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matlab sum function to python converstion

I am trying to convert this matlab code to python:
#T2 = (sum((log(X(1:m,:)))'));
Here is my code in python:
T2 = sum(np.log(X[0:int(m),:]).T)
where m = 103 and X is a matrix:
f1 = np.float64(135)
f2 = np.float64(351)
X = np.float64(p[:, int(f1):int(f2)])
and p is dictionary (loaded data)
The problem is python gives me the exact same value with same dimension (216x103) like matlab before applying the sum function on (np.log(X[0:int(m), :]).T). However. after applying the sum function it gives me the correct value but wrong dimension (103x1). The correct dimension is (1x103). I have tried using transpose after getting the sum but it doesnt work. Any suggestions how to get my desired dimension?

A matrix in MATLAB consists of m rows and n columns, but a matrix in NumPy is an array of arrays. Each subarray is a flat vector having 1 dimension equal to the number of its elements n. MATLAB doesn't have flat vectors at all, a row is 1xn matrix, a column is mx1 matrix, and a scalar is 1x1 matrix.
So, back to the question, when you write T2 = sum(np.log(X[0:int(m),:]).T) in Python, it's neither 103x1 nor 1x103, it's a flat 103 vector. If you specifically want a 1x103 matrix like MATLAB, just reshape(1,-1) and you don't have to transpose since you can sum over the second axis.
import numpy as np
X = np.random.rand(216,103)
m = 103
T2 = np.sum(np.log(X[:m]), axis=1).reshape(1,-1)
T2.shape
# (1, 103)

Lets make a demo 2d array:
In [19]: x = np.arange(12).reshape(3,4)
In [20]: x
Out[20]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
And apply the base Python sum function (which isn't the same as numpy's own):
In [21]: sum(x)
Out[21]: array([12, 15, 18, 21])
The result is a (4,) shape array (not 4x1). Print sum(x).shape if you don't believe me.
The numpy.sum function adds all terms if no axis is given:
In [22]: np.sum(x)
Out[22]: 66
or with axis:
In [23]: np.sum(x, axis=0)
Out[23]: array([12, 15, 18, 21])
In [24]: np.sum(x, axis=1)
Out[24]: array([ 6, 22, 38])
The Python sum treats x as a list of arrays, and adds them together
In [25]: list(x)
Out[25]: [array([0, 1, 2, 3]), array([4, 5, 6, 7]), array([ 8, 9, 10, 11])]
In [28]: x[0]+x[1]+x[2]
Out[28]: array([12, 15, 18, 21])
Transpose, without parameter, switch axes. It does not add any dimensions:
In [29]: x.T # (4,3) shape
Out[29]:
array([[ 0, 4, 8],
[ 1, 5, 9],
[ 2, 6, 10],
[ 3, 7, 11]])
In [30]: sum(x).T
Out[30]: array([12, 15, 18, 21]) # still (4,) shape
Octave
>> x=reshape(0:11,4,3)'
x =
0 1 2 3
4 5 6 7
8 9 10 11
>> sum(x)
ans =
12 15 18 21
>> sum(x,1)
ans =
12 15 18 21
>> sum(x,2)
ans =
6
22
38
edit
The np.sum function has a keepdims parmeter:
In [32]: np.sum(x, axis=0, keepdims=True)
Out[32]: array([[12, 15, 18, 21]]) # (1,4) shape
In [33]: np.sum(x, axis=1, keepdims=True)
Out[33]:
array([[ 6], # (3,1) shape
[22],
[38]])
If I reshape the array to 3d, and sum, the result is 2d - unless I keepdims:
In [34]: np.sum(x.reshape(3,2,2), axis=0).shape
Out[34]: (2, 2)
In [36]: np.sum(x.reshape(3,2,2), axis=0,keepdims=True).shape
Out[36]: (1, 2, 2)
MATLAB/Octave on the other hand keeps the dims by default:
sum(reshape(x,3,2,2)) # (1,2,2)
unless I sum on that last, 3rd:
sum(reshape(x,3,2,2),3) # (3,2)
The key is that MATLAB everything is 2d, with the option of additional trailing dimensions, which aren't handled the same way. In numpy every number of dimensions, from 0 on up is handled the same way.

How to invert volume of matrices in numpy?

