Matrix multiplication of two vectors - numpy

I'm trying to do a matrix multiplication of two vectors in numpy which would result in an array.
Example
In [108]: b = array([[1],[2],[3],[4]])
In [109]: a =array([1,2,3])
In [111]: b.shape
Out[111]: (4, 1)
In [112]: a.shape
Out[112]: (3,)
In [113]: b.dot(a)
ValueError: objects are not aligned
As can be seen from the shapes, the array a isn't actually a matrix. The catch is to define a like this.
In [114]: a =array([[1,2,3]])
In [115]: a.shape
Out[115]: (1, 3)
In [116]: b.dot(a)
Out[116]:
array([[ 1, 2, 3],
[ 2, 4, 6],
[ 3, 6, 9],
[ 4, 8, 12]])
How to achieve the same result when acquiring the vectors as fields or columns of a matrix?
In [137]: mat = array([[ 1, 2, 3],
[ 2, 4, 6],
[ 3, 6, 9],
[ 4, 8, 12]])
In [138]: x = mat[:,0] #[1,2,3,4]
In [139]: y = mat[0,:] #[1,2,3]
In [140]: x.dot(y)
ValueError: objects are not aligned

You are computing the outer product of two vectors. You can use the function numpy.outer for this:
In [18]: a
Out[18]: array([1, 2, 3])
In [19]: b
Out[19]: array([10, 20, 30, 40])
In [20]: numpy.outer(b, a)
Out[20]:
array([[ 10, 20, 30],
[ 20, 40, 60],
[ 30, 60, 90],
[ 40, 80, 120]])

Use 2d arrays instead of 1d vectors and broadcasting with the * ...
In [8]: #your code from above
In [9]: y = mat[0:1,:]
In [10]: y
Out[10]: array([[1, 2, 3]])
In [11]: x = mat[:,0:1]
In [12]: x
Out[12]:
array([[1],
[2],
[3],
[4]])
In [13]: x*y
Out[13]:
array([[ 1, 2, 3],
[ 2, 4, 6],
[ 3, 6, 9],
[ 4, 8, 12]])

It's the similar catch as in the basic example.
Both x and y aren't perceived as matrices but as single dimensional arrays.
In [143]: x.shape
Out[143]: (4,)
In [144]: y.shape
Out[144]: (3,)
We have to add the second dimension to them, which will be 1.
In [171]: x = array([x]).transpose()
In [172]: x.shape
Out[172]: (4, 1)
In [173]: y = array([y])
In [174]: y.shape
Out[174]: (1, 3)
In [175]: x.dot(y)
Out[175]:
array([[ 1, 2, 3],
[ 2, 4, 6],
[ 3, 6, 9],
[ 4, 8, 12]])

Related

Can I select arbitrary windows from the last dimension of a numpy array?

