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Applying function on multiple dimensions of higher dimensional array

Suppose you have a higher dimensional array (3 or greater) which is composed of a series of 2d images. If this array is called x, then a 2d image will be represented as x[0,0,:,:]. Now what I want to do is apply a function that takes in a 2d image and outputs a scalar, on this higher dimensional array so that I would convert the dimension of the original array to one that is 2 dimensions lower. How would I do such a thing?
In other words, what is the faster numpy way of doing this: np.array([[f(x[i,j,:,:]) for i in range(x.shape[0])] for j in range(x.shape[1])]) for a list of axes and some function f that takes in an array.
I've looked at numpy.apply_along_axis but that only acts on a 1d array and the shape must be identical. numpy.apply_on_axes also doesn't work since it doesn't reduce the amount of dimensions which are given to the function (it gives my function a 4d array, not a 2d array which I can work with). numpy.vectorize doesn't work because it doesn't ever apply on more than one element at once.

Does np.dot automatically transpose vectors?

I am trying to calculate the first and second order moments for a portfolio of stocks (i.e. expected return and standard deviation).
expected_returns_annual
Out[54]:
ticker
adj_close CNP 0.091859
F -0.007358
GE 0.095399
TSLA 0.204873
WMT -0.000943
dtype: float64
type(expected_returns_annual)
Out[55]: pandas.core.series.Series
weights = np.random.random(num_assets)
weights /= np.sum(weights)
returns = np.dot(expected_returns_annual, weights)
So normally the expected return is calculated by
(x1,...,xn' * (R1,...,Rn)
with x1,...,xn are weights with a constraint that all the weights have to sum up to 1 and ' means that the vector is transposed.
Now I am wondering a bit about the numpy dot function, because
returns = np.dot(expected_returns_annual, weights)
and
returns = np.dot(expected_returns_annual, weights.T)
give the same results.
I tested also the shape of weights.T and weights.
weights.shape
Out[58]: (5,)
weights.T.shape
Out[59]: (5,)
The shape of weights.T should be (,5) and not (5,), but numpy displays them as equal (I also tried np.transpose, but there is the same result)
Does anybody know why numpy behave this way? In my opinion the np.dot product automatically shape the vector the right why so that the vector product work well. Is that correct?
Best regards
Tom

The semantics of np.dot are not great
As Dominique Paul points out, np.dot has very heterogenous behavior depending on the shapes of the inputs. Adding to the confusion, as the OP points out in his question, given that weights is a 1D array, np.array_equal(weights, weights.T) is True (array_equal tests for equality of both value and shape).
Recommendation: use np.matmul or the equivalent # instead
If you are someone just starting out with Numpy, my advice to you would be to ditch np.dot completely. Don't use it in your code at all. Instead, use np.matmul, or the equivalent operator #. The behavior of # is more predictable than that of np.dot, while still being convenient to use. For example, you would get the same dot product for the two 1D arrays you have in your code like so:
returns = expected_returns_annual # weights
You can prove to yourself that this gives the same answer as np.dot with this assert:
assert expected_returns_annual # weights == expected_returns_annual.dot(weights)
Conceptually, # handles this case by promoting the two 1D arrays to appropriate 2D arrays (though the implementation doesn't necessarily do this). For example, if you have x with shape (N,) and y with shape (M,), if you do x # y the shapes will be promoted such that:
x.shape == (1, N)
y.shape == (M, 1)
Complete behavior of matmul/#
Here's what the docs have to say about matmul/# and the shapes of inputs/outputs:
If both arguments are 2-D they are multiplied like conventional matrices.
If either argument is N-D, N > 2, it is treated as a stack of matrices residing in the last two indexes and broadcast accordingly.
If the first argument is 1-D, it is promoted to a matrix by prepending a 1 to its dimensions. After matrix multiplication the prepended 1 is removed.
If the second argument is 1-D, it is promoted to a matrix by appending a 1 to its dimensions. After matrix multiplication the appended 1 is removed.
Notes: the arguments for using # over dot
As hpaulj points out in the comments, np.array_equal(x.dot(y), x # y) for all x and y that are 1D or 2D arrays. So why do I (and why should you) prefer #? I think the best argument for using # is that it helps to improve your code in small but significant ways:
# is explicitly a matrix multiplication operator. x # y will raise an error if y is a scalar, whereas dot will make the assumption that you actually just wanted elementwise multiplication. This can potentially result in a hard-to-localize bug in which dot silently returns a garbage result (I've personally run into that one). Thus, # allows you to be explicit about your own intent for the behavior of a line of code.
Because # is an operator, it has some nice short syntax for coercing various sequence types into arrays, without having to explicitly cast them. For example, [0,1,2] # np.arange(3) is valid syntax.
To be fair, while [0,1,2].dot(arr) is obviously not valid, np.dot([0,1,2], arr) is valid (though more verbose than using #).
When you do need to extend your code to deal with many matrix multiplications instead of just one, the ND cases for # are a conceptually straightforward generalization/vectorization of the lower-D cases.