Please assume a vector of invertible matrices:
import numpy as np
a = np.arange(120).reshape((2, 2, 5, 6))
I want to invert the matrices over their defined axes:
b = np.linalg.inv(a, axis1=0, axis2=1)
but this does not seems supported.
How to achieve this?

inv docs specifies its array input as:
a : (..., M, M) array_like
Matrix to be inverted.
You have a
a = np.arange(120).reshape((2, 2, 5, 6))
(M,M,...)
The dimensions are in the wrong order - change them!
In [44]: a = np.arange(120).reshape((2, 2, 5, 6))
Change the axes to the order that inv accepts:
In [45]: A = a.transpose(2,3,0,1)
In [46]: Ai = np.linalg.inv(A)
In [47]: Ai.shape
Out[47]: (5, 6, 2, 2)
In [48]: ai = Ai.transpose(2,3,0,1) # and back
In [49]: ai.shape
Out[49]: (2, 2, 5, 6)
I was going to test the result, but got:
In [50]: x = a#ai
Traceback (most recent call last):
File "<ipython-input-50-9dfe3616745d>", line 1, in <module>
x = a#ai
ValueError: matmul: Input operand 1 has a mismatch in its core dimension 0, with gufunc signature (n?,k),(k,m?)->(n?,m?) (size 5 is different from 6)
Like inv, matmul treats the last 2 dimensions as the matrix, the first 2 as 'batch':
In [51]: x = A#Ai
In [52]: x[0,0]
Out[52]:
array([[1., 0.],
[0., 1.]])
In [53]: x[0,3]
Out[53]:
array([[1.00000000e+00, 1.38777878e-17],
[4.44089210e-16, 1.00000000e+00]])
We can do the equivalent with einsum:
In [55]: x = np.einsum('ijkl,jmkl->imkl',a,ai)
In [56]: x[:,:,0,0]
Out[56]:
array([[1., 0.],
[0., 1.]])
You might want to change the original specification to match the inv and matmul usage. It could make life easier for you. Also remember that in numpy the trailing dimensions are the inner most ones.

If you know that the matrices are 2x2 you can do that easily using the standard formula for inverting such matrices; otherwise, I fear the only reasonable solution would be to do it with for loops? For example, the following works for any shape (modifying the sizes adequately):
b = np.stack([np.linalg.inv(a[:, :, i, j]) for i in range(a.shape[2]) for j in range(a.shape[3])], axis=2)
b = b.reshape(2, 2, 5, 6)
as checked by
for i in range(a.shape[2]):
for j in range(a.shape[3]):
assert np.allclose(np.dot(a[:,:,i,j], b[:,:,i,j]), np.eye(2))
In the specific 2x2 case you can do the following, which is fully vectorized hence probably faster:
determinants = a[0, 0] * a[1, 1] - a[0, 1] * a[1, 0]
b = 1 / determinants * np.stack([
np.stack([a[1, 1], -a[0, 1]]),
np.stack([-a[1, 0], a[0, 0]]),
])
On the specific (small) input size, the second solution is about 10 times faster in my tests (43us vs. 537us).

How to reshape matrix with numpy without using explicit values for argument?

I am trying to create a function and calculate the inner product using numpy.
I got the function to work however I am using explicit numbers in my np.reshape function and I need to make it to use based on input.
my code looks like this:
import numpy as np
X = np.array([[1,2],[3,4]])
Z = np.array([[1,4],[2,5],[3,6]])
# Calculating S
def calculate_S(X, n, m):
assert n == X.shape[0]
n,d1=X.shape
m,d2=X.shape
S = np.diag(np.inner(X,X))
return S
S= calculate_S(X,n,m)
S = S.reshape(2,1)
print(s)
output:
---------------------------------
[[ 5]
[25]]
So the output is correct however instead of specifying 2,1 I need that those values to be automatically placed there based on the shape of my matrix.
How do I do that?

In [163]: X = np.array([[1,2],[3,4]])
In [164]: np.inner(X,X)
Out[164]:
array([[ 5, 11],
[11, 25]])
In [165]: np.diag(np.inner(X,X))
Out[165]: array([ 5, 25])
reshape with -1 gets around having to specify the 2:
In [166]: np.diag(np.inner(X,X)).reshape(-1,1)
Out[166]:
array([[ 5],
[25]])
another way of adding dimension:
In [167]: np.diag(np.inner(X,X))[:,None]
Out[167]:
array([[ 5],
[25]])
You can get the "diagonal" directly with:
In [175]: np.einsum('ij,ij->i',X,X)
Out[175]: array([ 5, 25])
another
In [177]: (X[:,None,:]#X[:,:,None])[:,0,:]
Out[177]:
array([[ 5],
[25]])

Tensordot Explanation from Numpy documentation

I don't udnerstand how tensordot works and I was reading the official documentation but I don't understand at all what is happening there.
a = np.arange(60.).reshape(3,4,5)
b = np.arange(24.).reshape(4,3,2)
c = np.tensordot(a,b, axes=([1,0],[0,1]))
c.shape
(5, 2)
Why is the shape (5, 2)? What exactly is happening?
I also read this article but the answer is confusing me.
In [7]: A = np.random.randint(2, size=(2, 6, 5))
...: B = np.random.randint(2, size=(3, 2, 4))
...:
In [9]: np.tensordot(A, B, axes=((0),(1))).shape
Out[9]: (6, 5, 3, 4)
A : (2, 6, 5) -> reduction of axis=0
B : (3, 2, 4) -> reduction of axis=1
Output : `(2, 6, 5)`, `(3, 2, 4)` ===(2 gone)==> `(6,5)` + `(3,4)` => `(6,5,3,4)`
Why is the shape (6, 5, 3, 4)?

