如何有效地创建一个多维numpy数组，其条目仅取决于一维索引？

Question

如何使用内置函数更有效地创建下列numpy数组？

import numpy as np 
MyArray = np.zeros([10,10,10,10,10]) 
for i in range(10): 
    for j in range(10): 
     for k in range(10): 
      for l in range(10): 
       for m in range(10): 
        MyArray[i,j,k,l,m] = l 

正如你所看到的，元素只应该依赖于一个维度指数。我尝试使用numpy.tile，但到目前为止还不能确定。

Answer 1

看来你np.broadcast_to后：

# build an array where l_vals[0,0,0,i,0] = i 
l_vals = np.arange(10).reshape(1, 1, 1, -1, 1) 

# duplicate that array, without copies, in the other axes 
l_grid = np.broadcast_to(l_vals, (10, 10, 10, 10, 10)) 

值得一提的是，broadcast_to返回一个只读数组，因为元素实际上共享内存位置。如果你想创建后写入到这一点，那么你可以调用np.copy，或使用tile代替：

l_grid = np.tile(l_vals, (10, 10, 10, 1, 10)) 

你也只是具有扁平您的循环：

MyArray = np.zeros([10,10,10,10,10]) 
for l in range(10): 
    MyArray[:,:,:,l,:] = l 

Answer 2

尝试np.fromfunction，你的情况这将是类似的东西： MyArray = np.fromfunction(lambda i, j, k, l, m : l, (10, 10, 10, 10, 10))

Answer 3

你可以这样做一个循环：

import numpy as np 
MyArray = np.zeros([10,10,10,10,10]) 

for l in range(10): 
    MyArray[:, :, :, l, :] = l 

很明显，你也可以做到这一点作为一个列表理解。

Answer 4

下面是与initialization的方法 -

n = 10 
a = np.empty((n,n,n,n,n),dtype=int) 
a[...] = np.arange(n)[:,None] 

这里还有一个NumPy strides基础的方法 -

r = np.arange(n) 
s = r.strides[0] 
shp = (n,n,n,n,n) 
out = np.lib.index_tricks.as_strided(r, shape=shp, strides=(0,0,0,s,0)) 

运行测试

途径 -

# @Eric's soln1 
def broadcast_to_based(n): # Creates a read-only array 
    l_vals = np.arange(n).reshape(1, 1, 1, -1, 1) 
    return np.broadcast_to(l_vals, (n, n, n, n, n)) 

# @Eric's soln2  
def tile_based(n): 
    l_vals = np.arange(n).reshape(1, 1, 1, -1, 1) 
    return np.tile(l_vals, (n, n, n, 1, n)) 

# @kmichael08's soln   
def fromfunc_based(n): 
    return np.fromfunction(lambda i, j, k, l, m : l, (n, n, n, n, n)) 

# @Tw UxTLi51Nus's soln 
def loop_based(n): 
    MyArray = np.zeros([n,n,n,n,n],dtype=int) 
    for l in range(n): 
     MyArray[:, :, :, l, :] = l 
    return MyArray 

# Proposed-1 in this post  
def initialization_based(n): 
    a = np.empty((n,n,n,n,n),dtype=int) 
    a[...] = np.arange(n)[:,None] 
    return a 

# Proposed-2 in this post  
def strided_based(n): 
    r = np.arange(n) 
    s = r.strides[0] 
    shp = (n,n,n,n,n) 
    return np.lib.index_tricks.as_strided(r, shape=shp, strides=(0,0,0,s,0)) 

个

计时 -

In [153]: n = 10 
    ...: %timeit broadcast_to_based(n) 
    ...: %timeit tile_based(n) 
    ...: %timeit fromfunc_based(n) 
    ...: %timeit loop_based(n) 
    ...: %timeit initialization_based(n) 
    ...: %timeit strided_based(n) 
    ...: 
100000 loops, best of 3: 4.1 µs per loop 
1000 loops, best of 3: 236 µs per loop 
1000 loops, best of 3: 645 µs per loop 
10000 loops, best of 3: 180 µs per loop 
10000 loops, best of 3: 89.1 µs per loop 
100000 loops, best of 3: 5.44 µs per loop 

In [154]: n = 20 
    ...: %timeit broadcast_to_based(n) 
    ...: %timeit tile_based(n) 
    ...: %timeit fromfunc_based(n) 
    ...: %timeit loop_based(n) 
    ...: %timeit initialization_based(n) 
    ...: %timeit strided_based(n) 
    ...: 
100000 loops, best of 3: 4.05 µs per loop 
100 loops, best of 3: 8.16 ms per loop 
10 loops, best of 3: 24.1 ms per loop 
100 loops, best of 3: 6.07 ms per loop 
100 loops, best of 3: 2.31 ms per loop 
100000 loops, best of 3: 5.48 µs per loop