I had the same question some time ago. It seems that when one of your matrices is one dimensional, then numpy will figure out automatically what you are trying to do.
The documentation for the dot function has a more specific explanation of the logic applied:
If both a and b are 1-D arrays, it is inner product of vectors
(without complex conjugation).
If both a and b are 2-D arrays, it is matrix multiplication, but using
matmul or a # b is preferred.
If either a or b is 0-D (scalar), it is equivalent to multiply and
using numpy.multiply(a, b) or a * b is preferred.
If a is an N-D array and b is a 1-D array, it is a sum product over
the last axis of a and b.
If a is an N-D array and b is an M-D array (where M>=2), it is a sum
product over the last axis of a and the second-to-last axis of b:

In NumPy, a transpose .T reverses the order of dimensions, which means that it doesn't do anything to your one-dimensional array weights.
This is a common source of confusion for people coming from Matlab, in which one-dimensional arrays do not exist. See Transposing a NumPy Array for some earlier discussion of this.
np.dot(x,y) has complicated behavior on higher-dimensional arrays, but its behavior when it's fed two one-dimensional arrays is very simple: it takes the inner product. If we wanted to get the equivalent result as a matrix product of a row and column instead, we'd have to write something like
np.asscalar(x # y[:, np.newaxis])
adding a trailing dimension to y to turn it into a "column", multiplying, and then converting our one-element array back into a scalar. But np.dot(x,y) is much faster and more efficient, so we just use that.
Edit: actually, this was dumb on my part. You can, of course, just write matrix multiplication x # y to get equivalent behavior to np.dot for one-dimensional arrays, as tel's excellent answer points out.

The shape of weights.T should be (,5) and not (5,),
suggests some confusion over the shape attribute. shape is an ordinary Python tuple, i.e. just a set of numbers, one for each dimension of the array. That's analogous to the size of a MATLAB matrix.
(5,) is just the way of displaying a 1 element tuple. The , is required because of older Python history of using () as a simple grouping.
In [22]: tuple([5])
Out[22]: (5,)
Thus the , in (5,) does not have a special numpy meaning, and
In [23]: (,5)
File "<ipython-input-23-08574acbf5a7>", line 1
(,5)
^
SyntaxError: invalid syntax
A key difference between numpy and MATLAB is that arrays can have any number of dimensions (upto 32). MATLAB has a lower boundary of 2.
The result is that a 5 element numpy array can have shapes (5,), (1,5), (5,1), (1,5,1)`, etc.
The handling of a 1d weight array in your example is best explained the np.dot documentation. Describing it as inner product seems clear enough to me. But I'm also happy with the
sum product over the last axis of a and the second-to-last axis of b
description, adjusted for the case where b has only one axis.
(5,) with (5,n) => (n,) # 5 is the common dimension
(n,5) with (5,) => (n,)
(n,5) with (5,1) => (n,1)
In:
(x1,...,xn' * (R1,...,Rn)
are you missing a )?
(x1,...,xn)' * (R1,...,Rn)
And the * means matrix product? Not elementwise product (.* in MATLAB)? (R1,...,Rn) would have size (n,1). (x1,...,xn)' size (1,n). The product (1,1).
By the way, that raises another difference. MATLAB expands dimensions to the right (n,1,1...). numpy expands them to the left (1,1,n) (if needed by broadcasting). The initial dimensions are the outermost ones. That's not as critical a difference as the lower size 2 boundary, but shouldn't be ignored.

Seamlessly and efficiently flipping numpy array or sparse matrix along all axes

Let's say we have a matrix (numpy array) of unknown shape, the shape can be for example (1,5) (row), (5,1) (column), (5,5) (square), (5,6) (non-square) or (5,) (degenerated) (ok the last case isn't a matrix but is a valid input).
I would like to given a matrix of any shape (column, row, square, nonsquare, degenerated). I will return a flipped up/down left/right version of it.
Since np.flip has some issues with 1d arrays. My approach was:
def flipit(M):
return M.ravel()[::-1].reshape(M.shape)
It works, but is that acceptable? Any faster ways to do it?
In the other hand, how can I do the same for sparse matrices (for example if M is scipy.sparse.csr_matrix).

We can use slice notation with a step-size of -1 for the number of dims in the input to flip along all the axes, as that's what the original code is essentially doing. This would cover both arrays and sparse matrices -
def flip_allaxes(a): # a can be array or sparse matrix
# generate flipping slice
sl = slice(None,None,-1) # or np.s_[::-1] suggested by #kmario23
return a[tuple([sl]*a.ndim)]
Simplified on newer NumPy versions (15.1 onwards)
On newer NumPy versions : Version 15.1 and newer, that allows us to specify tuple of ints for the axes along which the flipping is needed. For the default case with axis=None from the docs, it flips along all axes. Thus, to solve our case, it would be simply np.flip(a) and this would again cover both generic ndarrays and sparse matrices.

