具有重复索引的numpy数组的矢量化赋值(d[i，j，i，j] = s[i，j])

Question

如何设置

d[i,j,i,j] = s[i,j]

使用"NumPy“而不使用for循环？

我尝试过以下几种方法：

l1=range(M)
l2=range(N)
d[l1,l2,l1,l2] = s[l1,l2]

Answer 1

您可以使用integer array indexing (使用np.ix_创建广播索引)：

d[np.ix_(l1,l2)*2] = s[np.ix_(l1,l2)]

第一次必须复制索引(您需要[i, j, i, j]而不仅仅是[i, j])，这就是我将np.ix_返回的tuple乘以2的原因。

例如：

>>> d = np.zeros((10, 10, 10, 10), dtype=int)
>>> s = np.arange(100).reshape(10, 10)
>>> l1 = range(3)
>>> l2 = range(5)
>>> d[np.ix_(l1,l2)*2] = s[np.ix_(l1,l2)]

并确保分配了正确的值：

>>> # Assert equality for the given condition
>>> for i in l1:
...     for j in l2:
...         assert d[i, j, i, j] == s[i, j]

>>> # Interactive tests
>>> d[0, 0, 0, 0], s[0, 0]
(0, 0)
>>> d[1, 2, 1, 2], s[1, 2]
(12, 12)
>>> d[2, 0, 2, 0], s[2, 0]
(20, 20)
>>> d[2, 4, 2, 4], s[2, 4]
(24, 24)

Answer 2

如果你想一想，这就等同于创建一个形状为(m*n, m*n)的2D数组，并将s中的值赋给对角线位置。为了让最终的输出成为4D，我们只需要在最后重塑一下。这基本上在下面实现-

m,n = s.shape
d = np.zeros((m*n,m*n),dtype=s.dtype)
d.ravel()[::m*n+1] = s.ravel()
d.shape = (m,n,m,n)

运行时测试

方法-

# @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