# @MSeifert's solution
def assign_vals_ix(s):
d = np.zeros((m, n, m, n), dtype=s.dtype)
l1 = range(m)
l2 = range(n)
d[np.ix_(l1,l2)*2] = s[np.ix_(l1,l2)]
return d
# Proposed in this post
def assign_vals(s):
m,n = s.shape
d = np.zeros((m*n,m*n),dtype=s.dtype)
d.ravel()[::m*n+1] = s.ravel()
return d.reshape(m,n,m,n)
# Using a strides based approach
def assign_vals_strides(a):
m,n = a.shape
p,q = a.strides
d = np.zeros((m,n,m,n),dtype=a.dtype)
out_strides = (q*(n*m*n+n),(m*n+1)*q)
d_view = np.lib.stride_tricks.as_strided(d, (m,n), out_strides)
d_view[:] = a
return d
计时-
In [285]: m,n = 10,10
...: s = np.random.rand(m,n)
...: d = np.zeros((m,n,m,n))
...:
In [286]: %timeit assign_vals_ix(s)
10000 loops, best of 3: 21.3 µs per loop
In [287]: %timeit assign_vals_strides(s)
100000 loops, best of 3: 9.37 µs per loop
In [288]: %timeit assign_vals(s)
100000 loops, best of 3: 4.13 µs per loop
In [289]: m,n = 20,20
...: s = np.random.rand(m,n)
...: d = np.zeros((m,n,m,n))
In [290]: %timeit assign_vals_ix(s)
10000 loops, best of 3: 60.2 µs per loop
In [291]: %timeit assign_vals_strides(s)
10000 loops, best of 3: 41.8 µs per loop
In [292]: %timeit assign_vals(s)
10000 loops, best of 3: 35.5 µs per loop
仔细想想,这与创建形状
(m*n, m*n)
的2D
数组并将值从s
分配到对角线位置是一样的。要使最终输出为4D
,我们只需要在最后进行重塑。基本上是在下面实施的-运行时测试
接近-
计时-
可以使用integer array indexing(使用^{} 创建广播索引):
第一次必须复制索引(您希望
[i, j, i, j]
而不是仅仅[i, j]
),这就是为什么我将np.ix_
返回的tuple
乘以2。你知道吗例如:
并确保分配了正确的值:
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