I found NumPy: The tricks of the trade (Part II) instructive

We use -> to indicate the order of the output array. So think of ‘ij, i->j’ as having left hand side (LHS) and right hand side (RHS). Any repetition of labels on the LHS computes the product element wise and then sums over. By changing the label on the RHS (output) side, we can define the axis in which we want to proceed with respect to the input array, i.e. summation along axis 0, 1 and so on.

import numpy as np

>>> a
array([[1, 1, 1],
       [2, 2, 2],
       [3, 3, 3]])
>>> b
array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])
>>> d = np.einsum('ij, jk->ki', a, b)

Notice there are three axes, i, j, k, and that j is repeated (on the left-hand-side). i,j represent rows and columns for a. j,k for b.

In order to calculate the product and align the j axis we need to add an axis to a. (b will be broadcast along(?) the first axis)

a[i, j, k]
   b[j, k]

>>> c = a[:,:,np.newaxis] * b
>>> c
array([[[ 0,  1,  2],
        [ 3,  4,  5],
        [ 6,  7,  8]],

       [[ 0,  2,  4],
        [ 6,  8, 10],
        [12, 14, 16]],

       [[ 0,  3,  6],
        [ 9, 12, 15],
        [18, 21, 24]]])

j is absent from the right-hand-side so we sum over j which is the second axis of the 3x3x3 array

>>> c = c.sum(1)
>>> c
array([[ 9, 12, 15],
       [18, 24, 30],
       [27, 36, 45]])

Finally, the indices are (alphabetically) reversed on the right-hand-side so we transpose.

>>> c.T
array([[ 9, 18, 27],
       [12, 24, 36],
       [15, 30, 45]])

>>> np.einsum('ij, jk->ki', a, b)
array([[ 9, 18, 27],
       [12, 24, 36],
       [15, 30, 45]])
>>>

Lets make 2 arrays, with different, but compatible dimensions to highlight their interplay

In [43]: A=np.arange(6).reshape(2,3)
Out[43]: 
array([[0, 1, 2],
       [3, 4, 5]])


In [44]: B=np.arange(12).reshape(3,4)
Out[44]: 
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])

Your calculation, takes a ‘dot’ (sum of products) of a (2,3) with a (3,4) to produce a (4,2) array. i is the 1st dim of A, the last of C; k the last of B, 1st of C. j is ‘consumed’ by the summation.

In [45]: C=np.einsum('ij,jk->ki',A,B)
Out[45]: 
array([[20, 56],
       [23, 68],
       [26, 80],
       [29, 92]])

This is the same as np.dot(A,B).T – it’s the final output that’s transposed.

To see more of what happens to j, change the C subscripts to ijk:

In [46]: np.einsum('ij,jk->ijk',A,B)
Out[46]: 
array([[[ 0,  0,  0,  0],
        [ 4,  5,  6,  7],
        [16, 18, 20, 22]],

       [[ 0,  3,  6,  9],
        [16, 20, 24, 28],
        [40, 45, 50, 55]]])

This can also be produced with:

A[:,:,None]*B[None,:,:]

That is, add a k dimension to the end of A, and an i to the front of B, resulting in a (2,3,4) array.

0 + 4 + 16 = 20, 9 + 28 + 55 = 92, etc; Sum on j and transpose to get the earlier result:

np.sum(A[:,:,None] * B[None,:,:], axis=1).T

# C[k,i] = sum(j) A[i,j (,k) ] * B[(i,)  j,k]

Grasping the idea of numpy.einsum() is very easy if you understand it intuitively. As an example, let’s start with a simple description involving matrix multiplication.

---

To use numpy.einsum(), all you have to do is to pass the so-called subscripts string as an argument, followed by your input arrays.

Let’s say you have two 2D arrays, A and B, and you want to do matrix multiplication. So, you do:

np.einsum("ij, jk -> ik", A, B)

Here the subscript string ij corresponds to array A while the subscript string jk corresponds to array B. Also, the most important thing to note here is that the number of characters in each subscript string must match the dimensions of the array (i.e., two chars for 2D arrays, three chars for 3D arrays, and so on). And if you repeat the chars between subscript strings (j in our case), then that means you want the einsum to happen along those dimensions. Thus, they will be sum-reduced (i.e., that dimension will be gone).

