Difference between numpy.array shape (R, 1) and (R,)

Question

In numpy, some of the operations return in shape (R, 1) but some return (R,). This will make matrix multiplication more tedious since explicit reshape is required. For example, given a matrix M, if we want to do numpy.dot(M[:,0], numpy.ones((1, R))) where R is the number of rows (of course, the same issue also occurs column-wise). We will get matrices are not aligned error since M[:,0] is in shape (R,) but numpy.ones((1, R)) is in shape (1, R).

So my questions are:

1. What's the difference between shape (R, 1) and (R,). I know literally it's list of numbers and list of lists where all list contains only a number. Just wondering why not design numpy so that it favors shape (R, 1) instead of (R,) for easier matrix multiplication.

2. Are there better ways for the above example? Without explicitly reshape like this: numpy.dot(M[:,0].reshape(R, 1), numpy.ones((1, R)))

Genny

This might help. Not with finding a practical solution though. – Jocosity

Proper solution: numpy.ravel( M[ : , 0] ) -- converts shape from (R, 1) to (R,) – Firecrest

A tuple is not determined by the parentheses, they are not part of it, but by the comma. x=4, assigns a tuple, x=(4) assigns an int, creating a usual confusion. Shape n, expresses the shape of a 1D array with n items, and n, 1 the shape of a n-row x 1-column array. (R,) and (R,1) just add (useless) parentheses but still express respectively 1D and 2D array shapes, Parentheses around a tuple force the evaluation order and prevent it to be read as a list of values (e.g. in function calls). This tuple oddity in mind, things get clearer, NumPy returns the shape which makes sense. – Bello

Answer 1

1. The meaning of shapes in NumPy

You write, "I know literally it's list of numbers and list of lists where all list contains only a number" but that's a bit of an unhelpful way to think about it.

The best way to think about NumPy arrays is that they consist of two parts, a data buffer which is just a block of raw elements, and a view which describes how to interpret the data buffer.

For example, if we create an array of 12 integers:

>>> a = numpy.arange(12)
>>> a
array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11])

Then a consists of a data buffer, arranged something like this:

┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
│  0 │  1 │  2 │  3 │  4 │  5 │  6 │  7 │  8 │  9 │ 10 │ 11 │
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

and a view which describes how to interpret the data:

>>> a.flags
  C_CONTIGUOUS : True
  F_CONTIGUOUS : True
  OWNDATA : True
  WRITEABLE : True
  ALIGNED : True
  UPDATEIFCOPY : False
>>> a.dtype
dtype('int64')
>>> a.itemsize
8
>>> a.strides
(8,)
>>> a.shape
(12,)

Here the shape (12,) means the array is indexed by a single index which runs from 0 to 11. Conceptually, if we label this single index i, the array a looks like this:

i= 0    1    2    3    4    5    6    7    8    9   10   11
┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
│  0 │  1 │  2 │  3 │  4 │  5 │  6 │  7 │  8 │  9 │ 10 │ 11 │
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

If we reshape an array, this doesn't change the data buffer. Instead, it creates a new view that describes a different way to interpret the data. So after:

>>> b = a.reshape((3, 4))

the array b has the same data buffer as a, but now it is indexed by two indices which run from 0 to 2 and 0 to 3 respectively. If we label the two indices i and j, the array b looks like this:

i= 0    0    0    0    1    1    1    1    2    2    2    2
j= 0    1    2    3    0    1    2    3    0    1    2    3
┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
│  0 │  1 │  2 │  3 │  4 │  5 │  6 │  7 │  8 │  9 │ 10 │ 11 │
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

which means that:

>>> b[2,1]
9

You can see that the second index changes quickly and the first index changes slowly. If you prefer this to be the other way round, you can specify the order parameter:

>>> c = a.reshape((3, 4), order='F')

which results in an array indexed like this:

i= 0    1    2    0    1    2    0    1    2    0    1    2
j= 0    0    0    1    1    1    2    2    2    3    3    3
┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
│  0 │  1 │  2 │  3 │  4 │  5 │  6 │  7 │  8 │  9 │ 10 │ 11 │
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

which means that:

>>> c[2,1]
5

It should now be clear what it means for an array to have a shape with one or more dimensions of size 1. After:

