我有一组值,x。给定“开始”和“停止”索引,我需要使用 x 的子数组构造一个数组 y。
import numpy as np
x = np.arange(20)
start = np.array([2, 8, 15])
stop = np.array([5, 10, 20])
nsubarray = len(start)
我希望 y 的位置:
y = array([ 2, 3, 4, 8, 9, 15, 16, 17, 18, 19])
(实际上我使用的数组要大得多)。
构造 y 的一种方法是使用列表理解,但之后需要将列表展平:
import itertools as it
y = [x[start[i]:stop[i]] for i in range(nsubarray)]
y = np.fromiter(it.chain.from_iterable(y), dtype=int)
我发现使用 for 循环实际上更快:
y = np.empty(sum(stop - start), dtype = int)
a = 0
for i in range(nsubarray):
b = a + stop[i] - start[i]
y[a:b] = x[start[i]:stop[i]]
a = b
我想知道是否有人知道我可以优化它的方法?非常感谢!
编辑
以下始终测试:
import numpy as np
import numpy.random as rd
import itertools as it
def get_chunks(arr, start, stop):
rng = stop - start
rng = rng[rng!=0] #Need to add this in case of zero sized ranges
np.cumsum(rng, out=rng)
inds = np.ones(rng[-1], dtype=np.int)
inds[rng[:-1]] = start[1:]-stop[:-1]+1
inds[0] = start[0]
np.cumsum(inds, out=inds)
return np.take(arr, inds)
def for_loop(arr, start, stop):
y = np.empty(sum(stop - start), dtype = int)
a = 0
for i in range(nsubarray):
b = a + stop[i] - start[i]
y[a:b] = arr[start[i]:stop[i]]
a = b
return y
xmax = 1E6
nsubarray = 100000
x = np.arange(xmax)
start = rd.randint(0, xmax - 10, nsubarray)
stop = start + 10
结果是:
In [379]: %timeit np.hstack([x[i:j] for i,j in it.izip(start, stop)])
1 loops, best of 3: 410 ms per loop
In [380]: %timeit for_loop(x, start, stop)
1 loops, best of 3: 281 ms per loop
In [381]: %timeit np.concatenate([x[i:j] for i,j in it.izip(start, stop)])
10 loops, best of 3: 97.8 ms per loop
In [382]: %timeit get_chunks(x, start, stop)
100 loops, best of 3: 16.6 ms per loop
最佳答案
这有点复杂,但相当快。基本上我们所做的是基于向量加法创建索引列表,并使用 np.take
而不是任何 python 循环:
def get_chunks(arr, start, stop):
rng = stop - start
rng = rng[rng!=0] #Need to add this in case of zero sized ranges
np.cumsum(rng, out=rng)
inds = np.ones(rng[-1], dtype=np.int)
inds[rng[:-1]] = start[1:]-stop[:-1]+1
inds[0] = start[0]
np.cumsum(inds, out=inds)
return np.take(arr, inds)
检查它是否返回正确的结果:
xmax = 1E6
nsubarray = 100000
x = np.arange(xmax)
start = np.random.randint(0, xmax - 10, nsubarray)
stop = start + np.random.randint(1, 10, nsubarray)
old = np.concatenate([x[b:e] for b, e in izip(start, stop)])
new = get_chunks(x, start, stop)
np.allclose(old,new)
True
一些时间:
%timeit np.hstack([x[i:j] for i,j in zip(start, stop)])
1 loops, best of 3: 354 ms per loop
%timeit np.concatenate([x[b:e] for b, e in izip(start, stop)])
10 loops, best of 3: 119 ms per loop
%timeit get_chunks(x, start, stop)
100 loops, best of 3: 7.59 ms per loop
关于python - 从不同大小的较小数组构造单个 numpy 数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22326882/