Hypergeometric distribution

[Raven]

Hi to all, I need to calculate the hpergeometric distribution:
choose(r, x) * choose(b, n-x)
p(x; r,b,n) = -----------------------------
choose(r+b, n)

choose(r,x) is the binomial coefficient
I use the factorial to calculate the above formula but since I am using
large numbers, the result of choose(a,b) (ie: the binomial coefficient)
is too big even for large int. I've tried the scipy library, but this
library calculates
the hypergeometric using the factorials too, so the problem subsist. Is
there any other libray or an algorithm to calculate
the hypergeometric distribution? The statistical package R can handle
such calculations but I don't want to use python R binding since I want
a standalone app.
Thanks a lot
Ale

[Robert Kern]

Raven wrote:

Hi to all, I need to calculate the hpergeometric distribution:
choose(r, x) * choose(b, n-x)
p(x; r,b,n) = -----------------------------
choose(r+b, n)

choose(r,x) is the binomial coefficient
I use the factorial to calculate the above formula but since I am using
large numbers, the result of choose(a,b) (ie: the binomial coefficient)
is too big even for large int. I've tried the scipy library, but this
library calculates
the hypergeometric using the factorials too, so the problem subsist. Is
there any other libray or an algorithm to calculate
the hypergeometric distribution?

Use logarithms.

Specifically,

from scipy import special

def logchoose(n, k):
lgn1 = special.gammaln (n+1)
lgk1 = special.gammaln (k+1)
lgnk1 = special.gammaln (n-k+1)
return lgn1 - (lgnk1 + lgk1)

def gauss_hypergeom (x, r, b, n):
return exp(logchoose(r , x) +
logchoose(b, n-x) -
logchoose(r+b, n))

Or you could use gmpy if you need exact rational arithmetic rather than floating
point.

--
Robert Kern
ro*********@gma il.com

"In the fields of hell where the grass grows high
Are the graves of dreams allowed to die."
-- Richard Harter

[Gerard Flanagan]

Raven wrote:

Hi to all, I need to calculate the hpergeometric distribution:
choose(r, x) * choose(b, n-x)
p(x; r,b,n) = -----------------------------
choose(r+b, n)

choose(r,x) is the binomial coefficient
I use the factorial to calculate the above formula but since I am using
large numbers, the result of choose(a,b) (ie: the binomial coefficient)
is too big even for large int. I've tried the scipy library, but this
library calculates
the hypergeometric using the factorials too, so the problem subsist. Is
there any other libray or an algorithm to calculate
the hypergeometric distribution? The statistical package R can handle
such calculations but I don't want to use python R binding since I want
a standalone app.
Thanks a lot
Ale

Ale

I had this code lying about if it helps. I don't know if it's even
correct but it's non-recursive!

def Binomial( n, k ):
ret = 0
if k == 0:
ret = 1
elif k > 0:
a = range( n+1 )
a[0] = 1
for i in a[1:]:
a[i] = 1
for j in range(i-1,0,-1):
a[j] = a[j] + a[j-1]
ret = a[k]
return ret

Gerard

[Steven D'Aprano]

On Mon, 26 Dec 2005 12:18:55 -0800, Raven wrote:

Hi to all, I need to calculate the hpergeometric distribution:
choose(r, x) * choose(b, n-x)
p(x; r,b,n) = -----------------------------
choose(r+b, n)

choose(r,x) is the binomial coefficient
I use the factorial to calculate the above formula but since I am using
large numbers, the result of choose(a,b) (ie: the binomial coefficient)
is too big even for large int.
Are you sure about that? Python long ints can be as big as you have enough
memory for. My Python can print 10L**10000 to the console with a barely
detectable pause, and 10L**100000 with about a ten second delay. (Most of
that delay is printing it, not calculating it.)

25206 is the first integer whose factorial exceeds 10L**100000, so even if
you are calculating the binomial coefficient using the most naive
algorithm, calculating the factorials and dividing, you should easily be
able to generate it for a,b up to 20,000 unless you have a severe
shortage of memory.
I've tried the scipy library, but this
library calculates
the hypergeometric using the factorials too, so the problem subsist.

What exactly is your problem? What values of hypergeometric( x; r,b,n) fail
for you?

--
Steven.

[Alex Martelli]

Raven <ba********@gma il.com> wrote:

Hi to all, I need to calculate the hpergeometric distribution:
choose(r, x) * choose(b, n-x)
p(x; r,b,n) = -----------------------------
choose(r+b, n)

choose(r,x) is the binomial coefficient
I use the factorial to calculate the above formula but since I am using
large numbers, the result of choose(a,b) (ie: the binomial coefficient)
is too big even for large int. I've tried the scipy library, but this
library calculates
the hypergeometric using the factorials too, so the problem subsist. Is
there any other libray or an algorithm to calculate
the hypergeometric distribution? The statistical package R can handle
such calculations but I don't want to use python R binding since I want
a standalone app.