>>> d = a.reshape((12, 1))

the array d is indexed by two indices, the first of which runs from 0 to 11, and the second index is always 0:

i= 0    1    2    3    4    5    6    7    8    9   10   11
j= 0    0    0    0    0    0    0    0    0    0    0    0
┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
│  0 │  1 │  2 │  3 │  4 │  5 │  6 │  7 │  8 │  9 │ 10 │ 11 │
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

and so:

>>> d[10,0]
10

A dimension of length 1 is "free" (in some sense), so there's nothing stopping you from going to town:

>>> e = a.reshape((1, 2, 1, 6, 1))

giving an array indexed like this:

i= 0    0    0    0    0    0    0    0    0    0    0    0
j= 0    0    0    0    0    0    1    1    1    1    1    1
k= 0    0    0    0    0    0    0    0    0    0    0    0
l= 0    1    2    3    4    5    0    1    2    3    4    5
m= 0    0    0    0    0    0    0    0    0    0    0    0
┌────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┬────┐
│  0 │  1 │  2 │  3 │  4 │  5 │  6 │  7 │  8 │  9 │ 10 │ 11 │
└────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┴────┘

and so:

>>> e[0,1,0,0,0]
6

See the NumPy internals documentation for more details about how arrays are implemented.

2. What to do?

Since numpy.reshape just creates a new view, you shouldn't be scared about using it whenever necessary. It's the right tool to use when you want to index an array in a different way.

However, in a long computation it's usually possible to arrange to construct arrays with the "right" shape in the first place, and so minimize the number of reshapes and transposes. But without seeing the actual context that led to the need for a reshape, it's hard to say what should be changed.

The example in your question is:

numpy.dot(M[:,0], numpy.ones((1, R)))

but this is not realistic. First, this expression:

M[:,0].sum()

computes the result more simply. Second, is there really something special about column 0? Perhaps what you actually need is:

M.sum(axis=0)

Reube

This was extremely helpful in thinking about how arrays are stored. Thank you! Accessing a column (or row) of a (2-d) matrix for further matrix computation is inconvenient though since I always have to reshape the column appropriately. Everytime I have to change the shape from (n,) to (n,1). – Subgroup

@SammyLee: Use newaxis if you need another axis, for example a[:, j, np.newaxis] is the jth column of a, and a[np.newaxis, i] is the ith row. – Reube

i am trying to plot indices to get a better understanding on paper by this model and I dont seem to get it, if I had a shape 2 x 2 x 4 i understand the first 2 can be understood as 0000000011111111 and the last 4 can be understood as 0123012301230123 what happens to the middle one? – Fleurdelis

An easy way to think of this is that numpy is working exactly as expected here, but Python's printing of tuples can be misleading. In the (R, ) case, the shape of the ndarray is a tuple with a single elements, so is printed by Python with a trailing comma. Without the extra comma, it would be ambiguous with an expression in parenthesis. A ndarray with a single dimension can be though of as a column vector of length R. In the (R, 1) case, the tuple has two elements, so can be thought of as a row vector (or a matrix with 1 row of length R. – Whippletree

"If we reshape an array, this doesn't change the data buffer"... How could I check it? I can't see any methods that give me the same sequence 0, 1, 2, ... for both a and b – Refreshment

@Alex-droidAD: See this question and its answers. – Reube

@PirateApp: If i,j,k are the indices, i=00000000,11111111, j=000,111,000,111 and k=0123,0123,0123. Why? Because if you increment j by one unit, you advance by 3 elements on the data buffer. Similarly, if you increment k by 1 unit, you move by 2*3 units on the data buffer. – Lissalissak

Answer 2

The difference between (R,) and (1,R) is literally the number of indices that you need to use. ones((1,R)) is a 2-D array that happens to have only one row. ones(R) is a vector. Generally if it doesn't make sense for the variable to have more than one row/column, you should be using a vector, not a matrix with a singleton dimension.

For your specific case, there are a couple of options:

1) Just make the second argument a vector. The following works fine:

    np.dot(M[:,0], np.ones(R))

2) If you want matlab like matrix operations, use the class matrix instead of ndarray. All matricies are forced into being 2-D arrays, and operator * does matrix multiplication instead of element-wise (so you don't need dot). In my experience, this is more trouble that it is worth, but it may be nice if you are used to matlab.

