Generate a sequence of random numbers that sum up to 1?

Discussion in 'Python' started by Anthony Liu, Apr 22, 2006.

1. Anthony LiuGuest

I am at my wit's end.

I want to generate a certain number of random numbers.
This is easy, I can repeatedly do uniform(0, 1) for
example.

But, I want the random numbers just generated sum up
to 1 .

I am not sure how to do this. Any idea? Thanks.

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Anthony Liu, Apr 22, 2006

2. Mel WilsonGuest

Anthony Liu wrote:
> I am at my wit's end.
>
> I want to generate a certain number of random numbers.
> This is easy, I can repeatedly do uniform(0, 1) for
> example.
>
> But, I want the random numbers just generated sum up
> to 1 .
>
> I am not sure how to do this. Any idea? Thanks.

numbers.append (random.uniform (0, 1.0-sum(numbers)))

might help, perhaps.

or

scaled = [x/sum(numbers) for x in numbers]

Mel.
>

Mel Wilson, Apr 22, 2006

3. Edward ElliottGuest

Anthony Liu wrote:
> But, I want the random numbers just generated sum up
> to 1 .

This seems like an odd request. Might I ask what it's for?

Generating random numbers in [0,1) that are both uniform and sum to 1 looks
like an unsatisfiable task. Each number you generate restricts the
possibilities for future numbers. E.g. if the first number is 0.5, all
future numbers must be < 0.5 (indeed, must *sum* to 0.5). You'll end up
with a distribution increasingly skewed towards smaller numbers the more
you generate. I can't imagine what that would be useful for.

If that's not a problem, do this: generate the numbers, add them up, and
divide each by the sum.

nums = [random.uniform(0,1) for x in range(0,100)]
sum = reduce(lambda x,y: x+y, nums)
norm = [x/sum for x in nums]

Of course now the numbers aren't uniform over [0,1) anymore.

Also note that the sum of the normalized numbers will be very close to 1,
but slightly off due to representation issues. If that level of accuracy
matters, you might consider generating your rands as integers and then
fp-dividing by the sum (or just store them as integers/fractions).

Edward Elliott, Apr 22, 2006
4. Felipe Almeida LessaGuest

Em SÃ¡b, 2006-04-22 Ã s 03:16 +0000, Edward Elliott escreveu:
> If that level of accuracy
> matters, you might consider generating your rands as integers and then
> fp-dividing by the sum (or just store them as integers/fractions).

Or using decimal module: http://docs.python.org/lib/module-decimal.html

--
Felipe.

Felipe Almeida Lessa, Apr 22, 2006
5. Alex MartelliGuest

Anthony Liu <> wrote:
...
> As a matter of fact, given that we have to specify the
> number of states for an HMM, I would like to create a
> specified number of random floating numbers whose sum
> is 1.0.

def forAL(N):
N_randoms = [random.random() for x in xrange(N)]
total = sum(N_randoms)
return [x/total for x in N_randoms]

Does this do what you want? Of course, the resulting numbers are not
independent, but then the constraints you pose would contradict that.

Alex

Alex Martelli, Apr 22, 2006
6. Gerard FlanaganGuest

Anthony Liu wrote:
> I am at my wit's end.
>
> I want to generate a certain number of random numbers.
> This is easy, I can repeatedly do uniform(0, 1) for
> example.
>
> But, I want the random numbers just generated sum up
> to 1 .
>
> I am not sure how to do this. Any idea? Thanks.
>

--------------------------------------------------------------
import random

def partition(start=0,stop=1,eps=5):
d = stop - start
vals = [ start + d * random.random() for _ in range(2*eps) ]
vals = [start] + vals + [stop]
vals.sort()
return vals

P = partition()

intervals = [ P[i:i+2] for i in range(len(P)-1) ]

deltas = [ x[1] - x[0] for x in intervals ]

print deltas

print sum(deltas)
---------------------------------------------------------------