The subscript string after this -> symbol represent the dimensions of our resultant array.
If you leave it empty, then everything will be summed and a scalar value is returned as the result. Else the resultant array will have dimensions according to the subscript string. In our example, it’ll be ik. This is intuitive because we know that for the matrix multiplication to work, the number of columns in array A has to match the number of rows in array B which is what is happening here (i.e., we encode this knowledge by repeating the char j in the subscript string)

---

Here are some more examples illustrating the use/power of np.einsum() in implementing some common tensor or nd-array operations, succinctly.

Inputs

# a vector
In [197]: vec
Out[197]: array([0, 1, 2, 3])

# an array
In [198]: A
Out[198]: 
array([[11, 12, 13, 14],
       [21, 22, 23, 24],
       [31, 32, 33, 34],
       [41, 42, 43, 44]])

# another array
In [199]: B
Out[199]: 
array([[1, 1, 1, 1],
       [2, 2, 2, 2],
       [3, 3, 3, 3],
       [4, 4, 4, 4]])

1) Matrix multiplication (similar to np.matmul(arr1, arr2))

In [200]: np.einsum("ij, jk -> ik", A, B)
Out[200]: 
array([[130, 130, 130, 130],
       [230, 230, 230, 230],
       [330, 330, 330, 330],
       [430, 430, 430, 430]])

2) Extract elements along the main-diagonal (similar to np.diag(arr))

In [202]: np.einsum("ii -> i", A)
Out[202]: array([11, 22, 33, 44])

3) Hadamard product (i.e. element-wise product of two arrays) (similar to arr1 * arr2)

In [203]: np.einsum("ij, ij -> ij", A, B)
Out[203]: 
array([[ 11,  12,  13,  14],
       [ 42,  44,  46,  48],
       [ 93,  96,  99, 102],
       [164, 168, 172, 176]])

4) Element-wise squaring (similar to np.square(arr) or arr ** 2)

In [210]: np.einsum("ij, ij -> ij", B, B)
Out[210]: 
array([[ 1,  1,  1,  1],
       [ 4,  4,  4,  4],
       [ 9,  9,  9,  9],
       [16, 16, 16, 16]])

5) Trace (i.e. sum of main-diagonal elements) (similar to np.trace(arr))

In [217]: np.einsum("ii -> ", A)
Out[217]: 110

6) Matrix transpose (similar to np.transpose(arr))

In [221]: np.einsum("ij -> ji", A)
Out[221]: 
array([[11, 21, 31, 41],
       [12, 22, 32, 42],
       [13, 23, 33, 43],
       [14, 24, 34, 44]])

7) Outer Product (of vectors) (similar to np.outer(vec1, vec2))

In [255]: np.einsum("i, j -> ij", vec, vec)
Out[255]: 
array([[0, 0, 0, 0],
       [0, 1, 2, 3],
       [0, 2, 4, 6],
       [0, 3, 6, 9]])

8) Inner Product (of vectors) (similar to np.inner(vec1, vec2))

In [256]: np.einsum("i, i -> ", vec, vec)
Out[256]: 14

9) Sum along axis 0 (similar to np.sum(arr, axis=0))

In [260]: np.einsum("ij -> j", B)
Out[260]: array([10, 10, 10, 10])

10) Sum along axis 1 (similar to np.sum(arr, axis=1))

In [261]: np.einsum("ij -> i", B)
Out[261]: array([ 4,  8, 12, 16])

11) Batch Matrix Multiplication

In [287]: BM = np.stack((A, B), axis=0)

In [288]: BM
Out[288]: 
array([[[11, 12, 13, 14],
        [21, 22, 23, 24],
        [31, 32, 33, 34],
        [41, 42, 43, 44]],

       [[ 1,  1,  1,  1],
        [ 2,  2,  2,  2],
        [ 3,  3,  3,  3],
        [ 4,  4,  4,  4]]])

In [289]: BM.shape
Out[289]: (2, 4, 4)

