[英]Flawed random number generator?
我使用了這個加權隨機數發生器。
import random
def weighted_choice(weights):
totals = []
running_total = 0
for w in weights:
running_total += w
totals.append(running_total)
rnd = random.random() * running_total
for i, total in enumerate(totals):
if rnd < total:
return i
如下:
# The meaning of this dict is a little confusing, so here's the explanation:
# The keys are numbers and values are weights of its occurence and values - 1
# are weights of its disoccurence. You can imagine it like biased coins
# (except for 2 which is fair coin).
probabilities = { 0 : 1.0, 1 : 1.0, 2 : 0.5, 3 : 0.45, 4 : 0.4, 5 : 0.35,
6 : 0.3, 7 : 0.25, 8 : 0.2, 9 : 0.15, 10 : 0.1
}
numberOfDeactivations = []
for number in probabilities.keys():
x = weighted_choice([probabilities[number], 1 - probabilities[number]])
if x == 0:
numberOfDeactivations.append(number)
print "chance for ", repr(numberOfDeactivations)
我經常看到7
, 8
, 9
, 10
中的結果。
是否有一些證據或保證這對概率論是正確的?
編輯:作為旁注:我認為你的代碼相當於
import random
probabilities = { 0 : 1.0, 1 : 1.0, 2 : 0.5, 3 : 0.45, 4 : 0.4, 5 : 0.35,
6 : 0.3, 7 : 0.25, 8 : 0.2, 9 : 0.15, 10 : 0.1}
numberOfDeactivations=filter(
lambda kv:random.random()<=probabilities[kv] , probabilities)
原始答案:
方法是正確的。 下面是一個完整的示例,創建頻率表並將其與請求的權重進行比較。
通過100000次迭代,沒有任何跡象表明您沒有得到您的要求。 'expected'是您請求的概率,'got'是您實際獲得該值的一小部分。 比率應接近1,它是:
0, expected: 0.2128 got: 0.2107 ratio: 1.0100
1, expected: 0.2128 got: 0.2145 ratio: 0.9921
2, expected: 0.1064 got: 0.1083 ratio: 0.9825
3, expected: 0.0957 got: 0.0949 ratio: 1.0091
4, expected: 0.0851 got: 0.0860 ratio: 0.9900
5, expected: 0.0745 got: 0.0753 ratio: 0.9884
6, expected: 0.0638 got: 0.0635 ratio: 1.0050
7, expected: 0.0532 got: 0.0518 ratio: 1.0262
8, expected: 0.0426 got: 0.0418 ratio: 1.0179
9, expected: 0.0319 got: 0.0323 ratio: 0.9881
10, expected: 0.0213 got: 0.0209 ratio: 1.0162
A total of 469633 numbers where generated for this table.
這是代碼:
import random
def weighted_choice(weights):
totals = []
running_total = 0
for w in weights:
running_total += w
totals.append(running_total)
rnd = random.random() * running_total
for i, total in enumerate(totals):
if rnd < total:
return i
counts={ k:0 for k in range(11)}
probabilities = { 0 : 1.0, 1 : 1.0, 2 : 0.5, 3 : 0.45, 4 : 0.4, 5 : 0.35,
6 : 0.3, 7 : 0.25, 8 : 0.2, 9 : 0.15, 10 : 0.1
}
for x in range(100000):
numberOfDeactivations = []
for number in probabilities.keys():
x = weighted_choice([probabilities[number], 1 - probabilities[number]])
if x == 0:
numberOfDeactivations.append(number)
for k in numberOfDeactivations:
counts[k]+=1.0
sums=sum(counts.values())
counts2=[x*1.0/sums for x in counts.values()]
print "ratio expected frequency to requested:":
# make the probabilities real probabilities instead of weights:
psum=sum(probabilities.values())
for k in probabilities:
probabilities[k]=probabilities[k]/psum
for k in probabilities:
print "%3d, expected: %6.4f got: %6.4f ratio: %6.4f" %(k,probabilities[k],counts2[k], probabilities[k]/counts2[k])
這在數學上是正確的。 這是逆變換采樣的應用(盡管它在這種情況下工作的原因應該是相對直觀的)。
我不知道Python,所以我不能說是否有任何細微之處使得這個特定的實現無效。
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