I'd like to write a numpy function that takes an MxN array A, a window length L, and an MxP array idxs of starting indices into the M rows of A that selects P arbitrary slices of length L from each of the M rows of A. Except, I would love for this to work on the last dimension of A, and not necessarily care how many dimensions A has, so all dims of A and idxs match except the last one. Examples:
If A is just 1D:
A = np.array([1, 2, 3, 4, 5, 6])
window_len = 3
idxs = np.array([1, 3])
result = magical_routine(A, idxs, window_len)
Where result is a 2x3 array since I selected 2 slices of len 3:
np.array([[ 2, 3, 4],
[ 4, 5, 6]])
If A is 2D:
A = np.array([[ 1, 2, 3, 4, 5, 6],
[ 7, 8, 9,10,11,12],
[13,14,15,16,17,18]])
window_len = 3
idxs = np.array([[1, 3],
[0, 1],
[2, 2]])
result = magical_routine(A, idxs, window_len)
Where result is a 3x2x3 array since there are 3 rows of A, and I selected 2 slices of len 3 from each row:
np.array([[[ 2, 3, 4], [ 4, 5, 6]],
[[ 7, 8, 9], [ 8, 9,10]],
[[15,16,17], [15,16,17]]])
And so on.
I have discovered an number of inefficient ways to do this, along with ways that work for a specific number of dimensions of A. For 2D, the following is pretty tidy:
col_idxs = np.add.outer(idxs, np.arange(window_len))
np.take_along_axis(A[:, np.newaxis], col_idxs, axis=-1)
I can't see a nice way to generalize this for 1D and other D's though...
Is anyone aware of an efficient way that generalizes to any number of dims?
For your 1d case
In [271]: A=np.arange(1,7)
In [272]: idxs = np.array([1,3])
Using the kind of iteration that this questions usually gets:
In [273]: np.vstack([A[i:i+3] for i in idxs])
Out[273]:
array([[2, 3, 4],
[4, 5, 6]])
Alternatively generate all indices, and one indexing. linspace is handy for this (though it's not the only option):
In [278]: j = np.linspace(idxs,idxs+3,3,endpoint=False)
In [279]: j
Out[279]:
array([[1., 3.],
[2., 4.],
[3., 5.]])
In [282]: A[j.T.astype(int)]
Out[282]:
array([[2, 3, 4],
[4, 5, 6]])
for the 2d
In [284]: B
Out[284]:
array([[ 1, 2, 3, 4, 5, 6],
[ 7, 8, 9, 10, 11, 12],
[13, 14, 15, 16, 17, 18]])
In [285]: idxs = np.array([[1, 3],
...: [0, 1],
...: [2, 2]])
In [286]: j = np.linspace(idxs,idxs+3,3,endpoint=False)
In [287]: j
Out[287]:
array([[[1., 3.],
[0., 1.],
[2., 2.]],
[[2., 4.],
[1., 2.],
[3., 3.]],
[[3., 5.],
[2., 3.],
[4., 4.]]])
With a bit of trial and error, pair up the indices to get:
In [292]: B[np.arange(3)[:,None,None],j.astype(int).transpose(1,2,0)]
Out[292]:
array([[[ 2, 3, 4],
[ 4, 5, 6]],
[[ 7, 8, 9],
[ 8, 9, 10]],
[[15, 16, 17],
[15, 16, 17]]])
Or iterate as in the first case, but with an extra layer:
In [294]: np.array([[B[j,i:i+3] for i in idxs[j]] for j in range(3)])
Out[294]:
array([[[ 2, 3, 4],
[ 4, 5, 6]],
[[ 7, 8, 9],
[ 8, 9, 10]],
[[15, 16, 17],
[15, 16, 17]]])
With sliding windows:
In [295]: aa = np.lib.stride_tricks.sliding_window_view(A,3)
In [296]: aa.shape
Out[296]: (4, 3)
In [297]: aa
Out[297]:
array([[1, 2, 3],
[2, 3, 4],
[3, 4, 5],
[4, 5, 6]])
In [298]: aa[[1,3]]
Out[298]:
array([[2, 3, 4],
[4, 5, 6]])
and
In [300]: bb = np.lib.stride_tricks.sliding_window_view(B,(1,3))
In [301]: bb.shape
Out[301]: (3, 4, 1, 3)
In [302]: bb[np.arange(3)[:,None],idxs,0,:]
Out[302]:
array([[[ 2, 3, 4],
[ 4, 5, 6]],
[[ 7, 8, 9],
[ 8, 9, 10]],
[[15, 16, 17],
[15, 16, 17]]])
I got it! I was almost there:
def magical_routine(A, idxs, window_len=2000):
col_idxs = np.add.outer(idxs, np.arange(window_len))
return np.take_along_axis(A[..., np.newaxis, :], col_idxs, axis=-1)
I just needed to always add the new axis to A's second to last dim, and then leave remaining axes alone.

A question about axis in tensorflow.stack (tensorflow= 1.14)

Using tensorflow.stack what does it mean to have axis=-1 ?
I'm using tensorflow==1.14
Using axis=-1 simply means to stack the tensors along the last axis (as per the python list indexing syntax).
Let's take a look at how this looks like using these tensors of shape (2, 2):
>>> x = tf.constant([[1, 2], [3, 4]])
>>> y = tf.constant([[5, 6], [7, 8]])
>>> z = tf.constant([[9, 10], [11, 12]])
The default behavior for tf.stack as described in the documentation is to stack the tensors along the first axis (index 0) resulting in a tensor of shape (3, 2, 2)
>>> tf.stack([x, y, z], axis=0)
<tf.Tensor: shape=(3, 2, 2), dtype=int32, numpy=
array([[[ 1, 2],
[ 3, 4]],
[[ 5, 6],
[ 7, 8]],
[[ 9, 10],
[11, 12]]], dtype=int32)>
Using axis=-1, the three tensors are stacked along the last axis instead, resulting in a tensor of shape (2, 2, 3)
>>> tf.stack([x, y, z], axis=-1)
<tf.Tensor: shape=(2, 2, 3), dtype=int32, numpy=
array([[[ 1, 5, 9],
[ 2, 6, 10]],
[[ 3, 7, 11],
[ 4, 8, 12]]], dtype=int32)>

How to iterate through slices at the last dimension

For example, you have array
a = np.array([[[ 0, 1, 2],
[ 3, 4, 5]],
[[ 6, 7, 8],
[ 9, 10, 11]]])
We want to iterate through slices at the last dimension, i.e. [0,1,2], [3,4,5], [6,7,8], [9,10,11]. Any way to achieve this without the for loop? Thanks!
Tried this but it does not work, because numpy does not interpret the tuple in the way we wanted - a[(0, 0),:] is not the same as a[0, 0, :]
[a[i,:] for i in zip(*product(*(range(ii) for ii in a.shape[:-1])))]
More generally, any way for the last k dimensions? Something equivalent to looping through a[i,j,k, ...].
In [26]: a = np.array([[[ 0, 1, 2],
...: [ 3, 4, 5]],
...:
...: [[ 6, 7, 8],
...: [ 9, 10, 11]]])
In [27]: [a[i,j,:] for i in range(2) for j in range(2)]
Out[27]: [array([0, 1, 2]), array([3, 4, 5]), array([6, 7, 8]), array([ 9, 10, 11])]
or
In [31]: list(np.ndindex(2,2))
Out[31]: [(0, 0), (0, 1), (1, 0), (1, 1)]
In [32]: [a[i,j] for i,j in np.ndindex(2,2)]
another
list(a.reshape(-1,3))