# batch matrix multiply using einsum
In [292]: BMM = np.einsum("bij, bjk -> bik", BM, BM)

In [293]: BMM
Out[293]: 
array([[[1350, 1400, 1450, 1500],
        [2390, 2480, 2570, 2660],
        [3430, 3560, 3690, 3820],
        [4470, 4640, 4810, 4980]],

       [[  10,   10,   10,   10],
        [  20,   20,   20,   20],
        [  30,   30,   30,   30],
        [  40,   40,   40,   40]]])

In [294]: BMM.shape
Out[294]: (2, 4, 4)

---

12) Sum along axis 2 (similar to np.sum(arr, axis=2))

In [330]: np.einsum("ijk -> ij", BM)
Out[330]: 
array([[ 50,  90, 130, 170],
       [  4,   8,  12,  16]])

13) Sum all the elements in array (similar to np.sum(arr))

In [335]: np.einsum("ijk -> ", BM)
Out[335]: 480

14) Sum over multiple axes (i.e. marginalization)
(similar to np.sum(arr, axis=(axis0, axis1, axis2, axis3, axis4, axis6, axis7)))

# 8D array
In [354]: R = np.random.standard_normal((3,5,4,6,8,2,7,9))

# marginalize out axis 5 (i.e. "n" here)
In [363]: esum = np.einsum("ijklmnop -> n", R)

# marginalize out axis 5 (i.e. sum over rest of the axes)
In [364]: nsum = np.sum(R, axis=(0,1,2,3,4,6,7))

In [365]: np.allclose(esum, nsum)
Out[365]: True

15) Double Dot Products (similar to np.sum(hadamard-product) cf. 3)

In [772]: A
Out[772]: 
array([[1, 2, 3],
       [4, 2, 2],
       [2, 3, 4]])

In [773]: B
Out[773]: 
array([[1, 4, 7],
       [2, 5, 8],
       [3, 6, 9]])

In [774]: np.einsum("ij, ij -> ", A, B)
Out[774]: 124

16) 2D and 3D array multiplication

Such a multiplication could be very useful when solving linear system of equations (Ax = b) where you want to verify the result.

# inputs
In [115]: A = np.random.rand(3,3)
In [116]: b = np.random.rand(3, 4, 5)

# solve for x
In [117]: x = np.linalg.solve(A, b.reshape(b.shape[0], -1)).reshape(b.shape)

# 2D and 3D array multiplication :)
In [118]: Ax = np.einsum('ij, jkl', A, x)

# indeed the same!
In [119]: np.allclose(Ax, b)
Out[119]: True

On the contrary, if one has to use np.matmul() for this verification, we have to do couple of reshape operations to achieve the same result like:

# reshape 3D array `x` to 2D, perform matmul
# then reshape the resultant array to 3D
In [123]: Ax_matmul = np.matmul(A, x.reshape(x.shape[0], -1)).reshape(x.shape)

# indeed correct!
In [124]: np.allclose(Ax, Ax_matmul)
Out[124]: True

Bonus: Read more math here : Einstein-Summation and definitely here: Tensor-Notation

When reading einsum equations, I’ve found it the most helpful to just be able to
mentally boil them down to their imperative versions.

Let’s start with the following (imposing) statement:

C = np.einsum('bhwi,bhwj->bij', A, B)

Working through the punctuation first we see that we have two 4-letter comma-separated blobs – bhwi and bhwj, before the arrow,
and a single 3-letter blob bij after it. Therefore, the equation produces a rank-3 tensor result from two rank-4 tensor inputs.

Now, let each letter in each blob be the name of a range variable. The position at which the letter appears in the blob
is the index of the axis that it ranges over in that tensor.
The imperative summation that produces each element of C, therefore, has to start with three nested for loops, one for each index of C.

for b in range(...):
    for i in range(...):
        for j in range(...):
            # the variables b, i and j index C in the order of their appearance in the equation
            C[b, i, j] = ...

So, essentially, you have a for loop for every output index of C. We’ll leave the ranges undetermined for now.