Gerard

Gerard Flanagan, Apr 22, 2006
7. Gerard FlanaganGuest

Gerard Flanagan wrote:
> Anthony Liu wrote:
> > I am at my wit's end.
> >
> > I want to generate a certain number of random numbers.
> > This is easy, I can repeatedly do uniform(0, 1) for
> > example.
> >
> > But, I want the random numbers just generated sum up
> > to 1 .
> >
> > I am not sure how to do this. Any idea? Thanks.
> >

>
> --------------------------------------------------------------
> import random
>
> def partition(start=0,stop=1,eps=5):
> d = stop - start
> vals = [ start + d * random.random() for _ in range(2*eps) ]
> vals = [start] + vals + [stop]
> vals.sort()
> return vals
>
> P = partition()
>
> intervals = [ P[i:i+2] for i in range(len(P)-1) ]
>
> deltas = [ x[1] - x[0] for x in intervals ]
>
> print deltas
>
> print sum(deltas)
> ---------------------------------------------------------------
>

def partition(N=5):
vals = sorted( random.random() for _ in range(2*N) )
vals = [0] + vals + [1]
for j in range(2*N+1):
yield vals[j:j+2]

deltas = [ x[1]-x[0] for x in partition() ]

print deltas

print sum(deltas)

>>>

[0.10271966686994982, 0.13826576491042208, 0.064146913555132801,
0.11906452454467387, 0.10501198456091299, 0.011732423830768779,
0.11785369256442912, 0.065927165520102249, 0.098351305878176198,
0.077786747076205365, 0.099139810689226726]
1.0

Gerard Flanagan, Apr 22, 2006
8. Gerard FlanaganGuest

Gerard Flanagan wrote:
> Gerard Flanagan wrote:
> > Anthony Liu wrote:
> > > I am at my wit's end.
> > >
> > > I want to generate a certain number of random numbers.
> > > This is easy, I can repeatedly do uniform(0, 1) for
> > > example.
> > >
> > > But, I want the random numbers just generated sum up
> > > to 1 .
> > >
> > > I am not sure how to do this. Any idea? Thanks.
> > >

> >
> > --------------------------------------------------------------
> > import random
> >
> > def partition(start=0,stop=1,eps=5):
> > d = stop - start
> > vals = [ start + d * random.random() for _ in range(2*eps) ]
> > vals = [start] + vals + [stop]
> > vals.sort()
> > return vals
> >
> > P = partition()
> >
> > intervals = [ P[i:i+2] for i in range(len(P)-1) ]
> >
> > deltas = [ x[1] - x[0] for x in intervals ]
> >
> > print deltas
> >
> > print sum(deltas)
> > ---------------------------------------------------------------
> >

>
> def partition(N=5):
> vals = sorted( random.random() for _ in range(2*N) )
> vals = [0] + vals + [1]
> for j in range(2*N+1):
> yield vals[j:j+2]
>
> deltas = [ x[1]-x[0] for x in partition() ]
>
> print deltas
>
> print sum(deltas)
>

finally:

---------------------------------------------------------------
def distribution(N=2):
p = [0] + sorted( random.random() for _ in range(N-1) ) + [1]
for j in range(N):
yield p[j+1] - p[j]

---------------------------------------------------------------
Gerard

Gerard Flanagan, Apr 22, 2006
9. Nick Craig-WoodGuest

Gerard Flanagan <> wrote:
> def distribution(N=2):
> p = [0] + sorted( random.random() for _ in range(N-1) ) + [1]
> for j in range(N):
> yield p[j+1] - p[j]
>
>

This is simpler, easier to prove correct and most likely quicker.

def distribution(N=2):
L = [ random.uniform(0,1) for _ in xrange(N) ]
sumL = sum(L)
return [ l/sumL for l in L ]

--
Nick Craig-Wood <> -- http://www.craig-wood.com/nick

Nick Craig-Wood, Apr 23, 2006
10. Gerard FlanaganGuest

Nick Craig-Wood wrote:
> Gerard Flanagan <> wrote:
> > def distribution(N=2):
> > p = [0] + sorted( random.random() for _ in range(N-1) ) + [1]
> > for j in range(N):
> > yield p[j+1] - p[j]
> >
> >

>
> This is simpler, easier to prove correct and most likely quicker.
>
> def distribution(N=2):
> L = [ random.uniform(0,1) for _ in xrange(N) ]
> sumL = sum(L)
> return [ l/sumL for l in L ]
>

simpler:- ok

easier to prove correct:- in what sense?

quicker:- slightly slower in fact (using xrange in both functions).
This must be due to 'uniform' - using random() rather than uniform(0,1)
then yes, it's quicker. Roughly tested, I get yours (and Alex
Martelli's) to be about twice as fast. (2<=N<1000, probably greater
difference as N increases).

