I'm new to numpy & have a question about it :
according to docs.scipy.org, the "shape" method is "the dimensions of the array. For a matrix with n rows and m columns, shape will be (n,m)"
Suppose I am to create a simple array as below:
np.array([[0,2,4],[1,3,5]])
Using the "shape" method, it returns (2,3) (i.e. the array has 2 rows & 3 columns)
However, for an array ([0,2,4]), the shape method would return (3,) (which means it has 3 rows according to the definition above)
I'm confused : the array ([0,2,4]) should have 3 columns not 3 rows so I expect it to return (,3) instead.
Can anyone help to clarify ? Thanks a lot.
This is just notation - in Python, tuples are distinguished from expression grouping (or order of operations stuff) by the use of commas - that is, (1,2,3) is a tuple and (2x + 4) ** 5 contains an expression 2x + 4. In order to keep single-element tuples distinct from single-element expressions, which would otherwise be ambiguous ((1) vs (1) - which is the single-element tuple and which a simple expression that evaluates to 1?), we use a trailing comma to denote tuple-ness.
What you're getting is a single dimension response, since there's only one dimension to measure, packed into a tuple type.
Numpy supports not only 2-dimensional arrays, but multi-dimensional arrays, and by multi-dimension I mean 1-D, 2-D, 3-D .... n-D, And there is a format for representing respective dimension array. The len of array.shape would get you the number of dimensions of that array. If the array is 1-D, the there is no need to represent as (m, n) or if the array is 3-D then it (m, n) would not be sufficient to represent its dimensions.
So the output of array.shape would not always be in (m, n) format, it would depend upon the array itself and you will get different outputs for different dimensions.
Related
I'm having some trouble understanding the rules for array broadcasting in Numpy.
Obviously, if you perform element-wise multiplication on two arrays of the same dimensions and shape, everything is fine. Also, if you multiply a multi-dimensional array by a scalar it works. This I understand.
But if you have two N-dimensional arrays of different shapes, it's unclear to me exactly what the broadcasting rules are. This documentation/tutorial explains that: In order to broadcast, the size of the trailing axes for both arrays in an operation must either be the same size or one of them must be one.
Okay, so I assume by trailing axis they are referring to the N in a M x N array. So, that means if I attempt to multiply two 2D arrays (matrices) with equal number of columns, it should work? Except it doesn't...
>>> from numpy import *
>>> A = array([[1,2],[3,4]])
>>> B = array([[2,3],[4,6],[6,9],[8,12]])
>>> print(A)
[[1 2]
[3 4]]
>>> print(B)
[[ 2 3]
[ 4 6]
[ 6 9]
[ 8 12]]
>>>
>>> A * B
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: shape mismatch: objects cannot be broadcast to a single shape
Since both A and B have two columns, I would have thought this would work. So, I'm probably misunderstanding something here about the term "trailing axis", and how it applies to N-dimensional arrays.
Can someone explain why my example doesn't work, and what is meant by "trailing axis"?
Well, the meaning of trailing axes is explained on the linked documentation page.
If you have two arrays with different dimensions number, say one 1x2x3 and other 2x3, then you compare only the trailing common dimensions, in this case 2x3. But if both your arrays are two-dimensional, then their corresponding sizes have to be either equal or one of them has to be 1. Dimensions along which the array has size 1 are called singular, and the array can be broadcasted along them.
In your case you have a 2x2 and 4x2 and 4 != 2 and neither 4 or 2 equals 1, so this doesn't work.
From http://cs231n.github.io/python-numpy-tutorial/#numpy-broadcasting:
Broadcasting two arrays together follows these rules:
If the arrays do not have the same rank, prepend the shape of the lower rank array with 1s until both shapes have the same length.
The two arrays are said to be compatible in a dimension if they have the same size in the dimension, or if one of the arrays has size 1 in that dimension.
The arrays can be broadcast together if they are compatible in all dimensions.
After broadcasting, each array behaves as if it had shape equal to the elementwise maximum of shapes of the two input arrays.
In any dimension where one array had size 1 and the other array had size greater than 1, the first array behaves as if it were copied along that dimension
If this explanation does not make sense, try reading the explanation from the documentation or this explanation.
we should consider two points about broadcasting. first: what is possible. second: how much of the possible things is done by numpy.
I know it might look a bit confusing, but I will make it clear by some example.
lets start from the zero level.
suppose we have two matrices. first matrix has three dimensions (named A) and the second has five (named B). numpy tries to match last/trailing dimensions. so numpy does not care about the first two dimensions of B. then numpy compares those trailing dimensions with each other. and if and only if they be equal or one of them be 1, numpy says "O.K. you two match". and if it these conditions don't satisfy, numpy would "sorry...its not my job!".
But I know that you may say comparison was better to be done in way that can handle when they are devisable(4 and 2 / 9 and 3). you might say it could be replicated/broadcasted by a whole number(2/3 in out example). and i am agree with you. and this is the reason I started my discussion with a distinction between what is possible and what is the capability of numpy.
