[Numpy-discussion] random number genration
thanks Sturl, your second code works well although I had to divide the vector elements through 128 to get just the binaries ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] random number genration
On Tue, Mar 29, 2011 at 11:59 AM, Sturla Molden wrote:
> Den 29.03.2011 16:49, skrev Sturla Molden:
> >
> > This will not work. A boolean array is not compactly stored, but an
> > array of bytes. Only the first bit 0 is 1 with probability p, bits 1 to
> > 7 bits are 1 with probability 0. We thus have to do this 8 times for
> > each byte, shift left by range(8), and combine them with binary or.
> > Also the main use of random bits is crypto, which requires the use of
> > /dev/urandom or CrypGenRandom instead of Mersenne Twister
> (np.random.rand).
>
>
> Here's a cleaned-up one for those who might be interested :-)
>
How about adding it to http://www.scipy.org/Cookbook?
Warren
>
> Sturla
>
>
>
> import numpy as np
> import os
>
>
> def randombits(n, p, intention='numerical'):
> """
> Returns an array with packed bits drawn from n Bernoulli
> trials with successrate p.
> """
> assert (intention in ('numerical','crypto'))
> # number of bytes
> m = (n >> 3) + 1 if n % 8 else n >> 3
> if intention == 'numerical':
> # Mersenne Twister
> rflt = np.random.rand(m*8)
> else:
> # /dev/urandom on Linux, Apple, et al.,
> # CryptGenRandom on Windows
> rflt = np.frombuffer(os.urandom(m*64),dtype=np.uint64)
> rflt = rflt / float(2**64)
> b = (rflt.reshape((m,8)) # pack the bits
> b = (b< # zero the trailing m*8 - n bits
> b[-1] &= (0xFF >> (m*8 - n))
> return b
>
>
> def bitcount(a, bytewise=False):
> """
> Count the number of set bits in an array of np.uint8.
> """
> assert(a.dtype == np.uint8)
> c = a[:,np.newaxis].repeat(8, axis=1) >> range(8) & 0x01
> return (c.sum(axis=1) if bytewise else c.sum())
>
>
> if __name__ == '__main__':
>
> b = randombits(int(1e6), .1, intent='numerical')
> print bitcount(b) # should be close to 10
> b = randombits(int(1e6), .1, intent='crypto')
> print bitcount(b) # should be close to 10
>
>
>
>
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Re: [Numpy-discussion] random number genration
Den 29.03.2011 16:49, skrev Sturla Molden:
>
> This will not work. A boolean array is not compactly stored, but an
> array of bytes. Only the first bit 0 is 1 with probability p, bits 1 to
> 7 bits are 1 with probability 0. We thus have to do this 8 times for
> each byte, shift left by range(8), and combine them with binary or.
> Also the main use of random bits is crypto, which requires the use of
> /dev/urandom or CrypGenRandom instead of Mersenne Twister (np.random.rand).
Here's a cleaned-up one for those who might be interested :-)
Sturla
import numpy as np
import os
def randombits(n, p, intention='numerical'):
"""
Returns an array with packed bits drawn from n Bernoulli
trials with successrate p.
"""
assert (intention in ('numerical','crypto'))
# number of bytes
m = (n >> 3) + 1 if n % 8 else n >> 3
if intention == 'numerical':
# Mersenne Twister
rflt = np.random.rand(m*8)
else:
# /dev/urandom on Linux, Apple, et al.,
# CryptGenRandom on Windows
rflt = np.frombuffer(os.urandom(m*64),dtype=np.uint64)
rflt = rflt / float(2**64)
b = (rflt.reshape((m,8))> (m*8 - n))
return b
def bitcount(a, bytewise=False):
"""
Count the number of set bits in an array of np.uint8.
