Can't inderstand why numpy doesn't transpose matrix.
This doesn't work:
w=2
h=3
rc= np.array([[0,0,1],[0,h,1],[w,h,1],[w,0,1]])
array([[0, 0, 1],
[0, 3, 1],
[2, 3, 1],
[2, 0, 1]])
rc[0].T
array([0, 0, 1])
but this works:
v_x= np.array([[0,1,2,3]])
v_x.T
array([[0],
[1],
[2],
[3]])
Your rc[0] is not an 1x3 matrix, it's rather a vector of 3 items:
>>> rc[0].shape
(3,)
Fix its shape if you want to transpose it:
>>> np.reshape(rc[0], (1,3)).T
array([[0],
[0],
[1]])
np.array([0,1,2,3]) (which is its own transpose) and np.array([[0,1,2,3]]) (which, as noted in the OP, transposes into a 4x1 matrix).One of the first things I ask when confronted with a problem like this is: what's the shape?
In [14]: rc.shape
Out[14]: (4, 3)
In [15]: rc[0].shape
Out[15]: (3,)
Indexing has selected a row, and reduced the number of dimensions.
But if I index with a list (or array), the result is 2d
In [16]: rc[[0]].shape
Out[16]: (1, 3)
In [19]: v_x.shape
Out[19]: (1, 4)
There are other ways of getting that shape, or even the target (3,1).
rc[0].reshape(1,-1) # -1 stands in for the known 3
rc[0][np.newaxis,:] # useful when constructing outer products
rc[0].reshape(-1,1) # transposed
rc[0][:,np.newaxis]
np.matrix is a subclass of np.array that is always 2d (even after indexing with a scalar)
rcm = np.matrix(rc)
rcm[0] # (1,3)
rcm[0].T # (3,1)
np.matrix can ease the transition if you are coming to numpy from MATLAB, especially the older versions where everything was a 2d matrix. For others it is probably better to become familiar with ndarray, and use matrix only if it adds some functionality.
Transpose in fact works, but not as you expected, see docs:
By default, reverse the dimensions
So, as your array is 1d, reverse do nothing with its shape:
>>> np.array([0,0,1]).T.shape
(3,)
You can add more dimensions with reshape:
>>> np.array([0,0,1]).reshape(-1,1,1).shape
(3, 1, 1)
Now, nontrivial shape can be reversed:
>>> np.array([0,0,1]).reshape(-1,1,1).T.shape
(1, 1, 3)