Next we look at the left-hand side – are there any range variables there that don’t appear on the right-hand side? In our case – yes, h and w.
Add an inner nested for loop for every such variable:

for b in range(...):
    for i in range(...):
        for j in range(...):
            C[b, i, j] = 0
            for h in range(...):
                for w in range(...):
                    ...

Inside the innermost loop we now have all indices defined, so we can write the actual summation and
the translation is complete:

# three nested for-loops that index the elements of C
for b in range(...):
    for i in range(...):
        for j in range(...):

            # prepare to sum
            C[b, i, j] = 0

            # two nested for-loops for the two indexes that don't appear on the right-hand side
            for h in range(...):
                for w in range(...):
                    # Sum! Compare the statement below with the original einsum formula
                    # 'bhwi,bhwj->bij'

                    C[b, i, j] += A[b, h, w, i] * B[b, h, w, j]

If you’ve been able to follow the code thus far, then congratulations! This is all you need to be able to read einsum equations. Notice in particular how the original einsum formula maps to the final summation statement in the snippet above. The for-loops and range bounds are just fluff and that final statement is all you really need to understand what’s going on.

For the sake of completeness, let’s see how to determine the ranges for each range variable. Well, the range of each variable is simply the length of the dimension(s) which it indexes.
Obviously, if a variable indexes more than one dimension in one or more tensors, then the lengths of each of those dimensions have to be equal.
Here’s the code above with the complete ranges:

# C's shape is determined by the shapes of the inputs
# b indexes both A and B, so its range can come from either A.shape or B.shape
# i indexes only A, so its range can only come from A.shape, the same is true for j and B
assert A.shape[0] == B.shape[0]
assert A.shape[1] == B.shape[1]
assert A.shape[2] == B.shape[2]
C = np.zeros((A.shape[0], A.shape[3], B.shape[3]))
for b in range(A.shape[0]): # b indexes both A and B, or B.shape[0], which must be the same
    for i in range(A.shape[3]):
        for j in range(B.shape[3]):
            # h and w can come from either A or B
            for h in range(A.shape[1]):
                for w in range(A.shape[2]):
                    C[b, i, j] += A[b, h, w, i] * B[b, h, w, j]

I think the simplest example is in tensorflow docs

There are four steps to convert your equation to einsum notation. Lets take this equation as an example C[i,k] = sum_j A[i,j] * B[j,k]

1. First we drop the variable names. We get ik = sum_j ij * jk
2. We drop the sum_j term as it is implicit. We get ik = ij * jk
3. We replace * with ,. We get ik = ij, jk
4. The output is on the RHS and is separated with -> sign. We get ij, jk -> ik

The einsum interpreter just runs these 4 steps in reverse. All indices missing in the result are summed over.

Here are some more examples from the docs

# Matrix multiplication
einsum('ij,jk->ik', m0, m1)  # output[i,k] = sum_j m0[i,j] * m1[j, k]

# Dot product
einsum('i,i->', u, v)  # output = sum_i u[i]*v[i]

# Outer product
einsum('i,j->ij', u, v)  # output[i,j] = u[i]*v[j]

# Transpose
einsum('ij->ji', m)  # output[j,i] = m[i,j]

# Trace
einsum('ii', m)  # output[j,i] = trace(m) = sum_i m[i, i]

# Batch matrix multiplication
einsum('aij,ajk->aik', s, t)  # out[a,i,k] = sum_j s[a,i,j] * t[a, j, k]

Another view on np.einsum

Most answers here explain by example, I thought I’d give an additional point of view.

You can see einsum as a generalized matrix summation operator. The string given contains the subscripts which are labels representing axes. I like to think of it as your operation definition. The subscripts provide two apparent constraints:

1. the number of axes for each input array,

2. axis size equality between inputs.

Let’s take the initial example: np.einsum('ij,jk->ki', A, B). Here the constraints 1. translates to A.ndim == 2 and B.ndim == 2, and 2. to A.shape[1] == B.shape[0].

As you will see later down, there are other constraints. For instance:

3. labels in the output subscript must not appear more than once.

4. labels in the output subscript must appear in the input subscripts.

When looking at ij,jk->ki, you can think of it as:

which components from the input arrays will contribute to component [k, i] of the output array.

