.minimize twice with different objective functions.
First run optimization with an objective function that takes a as a parameter, and returns the objective value.
As the second step, run scipy.optimize.minimize for a second time on a second objective function that takes mu as a parameter.
When writing the objective functions, remember that Python has nested functions, which avoids the need for passing mu (in the first case) or a (in the second case) as additional arguments; although it can be done by minimize(..., args=[mu]) and minimize(..., args=[a]).
Repeat the two-step process in a for loop, until the answer is such that your convergence condition is satisfied.
Note:
This is for a homework assignment in my data mining class.
I'm going to put relevant code snippets on this SO post, but you can find my entire program at http://pastebin.com/CzNFbLJ2
The dataset I'm using for this program can be found at http://archive.ics.uci.edu/ml/datasets/Iris
So I'm getting: RuntimeWarning: invalid value encountered in sqrt
return np.sqrt(m)
I am attempting to find the average Mahalanobis distance of the given iris dataset (for both raw and normalized datasets). The error is only happening on the normalized version of the dataset which is making me wonder if I have incorrectly understood what normalization means (both in code and mathematically).
I thought that normalization means that each component of a vector is divided by it's vector length (causing the vector to add up to 1). I found this SO question How to normalize a 2-dimensional numpy array in python less verbose? and thought it matched up to my concept of normalization. But now my code is reporting that the Mahalanobis distance over the normalized dataset is NAN
def mahalanobis(data):
import numpy as np;
import scipy.spatial.distance;
avg = 0
count = 0
covar = np.cov(data, rowvar=0);
invcovar = np.linalg.inv(covar)
for i in range(len(data)):
for j in range(i + 1, len(data)):
if(j == len(data)):
break
avg += scipy.spatial.distance.mahalanobis(data[i], data[j], invcovar)
count += 1
return avg / count
def normalize(data):
import numpy as np
row_sums = data.sum(axis=1)
norm_data = np.zeros((50, 4))
for i, (row, row_sum) in enumerate(zip(data, row_sums)):
norm_data[i,:] = row / row_sum
return norm_data
Probably too late, but check out page 64-65 in our textbook "Introduction to Data Mining". There's a section called "Normalization or Standardization", which explains the concept of normalized data that Hearne is looking for.
Basically, standardized data set x' = (x - mean(x)) / standardDeviation(x)
Since I see you're using python, here's how to do it using SciPy:
normalizedData = (data - data.mean(axis=0)) / data.std(axis=0, ddof=1)
Source: http://mail.scipy.org/pipermail/numpy-discussion/2011-April/056023.html
You can use pdist() to do the distance calculation without for loop:
from sklearn import datasets
iris = datasets.load_iris()
from scipy.spatial.distance import pdist, squareform
print squareform(pdist(iris.data, 'mahalanobis'))
Normalization in this context probably does mean subtracting the mean and scaling so the data has a unit covariance matrix.
However, to scale every vector in your dataset to unit norm use: norm_data=data/np.sqrt(np.sum(data*data,1))[:,None].
You need to divide by the L2 norm of each vector, which means squaring the value of each element, then taking the square root of the sum. Broadcasting allows you to avoid explicitly coding the loop (see the answer to the question you cited: https://stackoverflow.com/a/8904762/1149913).