我正在用Python将用于创建Halton sequence的method转换为generator。我在Python2.7和Python3.3中遇到了同样的问题。你知道吗
当我使用生成器创建一个生成器对象并反复调用my_object.next()
时,我得到了预期的结果。然而,当我调用list(my_object)
或tuple(my_object)
时,我得到大小适当的列表/元组,但每个元素只是生成器输出的最终值的一个副本。我的Python 3.3代码如下:
def halton_gen(dim, num_pts):
sequence = np.empty(dim)
sequence.fill(np.nan)
primes = (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31) # first 11 prime #s
log_pts = log(num_pts + 1)
k = 1
# Generate the sequence
while k <= num_pts:
print("k = {}".format(k))
for i in range(dim):
print("i = {}".format(i))
prime = primes[i]
num_helper = int(ceil(log_pts / log(prime)))
kk = k
sum_ = 0
for j in range(num_helper):
sum_ += fmod(kk, prime) * pow(prime, -(j + 1))
kk = floor(kk / prime)
sequence[i] = sum_
print("sequence {}".format(sequence))
yield sequence
k += 1
注意,上面代码中的print语句给出了我所期望的结果,但是用list(my_object)
或tuple(my_object)
创建的列表/元组是不正确的:
my_object = halton_gen(2, 5)
tuple(my_object)
k = 1
i = 0
i = 1
sequence [ 0.5 0.33333333]
k = 2
i = 0
i = 1
sequence [ 0.25 0.66666667]
k = 3
i = 0
i = 1
sequence [ 0.75 0.11111111]
k = 4
i = 0
i = 1
sequence [ 0.125 0.44444444]
k = 5
i = 0
i = 1
sequence [ 0.625 0.77777778]
Out[86]:
(array([ 0.625 , 0.77777778]),
array([ 0.625 , 0.77777778]),
array([ 0.625 , 0.77777778]),
array([ 0.625 , 0.77777778]),
array([ 0.625 , 0.77777778]))
我期望(删除array()
说明符以提高可读性):
([0.5 0.333],
[0.25 0.667],
[0.75 0.111],
[0.125 0.444],
[0.625 0.778])
如何让生成器正确填充列表?或者我的清单,以正确地打开我的发电机?不管用什么恰当的方式来表达。你知道吗
正在回收您的
sequence
对象。一个解决办法是每次你屈服时重新创造它。你知道吗另一个修复方法是如注释所述,使用
sequence.copy()
。你知道吗相关问题 更多 >
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