Enfold

Yes. I was expecting a more matlab-like behavior. I'll take a look at matrix class. What's the trouble for matrix class BTW? – Genny

The problem with matrix is that it is only 2D, and also that because it overloads operator '*', functions written for ndarray may fail if used on a matrix. – Enfold

Answer 3

The shape is a tuple. If there is only 1 dimension the shape will be one number and just blank after a comma. For 2+ dimensions, there will be a number after all the commas.

# 1 dimension with 2 elements, shape = (2,). 
# Note there's nothing after the comma.
z=np.array([  # start dimension
    10,       # not a dimension
    20        # not a dimension
])            # end dimension
print(z.shape)

(2,)

# 2 dimensions, each with 1 element, shape = (2,1)
w=np.array([  # start outer dimension 
    [10],     # element is in an inner dimension
    [20]      # element is in an inner dimension
])            # end outer dimension
print(w.shape)

(2,1)

Word

Classic. So many complicated answers, and then way down I find this which exactly explains it. Thank you! – Carlinecarling

Answer 4

For its base array class, 2d arrays are no more special than 1d or 3d ones. There are some operations that preserve the dimensions, some that reduce them, other combine or even expand them.

M=np.arange(9).reshape(3,3)
M[:,0].shape # (3,) selects one column, returns a 1d array
M[0,:].shape # same, one row, 1d array
M[:,[0]].shape # (3,1), index with a list (or array), returns 2d
M[:,[0,1]].shape # (3,2)

In [20]: np.dot(M[:,0].reshape(3,1),np.ones((1,3)))

Out[20]: 
array([[ 0.,  0.,  0.],
       [ 3.,  3.,  3.],
       [ 6.,  6.,  6.]])

In [21]: np.dot(M[:,[0]],np.ones((1,3)))
Out[21]: 
array([[ 0.,  0.,  0.],
       [ 3.,  3.,  3.],
       [ 6.,  6.,  6.]])

Other expressions that give the same array

np.dot(M[:,0][:,np.newaxis],np.ones((1,3)))
np.dot(np.atleast_2d(M[:,0]).T,np.ones((1,3)))
np.einsum('i,j',M[:,0],np.ones((3)))
M1=M[:,0]; R=np.ones((3)); np.dot(M1[:,None], R[None,:])

MATLAB started out with just 2D arrays. Newer versions allow more dimensions, but retain the lower bound of 2. But you still have to pay attention to the difference between a row matrix and column one, one with shape (1,3) v (3,1). How often have you written [1,2,3].'? I was going to write row vector and column vector, but with that 2d constraint, there aren't any vectors in MATLAB - at least not in the mathematical sense of vector as being 1d.

Have you looked at np.atleast_2d (also _1d and _3d versions)?

In newer Python/numpy there's a matmul operator

In [358]: M[:,0,np.newaxis]@np.ones((1,3))
Out[358]: 
array([[0., 0., 0.],
       [3., 3., 3.],
       [6., 6., 6.]])

In numpy element-wise multiplication is in a sense more basic than matrix multiplication. With the sum-of-products on a size 1 dimension, there's no need to use dot/matmul:

In [360]: M[:,0,np.newaxis]*np.ones((1,3))
Out[360]: 
array([[0., 0., 0.],
       [3., 3., 3.],
       [6., 6., 6.]])

This uses broadcasting, a powerful feature that numpy has had all along. MATLAB only added it recently.

Senghor

Answer 5

There are a lot of good answers here already. But for me it was hard to find some example, where the shape or array can break all the program.

So here is the one:

import numpy as np
a = np.array([1,2,3,4])
b = np.array([10,20,30,40])


from sklearn.linear_model import LinearRegression
regr = LinearRegression()
regr.fit(a,b)

This will fail with error:

ValueError: Expected 2D array, got 1D array instead

but if we add reshape to a:

a = np.array([1,2,3,4]).reshape(-1,1)

this works correctly!

