tile函数在numpy库中
python文档如下:
tile(A, reps)
Construct an array by repeating A the number of times given by reps.
通过由reps给出的参数,重复A相应的次数,构造一个新的数组
If `reps` has length ``d``, the result will have dimension of
``max(d, A.ndim)``.
If ``A.ndim < d``, `A` is promoted to be d-dimensional by prepending new
axes. So a shape (3,) array is promoted to (1, 3) for 2-D replication,
or shape (1, 1, 3) for 3-D replication. If this is not the desired
behavior, promote `A` to d-dimensions manually before calling this
function.
If ``A.ndim > d``, `reps` is promoted to `A`.ndim by pre-pending 1's to it.
Thus for an `A` of shape (2, 3, 4, 5), a `reps` of (2, 2) is treated as
(1, 1, 2, 2).
Note : Although tile may be used for broadcasting, it is strongly
recommended to use numpy's broadcasting operations and functions.
Parameters
----------
A : array_like
The input array.
reps : array_like
The number of repetitions of `A` along each axis.
Returns
-------
c : ndarray
The tiled output array.
See Also
--------
repeat : Repeat elements of an array.
broadcast_to : Broadcast an array to a new shape
Examples
--------
>>> a = np.array([0, 1, 2])
>>> np.tile(a, 2)
array([0, 1, 2, 0, 1, 2])
>>> np.tile(a, (2, 2))
array([[0, 1, 2, 0, 1, 2],
[0, 1, 2, 0, 1, 2]])
>>> np.tile(a, (2, 1, 2))
array([[[0, 1, 2, 0, 1, 2]],
[[0, 1, 2, 0, 1, 2]]])
>>> b = np.array([[1, 2], [3, 4]])
>>> np.tile(b, 2)
array([[1, 2, 1, 2],
[3, 4, 3, 4]])
>>> np.tile(b, (2, 1))
array([[1, 2],
[3, 4],
[1, 2],
[3, 4]])
>>> c = np.array([1,2,3,4])
>>> np.tile(c,(4,1))
array([[1, 2, 3, 4],
[1, 2, 3, 4],
[1, 2, 3, 4],
[1, 2, 3, 4]])
自己添加几个示例:
>>> tile([1,2],(4,1))
array([[1, 2],
[1, 2],
[1, 2],
[1, 2]])
>>> tile([1,2],(4,2))
array([[1, 2, 1, 2],
[1, 2, 1, 2],
[1, 2, 1, 2],
[1, 2, 1, 2]])
>>> tile([[1,2],[3,4]],(4,2))
array([[1, 2, 1, 2],
[3, 4, 3, 4],
[1, 2, 1, 2],
[3, 4, 3, 4],
[1, 2, 1, 2],
[3, 4, 3, 4],
[1, 2, 1, 2],
[3, 4, 3, 4]])
>>> tile([[1,2],[3,4]],(4,1))
array([[1, 2],
[3, 4],
[1, 2],
[3, 4],
[1, 2],
[3, 4],
[1, 2],
[3, 4]])