在numpy中阻止标量向量乘法

Question

I have a large 1-dimensional array x which I've got by concatenation of smaller arrays x_0 ,..., x_m-1 of different length. I also know the list L of the length of each x_i . Given an array a of length m , the goal is to compute a flat array [a[0]*x0, a[1]*x1,...] .

For example, if I have x = np.array([1,2,3,4,5]) and a=np.array([2,-1]), L = [2,3] , then the result should be np.array([2,4,-3,-4,-5])

Is there any simpler (faster, more pythonic, etc.) way to do this in numpy, than this naive implementation?

L.insert(0,0)
cs = np.cumsum(L)
y = np.empty(x.shape) 
for i in range(m):
    y[cs[i]:cs[i+1]] = a[i] * x[cs[i]:cs[i+1]]

I can also do this in Numba.

m is of order of hundreds, the length of each x_i is around 1e6.

Answer 1

重复的元素a与np.repeat和执行的elementwise乘法-

y = x*np.repeat(a,L)

Answer 2

Out of place Numba version

@nb.njit(fastmath=True)
def mult(x,a,L):
  out=np.empty(x.shape[0],dtype=x.dtype)
  ii=0
  for i in range(L.shape[0]):
    for j in range(L[i]):
      out[ii]=x[ii]*a[i]
      ii+=1
  return out

In place Numba version

@nb.njit(fastmath=True)
def mult(x,a,L):
  ii=0
  for i in range(L.shape[0]):
    for j in range(L[i]):
      x[ii]=x[ii]*a[i]
      ii+=1
  return x

Timings

L=np.random.randint(low=1000,high=2000,size=500)
x=np.random.rand(L.sum())
a=np.random.rand(L.shape[0])

Divakar's version:          6.4ms
Out of place Numba version: 2.8ms
In place Numba version:     1.2ms

Please note that the first call to the Numba versions takes longer (compilation overhead).