In [196]: a = np.arange(60.).reshape(3,4,5)
...: b = np.arange(24.).reshape(4,3,2)
...: c = np.tensordot(a,b, axes=([1,0],[0,1]))
In [197]: c
Out[197]:
array([[4400., 4730.],
[4532., 4874.],
[4664., 5018.],
[4796., 5162.],
[4928., 5306.]])
I find the einsum equivalent to be easier to "read":
In [198]: np.einsum('ijk,jil->kl',a,b)
Out[198]:
array([[4400., 4730.],
[4532., 4874.],
[4664., 5018.],
[4796., 5162.],
[4928., 5306.]])
tensordot works by transposing and reshaping the inputs to reduce the problem to a simple dot:
In [204]: a1 = a.transpose(2,1,0).reshape(5,12)
In [205]: b1 = b.reshape(12,2)
In [206]: np.dot(a1,b1) # or a1#b1
Out[206]:
array([[4400., 4730.],
[4532., 4874.],
[4664., 5018.],
[4796., 5162.],
[4928., 5306.]])
tensordot can do further manipulation to the result, but that's not needed here.
I had to try several things before I got a1/b1 right. For example a.transpose(2,0,1).reshape(5,12) produces the right shape, but different values.
yet another version:
In [210]: (a.transpose(1,0,2)[:,:,:,None]*b[:,:,None,:]).sum((0,1))
Out[210]:
array([[4400., 4730.],
[4532., 4874.],
[4664., 5018.],
[4796., 5162.],
[4928., 5306.]])

broadcasting arrays in numpy

I got an array and reshaped it to the following dimentions: (-1,1,1,1) and (-1,1):
Array A:
[-0.888788523827 0.11842529285 0.319928774626 0.319928774626 0.378755429421 1.225877519716 3.830653798838]
A.reshape(-1,1,1,1):
[[[[-0.888788523827]]]
[[[ 0.11842529285 ]]]
[[[ 0.319928774626]]]
[[[ 0.319928774626]]]
[[[ 0.378755429421]]]
[[[ 1.225877519716]]]
[[[ 3.830653798838]]]]
A.reshape(-1,1):
[[-0.888788523827]
[ 0.11842529285 ]
[ 0.319928774626]
[ 0.319928774626]
[ 0.378755429421]
[ 1.225877519716]
[ 3.830653798838]]
Then I have done substractig and broadcasting came in, so my resulting matrix is 7x1x7x1.
I have a hard time to visualize the intermediate step what broadcasting does. I mean I cannot imagine what elements of arrays are repeated and what they look like while broadcasting.
Could somebody shed some light on this problem,please?

In [5]: arr = np.arange(4)
In [6]: A = arr.reshape(-1,1,1,1)
In [7]: B = arr.reshape(-1,1)
In [8]: C = A + B
In [9]: C.shape
Out[9]: (4, 1, 4, 1)
In [10]: A.shape
Out[10]: (4, 1, 1, 1)
In [11]: B.shape
Out[11]: (4, 1)
There are 2 basic broadcasting rules:
expand the dimensions to match - by adding size 1 dimensions at the start
adjust all size 1 dimensions to match
So in this example:
(4,1,1,1) + (4,1)
(4,1,1,1) + (1,1,4,1) # add 2 size 1's to B
(4,1,4,1) + (4,1,4,1) # adjust 2 of the 1's to 4
(4,1,4,1)
The first step is, perhaps, the most confusing. The (4,1) is expanded to (1,1,4,1), not (4,1,1,1). The rule is intended to avoid ambiguity - by expanding in a consistent manner, not necessarily what a human might intuitively want.
Imagine the case where both arrays need expansion to match, and it could add a dimension in either direction:
(4,) and (3,)
(1,4) and (3,1) or (4,1) and (1,3)
(3,4) or (4,3)
confusion
The rule requires that the programmer choose which one expands to the right (4,1) or (3,1). numpy can then unambiguously add the other.
For a simpler example:
In [22]: A=np.arange(3).reshape(-1,1)
In [23]: B=np.arange(3)
In [24]: C = A+B (3,1)+(3,) => (3,1)+(1,3) => (3,3)
In [25]: C
Out[25]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4]])
In [26]: C.shape
Out[26]: (3, 3)
The [0,2,4] are present, but on the diagonal of C.
When broadcasting like this, the result is a kind of outer sum:
In [27]: np.add.outer(B,B)
Out[27]:
array([[0, 1, 2],
[1, 2, 3],
[2, 3, 4]])