2 回答

TA贡献1834条经验 获得超8个赞
试试下面的代码,
def is_power_of_two (x):
return (x and (not(x & (x - 1))) )
def get_random(start, stop):
while True:
value = random.randint(start, stop)
if not is_power_of_two(value):
return value
return [i for i in range(limit) if not is_power_of_two(i)]
result = [get_random(0, 10000000000) for _ in range(8)]

TA贡献1827条经验 获得超8个赞
在这种情况下,生成器会派上用场。编写一个带有无限循环的函数,该函数返回满足您条件的随机数,然后yield在调用该函数时使用语句一次返回一个这些值。
这是一个例子。我添加了一些代码来检查输入参数,以确保在给定范围内有有效的结果。
def random_not_power_of_2(rmin, rmax):
# Returns a random number r such that rmin <= r < rmax,
# and r is not a power of 2
from random import randint
# Sanity check
if rmin < 0:
raise ValueError("rmin must be non-negative")
if rmax <= rmin:
raise ValueError("rmax must be greater than rmin")
# Abort if the given range contains no valid numbers
r = rmin
isValid = False
while r < rmax:
if r == 0 or (r & (r-1) > 0):
isValid = True
break
r += 1
if not isValid:
raise ValueError("no valid numbers in this range")
while True:
r = randint(rmin, rmax)
if r == 0 or (r & (r-1) > 0):
yield r
def get_random_list(rmin, rmax, n):
# Returns a list of n random numbers in the given range
gen = random_not_power_of_2(rmin, rmax)
return [next(gen) for i in range(n)]
get_random_list(0,17,10)
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