Numpy dot product problems

A=np.array([
[1,2],
[3,4]
])
B=np.ones(2)
A is clearly of shape 2X2
How does numpy allow me to compute a dot product np.dot(A,B)
[1,2] (dot) [1,1]
[3,4]
B has to have dimensions of 2X1 for a dot product or rather this
[1,2] (dot) [1]
[3,4] [1]
This is a very silly question but i am not able to figure out where i am going wrong here?
Earlier i used to think that np.ones(2) would give me this:
[1]
[1]
But it gives me this:
[1,1]

I'm copying part of an answer I wrote earlier today:
You should resist the urge to think of numpy arrays as having rows
and columns, but instead consider them as having dimensions and
shape. This is an important point which differentiates np.array and np.matrix:
x = np.array([1, 2, 3])
print(x.ndim, x.shape) # 1 (3,)
y = np.matrix([1, 2, 3])
print(y.ndim, y.shape) # 2 (1, 3)
An n-D array can only use n integer(s) to represent its shape.
Therefore, a 1-D array only uses 1 integer to specify its shape.
In practice, combining calculations between 1-D and 2-D arrays is not
a problem for numpy, and syntactically clean since # matrix
operation was introduced in Python 3.5. Therefore, there is rarely a
need to resort to np.matrix in order to satisfy the urge to see
expected row and column counts.

This behavior is by design. The NumPy docs state:
If a is an N-D array and b is a 1-D array, it is a sum product over the last axis of a and b.

Most of the rules for vector and matrix shapes relating to the dot product exist mostly in order to have a coherent method that scales up into higher tensor orders. But they aren't very important when dealing with 1st order (vectors) and 2nd order (matrix) tensors. And those orders are what the vast majority of numpy users need.
As a result, # and np.dot are optimized (both mathematically and input parsing) for those orders, always summing over the last axis of the first and the second to last axis (if applicable) of the second. The "if applicable" is sort of an idiot-proofing to assure the output is what is expected in the vast majority of cases, even if the shapes don't technically fit.
Those of us who use higher-order tensors, meanwhile, are relegated to np.tensordot or np.einsum, which come complete with all the niggling little rules about dimension matching.

What does (n,) mean in the context of numpy and vectors?

I've tried searching StackOverflow, googling, and even using symbolhound to do character searches, but was unable to find an answer. Specifically, I'm confused about Ch. 1 of Nielsen's Neural Networks and Deep Learning, where he says "It is assumed that the input a is an (n, 1) Numpy ndarray, not a (n,) vector."
At first I thought (n,) referred to the orientation of the array - so it might refer to a one-column vector as opposed to a vector with only one row. But then I don't see why we need (n,) and (n, 1) both - they seem to say the same thing. I know I'm misunderstanding something but am unsure.
For reference a refers to a vector of activations that will be input to a given layer of a neural network, before being transformed by the weights and biases to produce the output vector of activations for the next layer.
EDIT: This question equivocates between a "one-column vector" (there's no such thing) and a "one-column matrix" (does actually exist). Same for "one-row vector" and "one-row matrix".
A vector is only a list of numbers, or (equivalently) a list of scalar transformations on the basis vectors of a vector space. A vector might look like a matrix when we write it out, if it only has one row (or one column). Confusingly, we will sometimes refer to a "vector of activations" but actually mean "a single-row matrix of activation values transposed so that it is a single-column."
Be aware that in neither case are we discussing a one-dimensional vector, which would be a vector defined by only one number (unless, trivially, n==1, in which case the concept of a "column" or "row" distinction would be meaningless).

In numpy an array can have a number of different dimensions, 0, 1, 2 etc.
The typical 2d array has dimension (n,m) (this is a Python tuple). We tend to describe this as having n rows, m columns. So a (n,1) array has just 1 column, and a (1,m) has 1 row.
But because an array may have just 1 dimension, it is possible to have a shape (n,) (Python notation for a 1 element tuple: see here for more).
For many purposes (n,), (1,n), (n,1) arrays are equivalent (also (1,n,1,1) (4d)). They all have n terms, and can be reshaped to each other.
But sometimes that extra 1 dimension matters. A (1,m) array can multiply a (n,1) array to produce a (n,m) array. A (n,1) array can be indexed like a (n,m), with 2 indices, x[:,0] where as a (n,) only accepts x[0].
MATLAB matrices are always 2d (or higher). So people transfering ideas from MATLAB tend to expect 2 dimensions. There is a np.matrix subclass that supposed to imitate that.
For numpy programmers the distinctions between vector, row vector, column vector, matrix are loose and relatively unimportant. Or the use is derived from the application rather than from numpy itself. I think that's what's happening with this network book - the notation and expectations come from outside of numpy.
See as well this answer for how to interpret the shapes with respect to the data stored in ndarrays. It also provides insight on how to use .reshape: https://stackoverflow.com/a/22074424/3277902

(n,) is a tuple of length 1, whose only element is n. (The syntax isn't (n) because that's just n instead of making a tuple.)
If an array has shape (n,), that means it's a 1-dimensional array with a length of n along its only dimension. It's not a row vector or a column vector; it doesn't have rows or columns. It's just a vector.