import numpy as np
def gen_c():
c = np.ones(5, dtype=int)
j = 0
t = 10
while j < t:
c[0] = j
yield c.tolist()
j += 1
# What I did:
# res = np.array(list(gen_c())) <-- useless allocation of memory
# this line is what I'd like to do and it's killing me
res = np.fromiter(gen_c(), dtype=int) # dtype=list ?
错误说ValueError: setting an array element with a sequence.
这是一段非常愚蠢的代码。我想从生成器创建一个列表数组(最后是一个2D数组)......
虽然我到处搜索,但我仍然无法弄清楚如何使它工作。
答案 0 :(得分:3)
您只能使用numpy.fromiter()
创建一维数组(而不是二维数组),如documentation of numpy.fromiter
中所示 -
numpy.fromiter(iterable,dtype,count = -1)
从可迭代对象创建一个新的1维数组。
您可以做的一件事是将生成器函数转换为从c
中提供单个值,然后从中创建一个数组,然后将其重新整形为(-1,5)
。示例 -
import numpy as np
def gen_c():
c = np.ones(5, dtype=int)
j = 0
t = 10
while j < t:
c[0] = j
for i in c:
yield i
j += 1
np.fromiter(gen_c(),dtype=int).reshape((-1,5))
演示 -
In [5]: %paste
import numpy as np
def gen_c():
c = np.ones(5, dtype=int)
j = 0
t = 10
while j < t:
c[0] = j
for i in c:
yield i
j += 1
np.fromiter(gen_c(),dtype=int).reshape((-1,5))
## -- End pasted text --
Out[5]:
array([[0, 1, 1, 1, 1],
[1, 1, 1, 1, 1],
[2, 1, 1, 1, 1],
[3, 1, 1, 1, 1],
[4, 1, 1, 1, 1],
[5, 1, 1, 1, 1],
[6, 1, 1, 1, 1],
[7, 1, 1, 1, 1],
[8, 1, 1, 1, 1],
[9, 1, 1, 1, 1]])
答案 1 :(得分:0)
正如文档建议的那样,np.fromiter()
只接受一维迭代。
您可以先使用itertools.chain.from_iterable()
展开可迭代,然后再np.reshape()
将其展开:
import itertools
import numpy as np
def fromiter2d(it, dtype):
# clone the iterator to get its length
it, it2 = itertools.tee(it)
length = sum(1 for _ in it2)
flattened = itertools.chain.from_iterable(it)
array_1d = np.fromiter(flattened, dtype)
array_2d = np.reshape(array_1d, (length, -1))
return array_2d
演示:
>>> iter2d = (range(i, i + 4) for i in range(0, 12, 4))
>>> from_2d_iter(iter2d, int)
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
仅在Python 3.6上测试过,但也应该与Python 2一起使用。