All the best.

Gerard

Gerard Flanagan, Apr 23, 2006
11. fumanchuGuest

I'm surprised noone has pursued a course of subtraction rather than
division. Say you want 10 numbers:

>>> s = 1.0
>>> n = []
>>> for x in xrange(9):

.... value = random.random() * s
.... n.append(value)
.... s -= value
....
>>> n.append(s)
>>> n

[0.7279111122901516, 0.082128708606867745, 0.0080516733577621798,
0.12122060245902817, 0.0034460458833209676, 0.0021046234724371184,
0.054109424914363845, 0.00035750970249204185, 0.00051175075536832372,
0.00015854855820800087]
>>> sum(n)

1.0

Either:
1) Just because they're *ordered* doesn't mean they're not *random*,
or
2) You all now know why I'm not a mathematician.

It seems to me that the only constraint on the randomness of my results
is the OP's constraint: that they sum to 1. I'd be fascinated to learn
if and why that wouldn't work.

Robert Brewer
System Architect
Amor Ministries

fumanchu, Apr 23, 2006
12. Alex MartelliGuest

fumanchu <> wrote:

> I'm surprised noone has pursued a course of subtraction rather than
> division. Say you want 10 numbers:
>
> >>> s = 1.0
> >>> n = []
> >>> for x in xrange(9):

> ... value = random.random() * s
> ... n.append(value)
> ... s -= value
> ...
> >>> n.append(s)
> >>> n

> [0.7279111122901516, 0.082128708606867745, 0.0080516733577621798,
> 0.12122060245902817, 0.0034460458833209676, 0.0021046234724371184,
> 0.054109424914363845, 0.00035750970249204185, 0.00051175075536832372,
> 0.00015854855820800087]
> >>> sum(n)

> 1.0
>
>
> Either:
> 1) Just because they're *ordered* doesn't mean they're not *random*,
> or
> 2) You all now know why I'm not a mathematician.
>
> It seems to me that the only constraint on the randomness of my results
> is the OP's constraint: that they sum to 1. I'd be fascinated to learn
> if and why that wouldn't work.

n[0] is uniformly distributed between 0 and 1; n[1] is not -- not sure
how to characterize its distribution, but it's vastly skewed to favor
smaller values -- and further n[x] values for x>1 are progressively more
and more skewed similarly.

Such total disuniformity, where the very distribution of each value is
skewed by the preceding one, may still be "random" for some sufficiently
vague meaning of "random", but my intuition suggests it's unlikely to
prove satisfactory for the OP's purposes.

Alex

Alex Martelli, Apr 23, 2006
13. Robert KernGuest

Alex Martelli wrote:
> fumanchu <> wrote:
>
>>I'm surprised noone has pursued a course of subtraction rather than
>>division. Say you want 10 numbers:
>>
>>>>>s = 1.0
>>>>>n = []
>>>>>for x in xrange(9):

>>
>>... value = random.random() * s
>>... n.append(value)
>>... s -= value
>>...
>>
>>>>>n.append(s)
>>>>>n

>>
>>[0.7279111122901516, 0.082128708606867745, 0.0080516733577621798,
>>0.12122060245902817, 0.0034460458833209676, 0.0021046234724371184,
>>0.054109424914363845, 0.00035750970249204185, 0.00051175075536832372,
>>0.00015854855820800087]
>>
>>>>>sum(n)

>>
>>1.0
>>
>>Either:
>> 1) Just because they're *ordered* doesn't mean they're not *random*,
>>or
>> 2) You all now know why I'm not a mathematician.
>>
>>It seems to me that the only constraint on the randomness of my results
>>is the OP's constraint: that they sum to 1. I'd be fascinated to learn
>>if and why that wouldn't work.

