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[英]How to generate a certain number of random whole numbers that add up to a certain number and are at least a certain number or more?
[英]How to generate random numbers with each random number having a difference of at least x with all other elements?
我知道这违反了随机数的定义,但我仍然需要这个项目。 例如,我想生成一个在range(0, 200)
有5个随机元素的数组。
现在,我希望每个元素之间的差异至少为15。 所以随机数组看起来应该是这样的:
[15, 45, 99, 132, 199]
我可以使用numpy生成随机数:
np.random.uniform(low=0, high=200, size=5)
但是,我无法保持至少15的一致差异。
如果问题显示出更多努力来解决问题(例如来自Stack Overflow Tour :“不要问...问题,你没有试图找到答案(显示你的工作!)”,这将是很好的。) ,但有时候一个问题会引发痒,你只需要刮开......
这是你可以做到的一种方式,写成函数random_spaced
:
import numpy as np
def random_spaced(low, high, delta, n, size=None):
"""
Choose n random values between low and high, with minimum spacing delta.
If size is None, one sample is returned.
Set size=m (an integer) to return m samples.
The values in each sample returned by random_spaced are in increasing
order.
"""
empty_space = high - low - (n-1)*delta
if empty_space < 0:
raise ValueError("not possible")
if size is None:
u = np.random.rand(n)
else:
u = np.random.rand(size, n)
x = empty_space * np.sort(u, axis=-1)
return low + x + delta * np.arange(n)
例如,
In [27]: random_spaced(0, 200, 15, 5)
Out[27]: array([ 30.3524969 , 97.4773284 , 140.38221631, 161.9276264 , 189.3404236 ])
In [28]: random_spaced(0, 200, 15, 5)
Out[28]: array([ 81.01616136, 103.11710522, 118.98018499, 141.68196775, 169.02965952])
size
参数允许您一次生成多个样本:
In [29]: random_spaced(0, 200, 15, 5, size=3)
Out[29]:
array([[ 52.62401348, 80.04494534, 96.21983265, 138.68552066, 178.14784825],
[ 7.57714106, 33.05818556, 62.59831316, 81.86507168, 180.30946733],
[ 24.16367913, 40.37480075, 86.71321297, 148.24263974, 195.89405713]])
此代码使用100000个样本为每个组件生成直方图,并绘制每个组件的相应理论边缘PDF:
import matplotlib.pyplot as plt
from scipy.stats import beta
low = 0
high = 200
delta = 15
n = 5
s = random_spaced(low, high, delta, n, size=100000)
for k in range(s.shape[1]):
plt.hist(s[:, k], bins=100, density=True, alpha=0.25)
plt.title("Normalized marginal histograms and marginal PDFs")
plt.grid(alpha=0.2)
# Plot the PDFs of the marginal distributions of each component.
# These are beta distributions.
for k in range(n):
left = low + k*delta
right = high - (n - k - 1)*delta
xx = np.linspace(left, right, 400)
yy = beta.pdf(xx, k + 1, n - k, loc=left, scale=right - left)
plt.plot(xx, yy, 'k--', linewidth=1, alpha=0.25)
if n > 1:
# Mark the mode with a dot.
mode0 = k/(n-1)
mode = (right-left)*mode0 + left
plt.plot(mode, beta.pdf(mode, k + 1, n - k, loc=left, scale=right - left),
'k.', alpha=0.25)
plt.show()
这是它生成的情节:
从图中可以看出,边际分布是β分布 。 边际分布的模式对应于区间[low, high]
上的n
均匀间隔点的位置。
通过摆弄在random_spaced
生成u
方式,可以生成具有不同边缘的分布(这个答案的旧版本有一个例子),但random_spaced
当前生成的分布似乎是一个自然的选择。 如上所述,边缘的模式出现在“有意义”的位置。 此外,在n
为1的平凡情况下,分布简化为[ low
, high
]上的均匀分布。
反复试验怎么样? 例如,抛出一些随机数,排序,计算差异......如果重复得太小?
import random as r
def spreadRandom(theRange, howMany, minSpacing):
while True:
candidate = sorted([r.randint(*theRange) for _ in range(howMany)])
minDiff = min([ candidate[i+1]-candidate[i] for i, _ in enumerate(candidate[:-1])])
if minDiff >= minSpacing:
return candidate
spreadRandom([0,200], 5, 15)
您无法保证得到答案,但您根本不会偏离您的数字,因为您可能会强制执行基于相邻数字的范围。
尝试改组数字0-200:
import random
numbers = list(range(200))
random.shuffle(numbers)
distant_numbers = [numbers[0]]
for number in numbers:
if any(abs(number - x) < 15 for x in distant_numbers):
continue
distant_numbers.append(number)
if len(distant_numbers) >= 5: break
编辑:
这是一个使用z3
进行大规模过度z3
的解决方案:
def spaced_randoms(n, d, R, first=None):
solver = z3.SolverFor("QF_FD")
numbers = [z3.Int("x{}".format(x)) for x in range(n)]
for number in numbers:
solver.add(number >= 0)
solver.add(number <= R)
for ii in range(n):
for jj in range(ii+1,n):
solver.add(z3.Or(numbers[ii] - numbers[jj] > d, numbers[ii] - numbers[jj] < -d))
if first is not None:
solver.add(numbers[0] == first)
result = solver.check()
if str(result) != "sat":
raise Exception("Unsatisfiable")
model = solver.model()
return [model.get_interp(number) for number in numbers]
像这样称它为随机结果:
import random
spaced_randoms(n, d, R, random.randint(0,R))
我认为此代码可能有助于满足您的特定需求:
import random
import numpy as np
five_list = np.asarray([])
end = False
number = random.randint(0,200)
five_list = np.append(five_list,number)
while True:
new_number = random.randint(0,200)
if all(np.absolute(np.subtract(five_list, new_number)) >= 15):
five_list = np.append(five_list,new_number)
if np.size(five_list) == 5:
break
print(np.sort(five_list))
尝试“蛮力”:
l= [ i for i in range(201) ]
rslt= []
for i in range(5):
n=random.choice(l)
rslt.append(n)
l=[ k for k in l if abs(k-n)>=15 ]
#if not l:
# break
或巧妙地:
sgmnts= [(0,200)]
diff= 15
rslt= []
for i in range(5):
start,stop= sgmnts.pop( random.choice(range(len(sgmnts))) )
n= random.choice(range(start,stop+1))
rslt.append(n)
if n-diff > start:
sgmnts.append( (start,n-diff) )
if n+diff < stop:
sgmnts.append( (n+diff,stop) )
if not sgmnts:
break
“sgmnts”存储合适的范围。 我们也通过索引随机选择一个范围。
这将在200之间生成5个随机值,步长为5
import random
array = []
randomRange = 200
arrayRange = 5
valueStep = 15
for loop in range(arrayRange):
randomMaxValue = randomRange - valueStep * (arrayRange - loop) # First loop will the first randomMaxValue be 125 next will be 140, 155, 170, 185, 200
if not array: # Checks if the array is empty
array.append(random.randint(0, randomMaxValue)) # Appends a value between 0 and 125 (First will be 125 because 200 - 15 * 5)
else:
array.append(random.randint(array[-1] + 15, randomMaxValue)) # Appends the 4 next values
print(array)
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