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

 P: n/a 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. __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com Apr 22 '06 #1
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 P: n/a 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. Apr 22 '06 #2

 P: n/a 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). Apr 22 '06 #3

 P: n/a 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. Apr 22 '06 #4

 P: n/a 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 Apr 22 '06 #5

 P: n/a 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 - x for x in intervals ] print deltas print sum(deltas) --------------------------------------------------------------- Gerard Apr 22 '06 #6

 P: n/a 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 - x for x in intervals ] print deltas print sum(deltas) --------------------------------------------------------------- def partition(N=5): vals = sorted( random.random() for _ in range(2*N) ) vals =  + vals +  for j in range(2*N+1): yield vals[j:j+2] deltas = [ x-x 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 Apr 22 '06 #7

 P: n/a 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 - x for x in intervals ] print deltas print sum(deltas) --------------------------------------------------------------- def partition(N=5): vals = sorted( random.random() for _ in range(2*N) ) vals =  + vals +  for j in range(2*N+1): yield vals[j:j+2] deltas = [ x-x for x in partition() ] print deltas print sum(deltas) finally: --------------------------------------------------------------- def distribution(N=2): p =  + sorted( random.random() for _ in range(N-1) ) +  for j in range(N): yield p[j+1] - p[j] spread = list(distribution(10)) print spread print sum(spread) --------------------------------------------------------------- Gerard Apr 22 '06 #8

 P: n/a Gerard Flanagan wrote: def distribution(N=2): p =  + sorted( random.random() for _ in range(N-1) ) +  for j in range(N): yield p[j+1] - p[j] spread = list(distribution(10)) print spread print sum(spread) 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 ] spread = distribution(10) print spread print sum(spread) -- Nick Craig-Wood -- http://www.craig-wood.com/nick Apr 23 '06 #9

 P: n/a Nick Craig-Wood wrote: Gerard Flanagan wrote: def distribution(N=2): p =  + sorted( random.random() for _ in range(N-1) ) +  for j in range(N): yield p[j+1] - p[j] spread = list(distribution(10)) print spread print sum(spread) 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 Apr 23 '06 #10

 P: n/a 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 fu******@amor.org Apr 23 '06 #11

 P: n/a 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 is uniformly distributed between 0 and 1; n 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 Apr 23 '06 #12

 P: n/a Alex Martelli wrote: fumanchu wrote:I'm surprised noone has pursued a course of subtraction rather thandivision. 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)1.0Either: 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 resultsis the OP's constraint: that they sum to 1. I'd be fascinated to learnif and why that wouldn't work. n is uniformly distributed between 0 and 1; n 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 ro*********@gmail.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 Apr 23 '06 #13

 P: n/a 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. Apr 23 '06 #14

 P: n/a "fumanchu" wrote in message news:11**********************@j33g2000cwa.googlegr oups.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 Apr 24 '06 #15

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