Converting flat 1D matrix to a square matrix

Is there a way/code to convert the following n x 1 matrix,
x1
x2
x3
x4
x5
x6
x7
x8
x9
x10
etc.
into a square matrix of the form,
x1 x2 x4 x7
x2 x3 x5 x8
x4 x5 x6 x9
x7 x8 x9 x10
etc.
I have a 903 x 1 matrix (in a .csv format) that I hope to convert into a 42 x 42 matrix with the form as shown. Thanks!
I suppose I should wait until you edit the question, but I went ahead and looked at the figure. It looks like a symmetric matrix based on tri-upper and lower matrices. In what dicispline is that called a `full matrix'?
Anyhow's here one sequence that produces your figure:
In [93]: idx=np.tril_indices(4)
In [94]: idx
Out[94]: (array([0, 1, 1, 2, 2, 2, 3, 3, 3, 3]), array([0, 0, 1, 0, 1, 2, 0, 1, 2, 3]))
In [95]: arr = np.zeros((4,4),int)
In [96]: arr[idx] = np.arange(1,11)
In [97]: arr
Out[97]:
array([[ 1, 0, 0, 0],
[ 2, 3, 0, 0],
[ 4, 5, 6, 0],
[ 7, 8, 9, 10]])
In [98]: arr1 = arr + arr.T
In [99]: arr1
Out[99]:
array([[ 2, 2, 4, 7],
[ 2, 6, 5, 8],
[ 4, 5, 12, 9],
[ 7, 8, 9, 20]])
In [100]: dx = np.diag_indices(4)
In [101]: dx
Out[101]: (array([0, 1, 2, 3]), array([0, 1, 2, 3]))
In [102]: arr1[dx] = arr[dx]
In [103]: arr1
Out[103]:
array([[ 1, 2, 4, 7],
[ 2, 3, 5, 8],
[ 4, 5, 6, 9],
[ 7, 8, 9, 10]])
This is similar to what scipy.spatial calls a squareform for pairwise distances.
https://docs.scipy.org/doc/scipy-0.15.1/reference/generated/scipy.spatial.distance.squareform.html#scipy.spatial.distance.squareform
In [106]: from scipy.spatial import distance
In [107]: distance.squareform(np.arange(1,11))
Out[107]:
array([[ 0, 1, 2, 3, 4],
[ 1, 0, 5, 6, 7],
[ 2, 5, 0, 8, 9],
[ 3, 6, 8, 0, 10],
[ 4, 7, 9, 10, 0]])
It appears that this square_form uses compiled code, so I expect it will be quite a bit faster than my tril base code. But the order of elements isn't quite what you expect.
Numpy has a function to reshape arrays -
https://docs.scipy.org/doc/numpy/reference/generated/numpy.reshape.html
>>> np.reshape(a, (2, 3)) # C-like index ordering
array([[0, 1, 2],
[3, 4, 5]])

About reshaping numpy array

trainX.size == 43120000
trainX = trainX.reshape([-1, 28, 28, 1])
(1)Does reshape accept a list as an argment instead of a tuple?
(2)Are the following two statements equivalent?
trainX = trainX.reshape([-1, 28, 28, 1])
trainX = trainX.reshape((55000, 28, 28, 1))
Try the variations:
In [1]: np.arange(12).reshape(3,4)
Out[1]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
In [2]: np.arange(12).reshape([3,4])
Out[2]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
In [3]: np.arange(12).reshape((3,4))
Out[3]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
With the reshape method, the shape can be arguments, a tuple or a list. In the reshape function is has to be in a list or tuple, to separate them from the first array argument
In [4]: np.reshape(np.arange(12), (3,4))
Out[4]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
and yes, one -1 can be used. The total size of the reshape is fixed, so one value can be deduced from the others.
In [5]: np.arange(12).reshape(-1,4)
Out[5]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
The method documentation has this note:
Unlike the free function numpy.reshape, this method on ndarray allows
the elements of the shape parameter to be passed in as separate arguments.
For example, a.reshape(10, 11) is equivalent to
a.reshape((10, 11)).
It's a builtin function, but the signature looks like x.reshape(*shape), and it tries to be flexible as long as the values make sense.
From the numpy documentation:
newshape : int or tuple of ints
The new shape should be compatible with the original shape. If an
integer, then the result will be a 1-D array of that length. One shape
dimension can be -1. In this case, the value is inferred from the
length of the array and remaining dimensions.
So yes, -1 for one dimension is fine and your two statements are equivalent. About the tuple requirement,
>>> import numpy as np
>>> a = np.arange(9)
>>> a
array([0, 1, 2, 3, 4, 5, 6, 7, 8])
>>> a.reshape([3,3])
array([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
>>>
So apparently a list is good as well.