Does np.dot automatically transpose vectors?

Question

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

Answer 1

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.

Answer 2

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:

Answer 3

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.

Answer 4

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.