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)
Related
I've started to learn NumPy, when I create an array and then invoke the .shape function, I understand how it works for most cases. However, the result does not make sense to me for a single-dimensional array. Can someone please explain the outcome?
array = np.array([4,5,6])
print(array.shape)
The outcome is (3,)
Output tulpe of ints in the "np.shape" function gives the lengths of the corresponding array dimension, and this tuple will be (n,m), in which n and m indicate row and columns, respectively. For a single dimension array, this Tuple will be just (n,), in which n indicates the number of array elements.
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.
I have a numpy boolean vector of shape 1 x N, and an 2d array with shape 160 x N. What is a fast way of subsetting the columns of the 2d array such that for each index of the boolean vector that has True in it, the column is kept, and for each index of the boolean vector that has False in it, the column is discarded?
If you call the vector mask and the array features, i've found the following to be far too slow: np.array([f[mask] for f in features])
Is there a better way? I feel like there has to be, right?
You can try this,
new_array = 2d_array[:,bool_array==True]
So depending on the axes you can select which one you want to remove. In case you get a 1-d array, then you can just reshape it and get the required array. This method will be faster also.
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.
Consider the following simple example:
X = numpy.zeros([10, 4]) # 2D array
x = numpy.arange(0,10) # 1D array
X[:,0] = x # WORKS
X[:,0:1] = x # returns ERROR:
# ValueError: could not broadcast input array from shape (10) into shape (10,1)
X[:,0:1] = (x.reshape(-1, 1)) # WORKS
Can someone explain why numpy has vectors of shape (N,) rather than (N,1) ?
What is the best way to do the casting from 1D array into 2D array?
Why do I need this?
Because I have a code which inserts result x into a 2D array X and the size of x changes from time to time so I have X[:, idx1:idx2] = x which works if x is 2D too but not if x is 1D.
Do you really need to be able to handle both 1D and 2D inputs with the same function? If you know the input is going to be 1D, use
X[:, i] = x
If you know the input is going to be 2D, use
X[:, start:end] = x
If you don't know the input dimensions, I recommend switching between one line or the other with an if, though there might be some indexing trick I'm not aware of that would handle both identically.
Your x has shape (N,) rather than shape (N, 1) (or (1, N)) because numpy isn't built for just matrix math. ndarrays are n-dimensional; they support efficient, consistent vectorized operations for any non-negative number of dimensions (including 0). While this may occasionally make matrix operations a bit less concise (especially in the case of dot for matrix multiplication), it produces more generally applicable code for when your data is naturally 1-dimensional or 3-, 4-, or n-dimensional.
I think you have the answer already included in your question. Numpy allows the arrays be of any dimensionality (while afaik Matlab prefers two dimensions where possible), so you need to be correct with this (and always distinguish between (n,) and (n,1)). By giving one number as one of the indices (like 0 in 3rd row), you reduce the dimensionality by one. By giving a range as one of the indices (like 0:1 in 4th row), you don't reduce the dimensionality.
Line 3 makes perfect sense for me and I would assign to the 2-D array this way.
Here are two tricks that make the code a little shorter.
X = numpy.zeros([10, 4]) # 2D array
x = numpy.arange(0,10) # 1D array
X.T[:1, :] = x
X[:, 2:3] = x[:, None]