转置存储在 NumPy 中的一维数组中的矩阵的最快方法？

Question

Given a vector v of length N^2 that holds the entries of a NxN matrix M , what is the fastest way to compute the transpose of M in the same vector representation using NumPy?

I know this can be done by

v.reshape(N, N).T.flatten()

but is this the fastest way?

I am not interested in the intermediate explicit form of M .

Answer 1

Consider a test case:

In [207]: N=1000
In [208]: X = np.arange(N*N)

Your code:

In [209]: Y = X.reshape(N,N).T.flatten()
In [210]: timeit Y = X.reshape(N,N).T.flatten()
5.45 ms ± 13 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

A suggested alternative:

In [211]: Z = X.reshape(N,N).flatten('F')
In [212]: np.allclose(Y,Z)
Out[212]: True
In [213]: timeit Z = X.reshape(N,N).flatten('F')
5.46 ms ± 39.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

No real difference. reshape and transpose are views.