创建Numpy数组而不枚举数组 [英] Create Numpy array without enumerating array

问题描述

从此开始：

x = range(30,60,2)[::-1];
x = np.asarray(x); x

array([58, 56, 54, 52, 50, 48, 46, 44, 42, 40, 38, 36, 34, 32, 30])

创建像这样的数组：（注意，第一项重复）但是如果我可以在不重复第一项的情况下更快地得到它，我可以 np.hstack 第一项。

Create an array like this: (Notice, first item repeats) But if I can get this faster without the first item repeating, I can np.hstack first item.

[[58 58 56 54 52]
 [56 56 54 52 50]
 [54 54 52 50 48]
 [52 52 50 48 46]
 [50 50 48 46 44]
 [48 48 46 44 42]
 [46 46 44 42 40]
 [44 44 42 40 38]
 [42 42 40 38 36]
 [40 40 38 36 34]
 [38 38 36 34 32]
 [36 36 34 32 30]
 [34 34 32 30 None]
 [32 32 30 None None]
 [30 30 None None None]]

下面的代码有效，希望没有'for'循环并枚举更快。

The code below works, want it faster without 'for' loop and enumerate.

arr = np.empty((0,5), int)

for i,e in enumerate(x):
    arr2 = np.hstack((x[i], x[i:i+4], np.asarray([None]*5)))[:5]
    arr  = np.vstack((arr,arr2))

推荐答案

方法＃1

这里是使用 NumPy广播 -

Here's a vectorized approach using NumPy broadcasting -

N = 4 # width factor
x_ext = np.concatenate((x,[None]*(N-1)))
arr2D = x_ext[np.arange(N) + np.arange(x_ext.size-N+1)[:,None]]
out = np.column_stack((x,arr2D))

---

方法＃2

这里是另一个使用 hankel -

Here's another one using hankel -

from scipy.linalg import hankel

N = 4 # width factor
x_ext = np.concatenate((x,[None]*(N-1)))
out = np.column_stack((x,hankel(x_ext[:4], x_ext[3:]).T))

---

运行时测试

这是 @的修改版Aaron的基准测试脚本 使用的输入格式与此帖子中用于其帖子的输入格式相同，用于公平地进行基准测试，并只关注这些内容两种方法-

Here's a modified version of @Aaron's benchmarking script using an input format for this post identical to the one used for his post in that script for a fair benchmarking and focusing just on these two approaches -

upper_limit = 58 # We will edit this to vary the dataset sizes

print "Timings are : "
t = time()
for _ in range(1000):  #1000 iterations of @Aaron's soln.
    width = 3
    x = np.array(range(upper_limit,28,-2) + [float('nan')]*width)
    arr = np.empty([len(x)-width, width+2])
    arr[:,0] = x[:len(x)-width]
    for i in xrange(len(x)-width): 
        arr[i,1:] = x[i:i+width+1]
print(time()-t)

t = time()
for _ in range(1000): 
    N = 4 # width factor
    x_ext = np.array(range(upper_limit,28,-2) + [float('nan')]*(N-1))
    arr2D = x_ext[np.arange(N) + np.arange(x_ext.size-N+1)[:,None]]
    out = np.column_stack((x_ext[:len(x_ext)-N+1],arr2D))
print(time()-t)

案例1（upper_limit = 58）：

Case #1 (upper_limit = 58 ) :

Timings are : 
0.0316879749298
0.0322730541229

情况＃2（upper_limit = 1058）：

Case #2 (upper_limit = 1058 ) :

Timings are : 
0.680443048477
0.124517917633

Case＃3（upper_limit = 5058）：

Case #3 (upper_limit = 5058 ) :

Timings are : 
3.28129291534
0.47504901886

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