>
> n[0] is uniformly distributed between 0 and 1; n[1] is not -- not sure
> how to characterize its distribution, but it's vastly skewed to favor
> smaller values -- and further n[x] values for x>1 are progressively more
> and more skewed similarly.
>
> Such total disuniformity, where the very distribution of each value is
> skewed by the preceding one, may still be "random" for some sufficiently
> vague meaning of "random", but my intuition suggests it's unlikely to
> prove satisfactory for the OP's purposes.

[digression]

All of this discussion about whether the distribution of values is uniform or
not doesn't mean much until one has defined "uniformity," or equivalently,
"distance" in the space we're talking about. In this case, we're talking about
the unit n-simplex space (SS^n) which has elements S=(s_1, s_2, ... s_n) where
sum(S) = 1 and s_i >= 0. I favor the Aitchison distance:

import numpy as np

def aitchison_distance(x, y):
""" Compute the Aitchison distance between two vectors in simplex space.
"""
lx = np.log(x)
ly = np.log(y)
lgx = np.mean(lx, axis=-1)
lgy = np.mean(ly, axis=-1)
diff = (lx - lgx) - (ly - lgy)
return np.sqrt(np.sum(diff*diff))

Note that zeros yield inifinities, so the borders of the unit simplex are
infinitely farther away from other points in the interior. Consequently,
generating "uniform" random samples from this space is as impractical as it is
to draw "uniform" random samples from the entire infinite real number line. It's
also not very interesting. However, one can transform SS^n into RR^(n-1) and
back again, so drawing numbers from a multivariate normal of whatever mean and
covariance you like will give you "nice" simplicial data and quite possibly even
realistic data, too, depending on your model. I like using the isometric
log-ratio transform ("ilr" transform) for this.

Good Google search terms: "compositional data", Aitchison

But of course, for the OP's purpose of creating synthetic Markov chain
transition tables, generating some random vectors uniformly on [0, 1)^n and
normalizing them to sum to 1 works a treat. Don't bother with anything else.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
an underlying truth."
-- Umberto Eco

Robert Kern, Apr 23, 2006
14. Edward ElliottGuest

Alex Martelli wrote:
> Such total disuniformity, where the very distribution of each value is
> skewed by the preceding one, may still be "random" for some sufficiently
> vague meaning of "random", but my intuition suggests it's unlikely to
> prove satisfactory for the OP's purposes.

It does seem very odd. If you could restrict the range, you could get an
unskewed distribution. Set range = (0, 2*sum/cnt) and you can generate cnt
numbers whose sum will tend towards sum (because the average value will be
sum/cnt):

target_sum = 1
cnt = 100
max = 2.0 * target_sum / cnt # 0.02
nums = [random.uniform(0,max) for x in range(0,cnt)]
real_sum = sum(nums) # 0.975... in one sample run

If the sum has to be exact, you can set the last value to reach it:

nums[-1] = target_sum - sum(nums[:-1])
print sum(nums) # 1.0

which skews the sample ever so slightly. And check for negatives in case
the sum exceeded the target.

If the exact count doesn't matter, just generate random nums until you're
within some delta of the target sum.

Basically, there usually better options to the problem as originally posed.
Actually, now that I reread it the OP never said the range had be [0,1).
So maybe we read too much into the original phrasing. If you need
anything resembling a uniform distribution, scaling the results afterward
is not the way to go.

Edward Elliott, Apr 23, 2006
15. Terry ReedyGuest

"fumanchu" <> wrote in message
news:...
> I'm surprised noone has pursued a course of subtraction rather than
> division.

I believe someone did mention the subtraction method in one of the initial
responses. But the problem is this. If you independently sample n numbers
from a given distribution and then rescale, then the scaled numbers still
all have the same distribution (and are uniform in that sense). In the
subtraction method, each comes from a differnt distribution, as others
explained, with the nth being radically different from the first.

Terry Jan Reedy

Terry Reedy, Apr 24, 2006