I have two numpy arrays, one bigger than another, but both with the same number of dimensions.
I want to get a slice from the bigger array that matches the size of the smaller array. (Starting from 0,0,0....)
So, imagine the big array has shape (10,5,7).
And the small array has shape (10,4,6).
I want to get from the bigger array this slice:
biggerArray[:10,:4,:6]
The length of the shape tuple may vary, and I want to do it for any number of dimensions (Both will always have the same number of dimensions).
How to do that? Is there a way to use tuples as ranges in slices?
Construct the tuple of slice objects manually. biggerArray[:10, :4, :6] is syntactic sugar for biggerArray[(slice(10), slice(4), slice(6))], so:
biggerArray[tuple(map(slice, smallerArray.shape))]
or
biggerArray[tuple(slice(0, n) for n in smallerArray.shape)]
You may want to assert result.shape == smallerArray.shape afterwards, just in case the input shapes weren't what you thought they were.
I have an array of 3 dimensional vectors. The dimension of the array is arbitrary: it could be a single (N×3), double (M×N×3), triple (K×M×N×3) etc. I need to operate on two components of the vector while preserving the other dimensions.
For example, if I know it is three dimensionsional, I could do the following:
R = numpy.arctan2(A[:,:,:,1], A[:,:,:,0])
which gives me a three dimensional array of scalar values.
Now, to be able to do this on arbitrary number of dimensions. I need to slice over all other dimensions except the the last. So far, I'm able to do it with this:
s = [numpy.s_[:]] * (len(A.shape)-1)
R = numpy.arctan2(A[s+[1]], A[s+[0]])
which works even for single vectors. Is there a more numpythonic way of achieving the above?
I found an even nicer way. This here works for me
R = numpy.arctan2(A[...,1],A[...,0])
I'm messing around with 2-dimensional slicing and don't understand why leaving out some defaults grabs the same values from the original array but produces different output. What's going on with the double brackets and shape changing?
x = np.arange(9).reshape(3,3)
y = x[2]
z = x[2:,:]
print y
print z
print shape(y)
print shape(z)
[6 7 8]
[[6 7 8]]
(3L,)
(1L, 3L)
x is a two dimensional array, an instance of NumPy's ndarray object. You can index/slice these objects in essentially two ways: basic and advanced.
y[2] fetches the row at index 2 of the array, returning the array [6 7 8]. You're doing basic slicing because you've specified only an integer. You can also specify a tuple of slice objects and integers for basic slicing, e.g. x[:,2] to select the right-hand column.
With basic slicing, you're also reducing the number of dimensions of the returned object (in this case from two to just one):
An integer, i, returns the same values as i:i+1 except the dimensionality of the returned object is reduced by 1.
So when you ask for the shape of y, this is why you only get back one dimension (from your two-dimensional x).
Advanced slicing occurs when you specify an ndarray: or a tuple with at least one sequence object or ndarray. This is the case with x[2:,:] since 2: counts as a sequence object.
You get back an ndarray. When you ask for its shape, you will get back all of the dimensions (in this case two):
The shape of the output (or the needed shape of the object to be used for setting) is the broadcasted shape.
In a nutshell, as soon as you start slicing along any dimension of your array with :, you're doing advanced slicing and not basic slicing.
One brief point worth mentioning: basic slicing returns a view onto the original array (changes made to y will be reflected in x). Advanced slicing returns a brand new copy of the array.
You can read about array indexing and slicing in much more detail here.
I am using a package and it is returning me an array. When I print the shape it is (38845,). Just wondering why this ','.
I am wondering how to interpret this.
Thanks.
Python has tuples, which are like lists but of fixed size. A two-element tuple is (a, b); a three-element one is (a, b, c). However, (a) is just a in parentheses. To represent a one-element tuple, Python uses a slightly odd syntax of (a,). So there is only one dimension, and you have a bunch of elements in that one dimension.
It sounds like you're using Numpy. If so, the shape (38845,) means you have a 1-dimensional array, of size 38845.
It seems you're talking of a Numpy array.
shape returns a tuple with the same size as the number of dimensions of the array. Each value of the tuple is the size of the array along the corresponding dimensions, or, as the tutorial says:
An array has a shape given by the number of elements along each axis.
Here you have a 1D-array (as indicated with a 1-element tuple notation, with the coma (as #Amadan) said), and the size of the 1st (and only dimension) is 38845.
For example (3,4) would be a 2D-array of size 3 for the 1st dimension and 4 for the second.
You can check the documentation for shape here: http://docs.scipy.org/doc/numpy/reference/generated/numpy.ndarray.shape.html
Just wondering why this ','.
Because (38845) is the same thing as 38845, but a tuple is expected here, not an int (since in general, your array could have multiple dimensions). (38845,) is a 1-tuple.