在numpy.array中没有fortran命令

Question

I see no fortran order in:

import numpy as np
In [143]: np.array([[1,2],[3,4]],order='F')
Out[143]: 
array([[1, 2],
       [3, 4]])

But in the following it works:

In [139]: np.reshape(np.arange(9),newshape=(3,3),order='F')
Out[139]: 
array([[0, 3, 6],
       [1, 4, 7],
       [2, 5, 8]])

So what am I doing wrong in the first one?

Answer 1

When you call numpy.array to create an array from an existing Python object, it will give you an object with whatever shape that the original Python object has. So,

np.array([[1,2],[3,4]], ...)

Will always give you,

np.array([[1, 2],
          [3, 4]])

Which is exactly what you typed in, so it should not come as a surprise. Fortran order and C order do not describe the shape of the data, they describe the memory layout. When you print out an object, NumPy doesn't show you what the memory layout is, it only shows you the shape.

You can witness that the array truly is stored in Fortran order when you flatten it with the "K" order, which keeps the original order of the elements:

>>> a = np.array([[1,2],[3,4]], order="F")
>>> a.flatten(order="K")
array([1, 3, 2, 4])

This is what truly distinguishes Fortran from C order: the memory layout. Most NumPy functions do not force you to consider memory layout, instead, different layouts are handled transparently.

It sounds like what you want is to transpose, reversing the axis order. This can be done simply:

>>> b = numpy.transpose(a)
>>> b
array([[1, 3],
       [2, 4]])

This does not create a new array, but a new view of the same array:

>>> b.base is a
True

If you want the data to have the memory layout 1 2 3 4 and have a Fortran order view of that [[1, 3], [2, 4]], the efficient way to do this is to store the existing array with C order and then transpose it, which results in a Fortran-order array with the desired contents and requires no extra copies.

>>> a = np.array([[1, 2], [3, 4]]).transpose()
>>> a.flatten(order="K")
array([1, 2, 3, 4])
>>> a
array([[1, 3],
       [2, 4]])

If you store the original with Fortran order, the transposition will result in C order, so you don't want that (or maybe all you care about is the transposition, and memory order is not important?). In either case, the array will look the same in NumPy.

>>> a = np.array([[1, 2], [3, 4]], order="F").transpose()
>>> a.flatten(order="K")
array([1, 3, 2, 4])
>>> a
array([[1, 3],
       [2, 4]])

Answer 2

Your two means of constructing the 2D array are not at all equivalent. In the first, you specified the structure of the array. In the second, you formed an array and then reshaped to your liking.

>>> np.reshape([1,2,3,4],newshape=(2,2),order='F')
array([[1, 3],
       [2, 4]])

Again, for comparison, even if you ask for the reshape and format change to FORTRAN, you'll get your specified structure:

>>> np.reshape([[1,2],[3,4]],newshape=(2,2),order='F')
array([[1, 2],
       [3, 4]])