我试图遍历numpy中的数组,并使用对索引的一些计算对每个元素应用一个函数。所以我的代码看起来像这样:
# f takes in a matrix element and returns some calculation based on the
# element's 2D index i,j
def f(elt, i,j):
return elt*i + elt*j
# create a 2x3 matrix, A
A = np.array([[1,2,3]
[4,5,6]])
# Transform A by applying the function `f` over every element.
A_1 = applyFunction(f, A)
print(A_1)
# A_1 should now be a matrix that is transformed:
# [[0 2 6]
[4 10 18]
apply
或apply_along_axis
我还考虑过将矩阵转换为pandas DataFrame,然后也许将列名和行名用作索引。.但是我不知道如何在apply_along_axis
函数调用中访问它
任何帮助将不胜感激。谢谢!
答案 0 :(得分:2)
def f(elt, i,j):
return (i,j)
A = [[1,2,3],
[4,5,6]]
In [306]: [[f(None,i,j) for j in range(len(A[0]))] for i in range(len(A))]
Out[306]: [[(0, 0), (0, 1), (0, 2)], [(1, 0), (1, 1), (1, 2)]]
一个数组解决方案,速度可能大致相同:
In [309]: np.frompyfunc(f,3,1)(None, [[0],[1]],[0,1,2])
Out[309]:
array([[(0, 0), (0, 1), (0, 2)],
[(1, 0), (1, 1), (1, 2)]], dtype=object)
In [310]: _.shape
Out[310]: (2, 3)
最快的numpy方法,但不使用您的f
函数:
In [312]: I,J = np.meshgrid(range(2),range(3),indexing='ij')
In [313]: I
Out[313]:
array([[0, 0, 0],
[1, 1, 1]])
In [314]: J
Out[314]:
array([[0, 1, 2],
[0, 1, 2]])
In [315]: np.stack((I,J), axis=2)
Out[315]:
array([[[0, 0],
[0, 1],
[0, 2]],
[[1, 0],
[1, 1],
[1, 2]]])
In [316]: _.shape
Out[316]: (2, 3, 2)