尝试使用 np.tile 将重复的一维数组插入另一个数组的列时出错

Question

I need to fill an array column from top to bottom with a list that repeats. A toy example is shown below, with the various approaches I have tried.

The "reshape" approach was the one I thought would work, but I get the "could not broadcast input array from shape (12,1) into shape (12,)" error.

>>> x = np.zeros((12,4))
>>> #x[:,0] = np.tile(range(4),(3,1))
>>> #x[:,0] = np.tile(np.array(range(4)),(3,1))
>>> x[:,0] = np.tile(np.reshape(range(4),(4,1)),(3,1))

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
Input In [121], in <cell line: 4>()
      1 x = np.zeros((12,4))
      2 #x[:,0] = np.tile(range(4),(3,1))
      3 #x[:,0] = np.tile(range(4),(3,1))
----> 4 x[:,0] = np.tile(np.reshape(range(4),(4,1)),(3,1))

ValueError: could not broadcast input array from shape (12,1) into shape (12,)

Answer 1

It can be achieved without using np.tile :

x[:, 0] = np.array([*np.arange(4)] * 3)

Which will be faster than np.tile on such small arrays. numba usage will be the fastest for small and large arrays:

import numba as nb

@nb.njit
def numba_(x):
  rows_count = x.shape[0]
  range_ = np.arange(x.shape[1])
  for i in range(rows_count):
    x[i, 0] = range_[i % x.shape[1]]
  return x

Answer 2

it came to me moments after I posted. Just use a range of length 1 for the second index into x. Tried, worked.

x[:,0:1] = np.tile(np.reshape(range(4),(4,1)),(3,1))