Have you tried with gmpy?
Alex

[Raven]

Thanks to all of you guys, I could resolve my problem using the
logarithms as proposed by Robert. I needed to calculate the factorial
for genomic data, more specifically for the number of genes in the
human genome i.e. about 30.000 and that is a big number :-)
I didn't know gmpy
Thanks a lot, really

Ale

[Steven D'Aprano]

On Sat, 31 Dec 2005 16:24:02 -0800, Raven wrote:

Thanks to all of you guys, I could resolve my problem using the
logarithms as proposed by Robert. I needed to calculate the factorial
for genomic data, more specifically for the number of genes in the
human genome i.e. about 30.000 and that is a big number :-)
I didn't know gmpy
Thanks a lot, really

Are you *sure* the existing functions didn't work? Did you try them?

def log2(x): .... return math.log(x)/math.log(2)
.... n = 0.0
for i in range(1, 300000): # ten times bigger than you need .... n += log2(i)
.... n 5025564.6087276 665 t = time.time(); x = 2L**(int(n) + 1); time.time() - t 0.2664909362792 9688

That's about one quarter of a second to calculate 300,000 factorial
(approximately) , and it shows that the calculations are well within
Python's capabilities.

Of course, converting this 1.5 million-plus digit number to a string takes
a bit longer:
t = time.time(); len(str(x)); time.time() - t 1512846
6939.3762848377 228

A quarter of a second to calculate, and almost two hours to convert to a
string. Lesson one: calculations on longints are fast. Converting them to
strings is not.

As far as your original question goes, try something like this:

(formula from memory, may be wrong)
def bincoeff(n,r): .... x = 1
.... for i in range(r+1, n+1):
.... x *= i
.... for i in range(1, n-r+1):
.... x /= i
.... return x
.... bincoeff(10, 0) 1 bincoeff(10, 1) 10 bincoeff(10, 2) 45 bincoeff(10, 3) 120 import time
t = time.time(); L = bincoeff(30000, 7000); time.time() - t 28.317800045013 428

Less than thirty seconds to calculate a rather large binomial coefficient
exactly. How many digits?
len(str(L)) 7076

If you are calculating hundreds of hypergeometric probabilities, 30
seconds each could be quite painful, but it certainly shows that Python is
capable of doing it without resorting to logarithms which may lose some
significant digits. Although, in fairness, the log function doesn't seem
to lose much accuracy for arguments in the range you are dealing with.
How long does it take to calculate factorials?
def timefact(n): .... # calculate n! and return the time taken in seconds
.... t = time.time()
.... L = 1
.... for i in range(1, n+1):
.... L *= i
.... return time.time() - t
.... timefact(3000) 0.0549139976501 46484 timefact(30000) # equivalent to human genome 5.0699510574340 82 timefact(300000 ) # ten times bigger

4255.2370519638 062

Keep in mind, if you are calculating the hypergeometric probabilities
using raw factorials, you are doing way too much work.
--
Steven.

[Raven]

Thanks Steven for your very interesting post.

This was a critical instance from my problem:

from scipy import comb
comb(14354,174)

inf

The scipy.stats.dis tributions.hype rgeom function uses the scipy.comb
function, so it returned nan since it tries to divide an infinite. I
did not tried to write a self-made function using standard python as I
supposed that the scipy functions reached python's limits but I was
wrong, what a fool :-)
If you are calculating hundreds of hypergeometric probabilities, 30
seconds each could be quite painful, but it certainly shows that Python is
capable of doing it without resorting to logarithms which may lose some
significant digits. Although, in fairness, the log function doesn't seem
to lose much accuracy for arguments in the range you are dealing with.

Yes I am calculating hundreds of hypergeometric probabilities so I need
fast calculations

Ale

[Paul Rubin]

"Raven" <ba********@gma il.com> writes:

Yes I am calculating hundreds of hypergeometric probabilities so I need
fast calculations

Can you use Stirling's approximation to get the logs of the factorials?

[Steven D'Aprano]

On Sun, 01 Jan 2006 14:24:39 -0800, Raven wrote:

Thanks Steven for your very interesting post.

This was a critical instance from my problem:

from scipy import comb
comb(14354,174) inf
Curious. It wouldn't surprise me if scipy was using floats, because 'inf'
is usually a floating point value, not an integer.

Using my test code from yesterday, I got:

bincoeff(14354, 174)

111727771935623 249173533679580 244374733360180 53487854593870
070906374894056 044891924883461 446844023623444 09632515556732
335635231613081 458252082763952 387644418578294 54464446478336
901737770950418 910676375517833 240712336253706 19908633625448
310766773824486 162461253466677 378968915481668 98009878730510
574761395158405 427699564142041 306927336297233 05869285300247
645972456505830 620188961902165 086857407612722 931651840L

Took about three seconds on my system.
Yes I am calculating hundreds of hypergeometric probabilities so I
need fast calculations

Another possibility, if you want exact integer maths rather than floating
point with logarithms, is to memoise the binomial coefficients. Something
like this:

# untested
def bincoeff(n,r, \
cache={}):
try:
return cache((n,r))
except KeyError:
x = 1
for i in range(r+1, n+1):
x *= i
for i in range(1, n-r+1):
x /= i
cache((n,r)) = x
return x
--
Steven.