计算在大于0.99之前观察到的均匀分布随机数(0,1)的个数，不进行WHILE循环

Question

我需要编写一个代码来计算在大于0.99的值之前观察到的均匀分布随机数(0,1)的数量，而不使用WHILE循环。但我得到错误信息:'float'对象不能被解释为整数。有人知道问题在哪里吗?

---

import numpy as np
x = random.uniform(0,1)
count = 0
target = 0.99

for i in range(x):
x[i] = random.uniform(0,1)
if x[i] <= target:
count += 1
break
else: 
continue
print('%d uniformly random numbers observed before observing af value larger than 0.99' % count)

Answer 1

使用for-loop:

import itertools
import numpy as np
import random
count = 0
target = 0.99

for i in itertools.count(start=1):
x = random.uniform(0,1)
if x > target:
break
count += 1

print('%d uniformly random numbers observed before observing af value larger than 0.99' % count)

与while-loop使用这个:

import numpy as np
import random
count = 0
target = 0.99

while 1:
x = random.uniform(0,1)
if x <= target:
count += 1
else: 
break
print('%d uniformly random numbers observed before observing af value larger than 0.99' % count)

Answer 2

为什么不使用while循环呢?

import numpy as np
import random
count = 0
target = 0.99
x = random.uniform(0,1)
while x <= target:
count += 1
x = random.uniform(0,1)
print('%d uniformly random numbers observed before observing af value larger than 0.99' % count)