It's pretty simple really:
a[start:stop] # items start through stop-1
a[start:] # items start through the rest of the array
a[:stop] # items from the beginning through stop-1
a[:] # a copy of the whole array
There is also the step
value, which can be used with any of the above:
a[start:stop:step] # start through not past stop, by step
The key point to remember is that the :stop
value represents the first value that is not in the selected slice. So, the difference between stop
and start
is the number of elements selected (if step
is 1, the default).
The other feature is that start
or stop
may be a negative number, which means it counts from the end of the array instead of the beginning. So:
a[-1] # last item in the array
a[-2:] # last two items in the array
a[:-2] # everything except the last two items
Similarly, step
may be a negative number:
a[::-1] # all items in the array, reversed
a[1::-1] # the first two items, reversed
a[:-3:-1] # the last two items, reversed
a[-3::-1] # everything except the last two items, reversed
Python is kind to the programmer if there are fewer items than you ask for. For example, if you ask for a[:-2]
and a
only contains one element, you get an empty list instead of an error. Sometimes you would prefer the error, so you have to be aware that this may happen.
Relation to slice()
object
The slicing operator []
is actually being used in the above code with a slice()
object using the :
notation (which is only valid within []
), i.e.:
a[start:stop:step]
is equivalent to:
a[slice(start, stop, step)]
Slice objects also behave slightly differently depending on the number of arguments, similarly to range()
, i.e. both slice(stop)
and slice(start, stop[, step])
are supported.
To skip specifying a given argument, one might use None
, so that e.g. a[start:]
is equivalent to a[slice(start, None)]
or a[::-1]
is equivalent to a[slice(None, None, -1)]
.
While the :
-based notation is very helpful for simple slicing, the explicit use of slice()
objects simplifies the programmatic generation of slicing.
Here's a small kmeans that uses any of the 20-odd distances in
scipy.spatial.distance, or a user function.
Comments would be welcome (this has had only one user so far, not enough);
in particular, what are your N, dim, k, metric ?
#!/usr/bin/env python
# kmeans.py using any of the 20-odd metrics in scipy.spatial.distance
# kmeanssample 2 pass, first sample sqrt(N)
from __future__ import division
import random
import numpy as np
from scipy.spatial.distance import cdist # $scipy/spatial/distance.py
# http://docs.scipy.org/doc/scipy/reference/spatial.html
from scipy.sparse import issparse # $scipy/sparse/csr.py
__date__ = "2011-11-17 Nov denis"
# X sparse, any cdist metric: real app ?
# centres get dense rapidly, metrics in high dim hit distance whiteout
# vs unsupervised / semi-supervised svm
#...............................................................................
def kmeans( X, centres, delta=.001, maxiter=10, metric="euclidean", p=2, verbose=1 ):
""" centres, Xtocentre, distances = kmeans( X, initial centres ... )
in:
X N x dim may be sparse
centres k x dim: initial centres, e.g. random.sample( X, k )
delta: relative error, iterate until the average distance to centres
is within delta of the previous average distance
maxiter
metric: any of the 20-odd in scipy.spatial.distance
"chebyshev" = max, "cityblock" = L1, "minkowski" with p=
or a function( Xvec, centrevec ), e.g. Lqmetric below
p: for minkowski metric -- local mod cdist for 0 < p < 1 too
verbose: 0 silent, 2 prints running distances
out:
centres, k x dim
Xtocentre: each X -> its nearest centre, ints N -> k
distances, N
see also: kmeanssample below, class Kmeans below.
"""
if not issparse(X):
X = np.asanyarray(X) # ?
centres = centres.todense() if issparse(centres) \
else centres.copy()
N, dim = X.shape
k, cdim = centres.shape
if dim != cdim:
raise ValueError( "kmeans: X %s and centres %s must have the same number of columns" % (
X.shape, centres.shape ))
if verbose:
print "kmeans: X %s centres %s delta=%.2g maxiter=%d metric=%s" % (
X.shape, centres.shape, delta, maxiter, metric)
allx = np.arange(N)
prevdist = 0
for jiter in range( 1, maxiter+1 ):
D = cdist_sparse( X, centres, metric=metric, p=p ) # |X| x |centres|
xtoc = D.argmin(axis=1) # X -> nearest centre
distances = D[allx,xtoc]
avdist = distances.mean() # median ?
if verbose >= 2:
print "kmeans: av |X - nearest centre| = %.4g" % avdist
if (1 - delta) * prevdist <= avdist <= prevdist \
or jiter == maxiter:
break
prevdist = avdist
for jc in range(k): # (1 pass in C)
c = np.where( xtoc == jc )[0]
if len(c) > 0:
centres[jc] = X[c].mean( axis=0 )
if verbose:
print "kmeans: %d iterations cluster sizes:" % jiter, np.bincount(xtoc)
if verbose >= 2:
r50 = np.zeros(k)
r90 = np.zeros(k)
for j in range(k):
dist = distances[ xtoc == j ]
if len(dist) > 0:
r50[j], r90[j] = np.percentile( dist, (50, 90) )
print "kmeans: cluster 50 % radius", r50.astype(int)
print "kmeans: cluster 90 % radius", r90.astype(int)
# scale L1 / dim, L2 / sqrt(dim) ?
return centres, xtoc, distances
#...............................................................................
