How do I print out objects in an array in python?
- by Jonathan
I'm writing a code which performs a k-means clustering on a set of data. I'm actually using the code from a book called collective intelligence by O'Reilly. Everything works, but in his code he uses the command line and i want to write everything in notepad++. As a reference his line is
>>>kclust=clusters.kcluster(data,k=10)
>>>[rownames[r] for r in k[0]]
Here is my code:
from PIL import Image,ImageDraw
def readfile(filename):
lines=[line for line in file(filename)]
# First line is the column titles
colnames=lines[0].strip( ).split('\t')[1:]
rownames=[]
data=[]
for line in lines[1:]:
p=line.strip( ).split('\t')
# First column in each row is the rowname
rownames.append(p[0])
# The data for this row is the remainder of the row
data.append([float(x) for x in p[1:]])
return rownames,colnames,data
from math import sqrt
def pearson(v1,v2):
# Simple sums
sum1=sum(v1)
sum2=sum(v2)
# Sums of the squares
sum1Sq=sum([pow(v,2) for v in v1])
sum2Sq=sum([pow(v,2) for v in v2])
# Sum of the products
pSum=sum([v1[i]*v2[i] for i in range(len(v1))])
# Calculate r (Pearson score)
num=pSum-(sum1*sum2/len(v1))
den=sqrt((sum1Sq-pow(sum1,2)/len(v1))*(sum2Sq-pow(sum2,2)/len(v1)))
if den==0: return 0
return 1.0-num/den
class bicluster:
def __init__(self,vec,left=None,right=None,distance=0.0,id=None):
self.left=left
self.right=right
self.vec=vec
self.id=id
self.distance=distance
def hcluster(rows,distance=pearson):
distances={}
currentclustid=-1
# Clusters are initially just the rows
clust=[bicluster(rows[i],id=i) for i in range(len(rows))]
while len(clust)>1:
lowestpair=(0,1)
closest=distance(clust[0].vec,clust[1].vec)
# loop through every pair looking for the smallest distance
for i in range(len(clust)):
for j in range(i+1,len(clust)):
# distances is the cache of distance calculations
if (clust[i].id,clust[j].id) not in distances:
distances[(clust[i].id,clust[j].id)]=distance(clust[i].vec,clust[j].vec)
#print 'i'
#print i
#print
#print 'j'
#print j
#print
d=distances[(clust[i].id,clust[j].id)]
if d<closest:
closest=d
lowestpair=(i,j)
# calculate the average of the two clusters
mergevec=[
(clust[lowestpair[0]].vec[i]+clust[lowestpair[1]].vec[i])/2.0
for i in range(len(clust[0].vec))]
# create the new cluster
newcluster=bicluster(mergevec,left=clust[lowestpair[0]],
right=clust[lowestpair[1]],
distance=closest,id=currentclustid)
# cluster ids that weren't in the original set are negative
currentclustid-=1
del clust[lowestpair[1]]
del clust[lowestpair[0]]
clust.append(newcluster)
return clust[0]
def kcluster(rows,distance=pearson,k=4):
# Determine the minimum and maximum values for each point
ranges=[(min([row[i] for row in rows]),max([row[i] for row in rows]))
for i in range(len(rows[0]))]
# Create k randomly placed centroids
clusters=[[random.random( )*(ranges[i][1]-ranges[i][0])+ranges[i][0]
for i in range(len(rows[0]))] for j in range(k)]
lastmatches=None
for t in range(100):
print 'Iteration %d' % t
bestmatches=[[] for i in range(k)]
# Find which centroid is the closest for each row
for j in range(len(rows)):
row=rows[j]
bestmatch=0
for i in range(k):
d=distance(clusters[i],row)
if d<distance(clusters[bestmatch],row): bestmatch=i
bestmatches[bestmatch].append(j)
# If the results are the same as last time, this is complete
if bestmatches==lastmatches: break
lastmatches=bestmatches
# Move the centroids to the average of their members
for i in range(k):
avgs=[0.0]*len(rows[0])
if len(bestmatches[i])>0:
for rowid in bestmatches[i]:
for m in range(len(rows[rowid])):
avgs[m]+=rows[rowid][m]
for j in range(len(avgs)):
avgs[j]/=len(bestmatches[i])
clusters[i]=avgs
return bestmatches
