lower_bounds = torch.max(set_1[:, :2].unsqueeze(1),
set_2[:, :2].unsqueeze(0)) #(n1, n2, 2)
This code snippet uses unsqueeze(1) for one tensor, but unsqeeze(0) for another. What is the difference between them?
unsqueeze turns an n-dimensionsal tensor into an n+1-dimensional one, by adding an extra dimension of zero depth. However, since it is ambiguous which axis the new dimension should lie across (i.e. in which direction it should be "unsqueezed"), this needs to be specified by the dim argument.
Hence the resulting unsqueezed tensors have the same information, but the indices used to access them are different.
Here is a visual representation of what squeeze/unsqueeze do for an effectively 2d matrix, where it is going from a 2d tensor to a 3d one, and hence there are 3 choices for the new dimension's position:
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.
My goal is to to turn a row vector into a column vector and vice versa. The documentation for numpy.ndarray.transpose says:
For a 1-D array, this has no effect. (To change between column and row vectors, first cast the 1-D array into a matrix object.)
However, when I try this:
my_array = np.array([1,2,3])
my_array_T = np.transpose(np.matrix(myArray))
I do get the wanted result, albeit in matrix form (matrix([[66],[640],[44]])), but I also get this warning:
PendingDeprecationWarning: the matrix subclass is not the recommended way to represent matrices or deal with linear algebra (see https://docs.scipy.org/doc/numpy/user/numpy-for-matlab-users.html). Please adjust your code to use regular ndarray.
my_array_T = np.transpose(np.matrix(my_array))
How can I properly transpose an ndarray then?
A 1D array is itself once transposed, contrary to Matlab where a 1D array doesn't exist and is at least 2D.
What you want is to reshape it:
my_array.reshape(-1, 1)
Or:
my_array.reshape(1, -1)
Depending on what kind of vector you want (column or row vector).
The -1 is a broadcast-like, using all possible elements, and the 1 creates the second required dimension.
If your array is my_array and you want to convert it to a column vector you can do:
my_array.reshape(-1, 1)
For a row vector you can use
my_array.reshape(1, -1)
Both of these can also be transposed and that would work as expected.
IIUC, use reshape
my_array.reshape(my_array.size, -1)
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.
I'm having a trouble understanding the concept of Axis elimination in numpy. Suppose I have the following 2D matrix:
A =
1 2 3
3 4 5
6 7 8
Ok I understand that sum(A, axis=0) will sum each column down and will give a 1D array with 3 elements. I also understand that sum(A, axis=1) will sum each row.
But my trouble is when I read that axis=0 eliminates the 0th axis and axis=1 eliminates the 1th axis. Also sometime people mention "reduce" instead of "eliminate". I'm unable to understand what does that eliminate. For example sum(A, axis=0) will sum each column from top to bottom, but I don't see elimination or reduction here. What's the point? The same also for sum(A,axis=1).
AND how is it for higher dimensions?
p.s. I always confused between matrix dimensions and array dimensions. I wished that people who write the numpy documentation makes this distinction very clear.
http://docs.scipy.org/doc/numpy/reference/generated/numpy.ufunc.reduce.html
Reduces a‘s dimension by one, by applying ufunc along one axis.
For example, add.reduce() is equivalent to sum().
In numpy, the base class is ndarray - a multidimensional array (can 0d, 1d, or more)
http://docs.scipy.org/doc/numpy/reference/arrays.ndarray.html
Matrix is a subclass of array
http://docs.scipy.org/doc/numpy/reference/arrays.classes.html
Matrix objects are always two-dimensional
The history of the numpy Matrix is old, but basically it's meant to resemble the MATLAB matrix object. In the original MATLAB nearly everything was a matrix, which was always 2d. Later they generalized it to allow more dimensions. But it can't have fewer dimensions. MATLAB does have 'vectors', but they are just matrices with one dimension being 1 (row vector versus column vector).
'axis elimination' is not a common term when working with numpy. It could, conceivably, refer to any of several ways that reduce the number of dimensions of an array. Reduction, as in sum(), is one. Indexing is another: a[:,0,:]. Reshaping can also change the number of dimensions. np.squeeze is another.