"""
assert(a.dtype == np.uint8)
c = a[:,np.newaxis].repeat(8, axis=1) >> range(8) & 0x01
return (c.sum(axis=1) if bytewise else c.sum())
if __name__ == '__main__':
b = randombits(int(1e6), .1, intent='numerical')
print bitcount(b) # should be close to 10
b = randombits(int(1e6), .1, intent='crypto')
print bitcount(b) # should be close to 10
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Re: [Numpy-discussion] random number genration
On Tue, Mar 29, 2011 at 09:49, Sturla Molden wrote: > Den 29.03.2011 15:46, skrev Daniel Lepage: >> >> x = (np.random.random(size)< p) > > This will not work. A boolean array is not compactly stored, but an > array of bytes. Only the first bit 0 is 1 with probability p, bits 1 to > 7 bits are 1 with probability 0. We thus have to do this 8 times for > each byte, shift left by range(8), and combine them with binary or. It's not clear that the OP really meant bits rather than just bools. Judging by the example that he tried first, it's likely that he just wants bools (or even just 0s and 1s) and not a real string of bits compacted into bytes. -- Robert Kern "I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth." -- Umberto Eco ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] random number genration
Den 29.03.2011 16:49, skrev Sturla Molden: "Only the first bit 0 is 1 with probability p, bits 1 to 7 bits are 1 with probability 0." That should read: "Only bit 0 is 1 with probability p, bits 1 to 7 are 1 with probability 0." Sorry :) Sturla ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] random number genration
Den 29.03.2011 15:46, skrev Daniel Lepage: > > x = (np.random.random(size)< p) This will not work. A boolean array is not compactly stored, but an array of bytes. Only the first bit 0 is 1 with probability p, bits 1 to 7 bits are 1 with probability 0. We thus have to do this 8 times for each byte, shift left by range(8), and combine them with binary or. Also the main use of random bits is crypto, which requires the use of /dev/urandom or CrypGenRandom instead of Mersenne Twister (np.random.rand). Sturla ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] random number genration
On Tue, Mar 29, 2011 at 5:00 AM, Alex Ter-Sarkissov wrote: > If I want to generate a string of random bits with equal probability I run > > random.randint(0,2,size). > > What if I want a specific proportion of bits? In other words, each bit is 1 > with probability p<1/2 and 0 with probability q=1-p? x = (np.random.random(size) < p) Setting p = .5 should produce the same results as np.random.randint(0,2,size). Note that this gives you an array of bools, not ints; use x.astype(int) if integers are important (or x.astype(np.uint8) if memory is an issue). HTH, Dan Lepage ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] random number genration
Den 29.03.2011 14:56, skrev Sturla Molden: > import numpy as np > def randombits(n, p): > b = (np.random.rand(n*8).reshape((n,8))< p).astype(int) > return (b<< range(8)).sum(axis=1).astype(np.uint8) n is the number of bytes. If you prefer to count in bits: def randombits(n, p): assert(n%8 == 0) b = (np.random.rand(n).reshape((n>>3,8))< p).astype(int) return (b<< range(8)).sum(axis=1).astype(np.uint8) Sturla ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] random number genration
Den 29.03.2011 11:00, skrev Alex Ter-Sarkissov: > If I want to generate a string of random bits with equal probability I > run > > random.randint(0,2,size). > > What if I want a specific proportion of bits? In other words, each bit > is 1 with probability p<1/2 and 0 with probability q=1-p? Does this work you? import numpy as np def randombits(n, p): b = (np.random.rand(n*8).reshape((n,8)) < p).astype(int) return (b << range(8)).sum(axis=1).astype(np.uint8) Sturla ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] random number genration
On Tue, Mar 29, 2011 at 1:29 PM, eat wrote: > Hi, > > On Tue, Mar 29, 2011 at 12:00 PM, Alex Ter-Sarkissov > wrote: > >> If I want to generate a string of random bits with equal probability I run >> >> >> random.randint(0,2,size). >> >> What if I want a specific proportion of bits? In other words, each bit is >> 1 with probability p<1/2 and 0 with probability q=1-p? >> > Would it be sufficient to: > In []: bs= ones(1e6, dtype= int) > In []: bs[randint(0, 1e6, 1e5)]= 0 > In []: bs.sum()/ 1e6 > Out[]: 0.904706 > Or: In []: bs= ones(1e6, dtype= int) In []: bs[rand(1e6)< 1./ 10]= 0 In []: bs.sum()/ 1e6 Out[]: 0.89983 > > Regards, > eat > >> >> thanks >> >> ___ >> NumPy-Discussion mailing list >> NumPy-Discussion@scipy.org >> http://mail.scipy.org/mailman/listinfo/numpy-discussion >> >> > ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] random number genration
Hi, On Tue, Mar 29, 2011 at 12:00 PM, Alex Ter-Sarkissov wrote: > If I want to generate a string of random bits with equal probability I run > > random.randint(0,2,size). > > What if I want a specific proportion of bits? In other words, each bit is 1 > with probability p<1/2 and 0 with probability q=1-p? > Would it be sufficient to: In []: bs= ones(1e6, dtype= int) In []: bs[randint(0, 1e6, 1e5)]= 0 In []: bs.sum()/ 1e6 Out[]: 0.904706 Regards, eat > > thanks > > ___ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
[Numpy-discussion] random number genration
If I want to generate a string of random bits with equal probability I run random.randint(0,2,size). What if I want a specific proportion of bits? In other words, each bit is 1 with probability p<1/2 and 0 with probability q=1-p? thanks ___ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
