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

[Anthony Liu]

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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[fumanchu]

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 [0.7279111122901 516, 0.0821287086068 67745, 0.0080516733577 621798,
0.1212206024590 2817, 0.0034460458833 209676, 0.0021046234724 371184,
0.0541094249143 63845, 0.0003575097024 9204185, 0.0005117507553 6832372,
0.0001585485582 0800087] 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
fu******@amor.o rg

[Alex Martelli]

fumanchu <fu******@amor. org> 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 [0.7279111122901 516, 0.0821287086068 67745, 0.0080516733577 621798,
0.1212206024590 2817, 0.0034460458833 209676, 0.0021046234724 371184,
0.0541094249143 63845, 0.0003575097024 9204185, 0.0005117507553 6832372,
0.0001585485582 0800087] 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

[Robert Kern]

Alex Martelli wrote:

fumanchu <fu******@amor. org> 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

[0.7279111122901 516, 0.0821287086068 67745, 0.0080516733577 621798,
0.12122060245 902817, 0.0034460458833 209676, 0.0021046234724 371184,
0.05410942491 4363845, 0.0003575097024 9204185, 0.0005117507553 6832372,
0.00015854855 820800087]

>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_dista nce(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: "compositio nal 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
ro*********@gma il.com

"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

[Edward Elliott]

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.

[Terry Reedy]

"fumanchu" <fu******@amor. org> wrote in message
news:11******** **************@ j33g2000cwa.goo glegroups.com.. .

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