Rembrandt

Also TensorFlow 2.4 see e.g. #67663227 – Minutiae

Answer 6

The data structure of shape (n,) is called a rank 1 array. It doesn't behave consistently as a row vector or a column vector which makes some of its operations and effects non intuitive. If you take the transpose of this (n,) data structure, it'll look exactly same and the dot product will give you a number and not a matrix. The vectors of shape (n,1) or (1,n) row or column vectors are much more intuitive and consistent.

Firecracker

Your intuition has been shaped by linear algebra and/or MATLAB like languages, ones that work primarily with 2d arrays, matrices. In MATLAB everything is 2d, even "scalars". We use Python and numpy for a lot more than dot products:) – Senghor

I agree. Dot product helped me understand the structure better, I have mentioned it for the same reason :) – Firecracker

Answer 7

1) The reason not to prefer a shape of (R, 1) over (R,) is that it unnecessarily complicates things. Besides, why would it be preferable to have shape (R, 1) by default for a length-R vector instead of (1, R)? It's better to keep it simple and be explicit when you require additional dimensions.

2) For your example, you are computing an outer product so you can do this without a reshape call by using np.outer:

np.outer(M[:,0], numpy.ones((1, R)))

Bogard

Thanks for the answer. 1) M[:,0] is essentially getting all rows with first element, so it makes more sense to have (R, 1) than (1, R). 2) It's not always replaceable by np.outer, e.g., dot for matrix in shape (1, R) then (R, 1). – Genny

1) Yes, that could be the convention but that makes it less convenient in other circumstances. The convention could also be for M[1, 1] to return a shape (1, 1) array but that is also usually less convenient than a scalar. If you really want matrix-like behaviour, then you would be better of using a matrix object. 2) Actually, np.outer works regardless of whether the shapes are (1, R), (R, 1), or a combination of the two. – Bogard

Answer 8

To be clear, we are talking about:

• a NumPy array also known as numpy.ndarray
• the shape of an array known by numpy.ndarray.shape
• the question assumes some unknown numpy.ndarray with the shape (R,) where R should be understood as the length of its respective dimension

NumPy arrays have a shape. That .shape is represented by a tuple where each element in the tuple tells us the length of that dimension. To keep it simple, let's stick to rows and columns. While the values of our numpy.ndarray will not change in the following examples, the shape will.

Let's consider an array with the values 1, 2, 3, and 4.

Our examples will include the following .shape representations:

(4,)  # 1-dimensional array with length 4
(1,4) # 2-dimensional array with row length 1, column length 4
(4,1) # 2-dimensional array with row length 4, column length 1

We can think of this more abstractly with variables a and b.

(a,)  # 1-dimensional array with length a
(b,a) # 2-dimensional array with row length b, column length a
(a,b) # 2-dimensional array with row length a, column length b

For me, it is helpful to 'manually' build these out to get a better feel for what their dimensions mean.

>> # (4,)
>> one_dimensional_vector = np.array(
    [1, 2, 3, 4]
)

>> # (1,4)
>> row_vector = np.array(
    [
        [1, 2, 3, 4]
    ]
)

>> # (4,1)
>> column_vector = np.array(
    [
        [1], 
        [2], 
        [3], 
        [4]
    ]
)

So, the answer to the first question:

1. What's the difference between shape (R, 1) and (R,)?

Answer: They have different dimensions. a is the length of the one dimension and b the length of another, .shape is (a, b) and (a,) respectively. b just happens to be 1. One way to think of this is if a = 1 then the row has length 1 thus it is a row vector. If b = 1 then the column has length 1 so the numpy.ndarray it represents is a column vector.

2. Are there better ways for the above example?

Answer: Let's assume we have the array I used as example above with 1, 2, 3, and 4 as values. A convenient way to get (R,) to be (R, 1) is this:

>> one_dimensional_array = np.array([1,2,3,4])
>> one_dimensional_array.shape
(4,)
>> row_vector = one_dimensional_array[:, None]
>> row_vector.shape
(4, 1)

Resourses

1. NumPy — ndarrays — https://numpy.org/doc/stable/reference/arrays.ndarray.html
2. Cross Validated @unutbu — dimension trick — https://stats.stackexchange.com/a/285005

Amorino