def kmeanssample( X, k, nsample=0, **kwargs ):
""" 2-pass kmeans, fast for large N:
1) kmeans a random sample of nsample ~ sqrt(N) from X
2) full kmeans, starting from those centres
"""
# merge w kmeans ? mttiw
# v large N: sample N^1/2, N^1/2 of that
# seed like sklearn ?
N, dim = X.shape
if nsample == 0:
nsample = max( 2*np.sqrt(N), 10*k )
Xsample = randomsample( X, int(nsample) )
pass1centres = randomsample( X, int(k) )
samplecentres = kmeans( Xsample, pass1centres, **kwargs )[0]
return kmeans( X, samplecentres, **kwargs )
def cdist_sparse( X, Y, **kwargs ):
""" -> |X| x |Y| cdist array, any cdist metric
X or Y may be sparse -- best csr
"""
# todense row at a time, v slow if both v sparse
sxy = 2*issparse(X) + issparse(Y)
if sxy == 0:
return cdist( X, Y, **kwargs )
d = np.empty( (X.shape[0], Y.shape[0]), np.float64 )
if sxy == 2:
for j, x in enumerate(X):
d[j] = cdist( x.todense(), Y, **kwargs ) [0]
elif sxy == 1:
for k, y in enumerate(Y):
d[:,k] = cdist( X, y.todense(), **kwargs ) [0]
else:
for j, x in enumerate(X):
for k, y in enumerate(Y):
d[j,k] = cdist( x.todense(), y.todense(), **kwargs ) [0]
return d
def randomsample( X, n ):
""" random.sample of the rows of X
X may be sparse -- best csr
"""
sampleix = random.sample( xrange( X.shape[0] ), int(n) )
return X[sampleix]
def nearestcentres( X, centres, metric="euclidean", p=2 ):
""" each X -> nearest centre, any metric
euclidean2 (~ withinss) is more sensitive to outliers,
cityblock (manhattan, L1) less sensitive
"""
D = cdist( X, centres, metric=metric, p=p ) # |X| x |centres|
return D.argmin(axis=1)
def Lqmetric( x, y=None, q=.5 ):
# yes a metric, may increase weight of near matches; see ...
return (np.abs(x - y) ** q) .mean() if y is not None \
else (np.abs(x) ** q) .mean()
#...............................................................................
class Kmeans:
""" km = Kmeans( X, k= or centres=, ... )
in: either initial centres= for kmeans
or k= [nsample=] for kmeanssample
out: km.centres, km.Xtocentre, km.distances
iterator:
for jcentre, J in km:
clustercentre = centres[jcentre]
J indexes e.g. X[J], classes[J]
"""
def __init__( self, X, k=0, centres=None, nsample=0, **kwargs ):
self.X = X
if centres is None:
self.centres, self.Xtocentre, self.distances = kmeanssample(
X, k=k, nsample=nsample, **kwargs )
else:
self.centres, self.Xtocentre, self.distances = kmeans(
X, centres, **kwargs )
def __iter__(self):
for jc in range(len(self.centres)):
yield jc, (self.Xtocentre == jc)
#...............................................................................
if __name__ == "__main__":
import random
import sys
from time import time
N = 10000
dim = 10
ncluster = 10
kmsample = 100 # 0: random centres, > 0: kmeanssample
kmdelta = .001
kmiter = 10
metric = "cityblock" # "chebyshev" = max, "cityblock" L1, Lqmetric
seed = 1
exec( "\n".join( sys.argv[1:] )) # run this.py N= ...
np.set_printoptions( 1, threshold=200, edgeitems=5, suppress=True )
np.random.seed(seed)
random.seed(seed)
print "N %d dim %d ncluster %d kmsample %d metric %s" % (
N, dim, ncluster, kmsample, metric)
X = np.random.exponential( size=(N,dim) )
# cf scikits-learn datasets/
t0 = time()
if kmsample > 0:
centres, xtoc, dist = kmeanssample( X, ncluster, nsample=kmsample,
delta=kmdelta, maxiter=kmiter, metric=metric, verbose=2 )
else:
randomcentres = randomsample( X, ncluster )
centres, xtoc, dist = kmeans( X, randomcentres,
delta=kmdelta, maxiter=kmiter, metric=metric, verbose=2 )
print "%.0f msec" % ((time() - t0) * 1000)
# also ~/py/np/kmeans/test-kmeans.py
Some notes added 26mar 2012:
1) for cosine distance, first normalize all the data vectors to |X| = 1; then
cosinedistance( X, Y ) = 1 - X . Y = Euclidean distance |X - Y|^2 / 2
is fast. For bit vectors, keep the norms separately from the vectors
instead of expanding out to floats
(although some programs may expand for you).
For sparse vectors, say 1 % of N, X . Y should take time O( 2 % N ),
space O(N); but I don't know which programs do that.
2)
Scikit-learn clustering
gives an excellent overview of k-means, mini-batch-k-means ...
with code that works on scipy.sparse matrices.
3) Always check cluster sizes after k-means.
If you're expecting roughly equal-sized clusters, but they come out
[44 37 9 5 5] %
... (sound of head-scratching).
Best Answer
In the documentation it says:
To understand what that means you need to have a look at the k-means algorithm. What k-means essentially does is find cluster centers that minimize the sum of distances between data samples and their associated cluster centers.
It is a two-step process, where (a) each data sample is associated to its closest cluster center, (b) cluster centers are adjusted to lie at the center of all samples associated to them. These steps are repeated until a criterion (max iterations / min change between last two iterations) is met.
As you can see there remains a distance between the data samples and their associated cluster centers, and the objective of our minimization is that distance (sum of all distances).
You naturally get large distances if you have a big variety in data samples, if the number of data samples is significantly higher than the number of clusters, which in your case is only two. On the contrary, if all data samples were the same, you would always get a zero distance regardless of number of clusters.
From the documentation I would expect that all values are negative, though. If you observe both negative and positive values, maybe there is more to the score than that.
I wonder how you got the idea of clustering into two clusters though.