The subscripts contain the exact definition of the operation for each component of the output array.

We will stick with operation ij,jk->ki, and the following definitions of A and B:

>>> A = np.array([[1,4,1,7], [8,1,2,2], [7,4,3,4]])
>>> A.shape
(3, 4)

>>> B = np.array([[2,5], [0,1], [5,7], [9,2]])
>>> B.shape
(4, 2)

The output, Z, will have a shape of (B.shape[1], A.shape[0]) and could naively be constructed in the following way. Starting with a blank array for Z:

Z = np.zeros((B.shape[1], A.shape[0]))
for i in range(A.shape[0]):
    for j in range(A.shape[1]):
        for k range(B.shape[0]):
           Z[k, i] += A[i, j]*B[j, k] # ki <- ij*jk

np.einsum is about accumulating contributions in the output array. Each (A[i,j], B[j,k]) pair is seen contributing to each Z[k, i] component.

You might have noticed, it looks extremely similar to how you would go about computing general matrix multiplications…

---

Minimal implementation

Here is a minimal implementation of np.einsum in Python. This should help understand what is really going on under the hood.

As we go along I will keep referring to the previous example. Defining inputs as [A, B].

np.einsum can actually take more than two inputs. In the following, we will focus on the general case: n inputs and n input subscripts. The main goal is to find the domain of iteration, i.e. the cartesian product of all our ranges.

We can’t rely on manually writing for loops, simply because we don’t know how many there will be. The main idea is this: we need to find all unique labels (I will use key and keys to refer to them), find the corresponding array shape, then create ranges for each one, and compute the product of the ranges using itertools.product to get the domain of study.

The domain of study is the cartesian product: range(0, 2) x range(0, 3) x range(0, 4).

1. Subscripts processing:

   >>> expr = 'ij,jk->ki'
   >>> qry_expr, res_expr = expr.split('->')
   >>> inputs_expr = qry_expr.split(',')
   >>> inputs_expr, res_expr
   (['ij', 'jk'], 'ki')
   

2. Find the unique keys (labels) in the input subscripts:

   >>> keys = set([(key, size) for keys, input in zip(inputs_expr, inputs) 
                  for key, size in list(zip(keys, input.shape))])
   {('i', 3), ('j', 4), ('k', 2)}
   

   We should be checking for constraints (as well as in the output subscript)! Using set is a bad idea but it will work for the purpose of this example.

3. Get the associated sizes (used to initialize the output array) and construct the ranges (used to create our domain of iteration):

   >>> sizes = dict(keys)
   {'i': 3, 'j': 4, 'k': 2}
   
   >>> ranges = [range(size) for _, size in keys]
   [range(0, 2), range(0, 3), range(0, 4)]
   

4. We need an list containing the keys (labels):

   >>> to_key = sizes.keys()
   ['k', 'i', 'j']
   

5. Compute the cartesian product of the ranges

   >>> domain = product(*ranges)
   

   Note: [itertools.product][1] returns an iterator which gets consumed over time.

6. Initialize the output tensor as:

   >>> res = np.zeros([sizes[key] for key in res_expr])
   

7. We will be looping over domain:

   >>> for indices in domain:
   ...     pass
   

   For each iteration, indices will contain the values on each axis. In our example, that would provide i, j, and k as a tuple: (k, i, j). For each input (A and B) we need to determine which component to fetch. That’s A[i, j] and B[j, k], yes! However, we don’t have variables i, j, and k, literally speaking.

   We can zip indices with to_key to create a mapping between each key (label) and its current value:

   >>> vals = dict(zip(to_key, indices))
   

   To get the coordinates for the output array, we use vals and loop over the keys: [vals[key] for key in res_expr]. However, to use these to index the output array, we need to wrap it with tuple and zip to separate the indices along each axis:

   >>> res_ind = tuple(zip([vals[key] for key in res_expr]))
   

   Same for the input indices (although there can be several):

   >>> inputs_ind = [tuple(zip([vals[key] for key in expr])) for expr in inputs_expr]
   

8. We will use a itertools.reduce to compute the product of all contributing components:

   >>> def reduce_mult(L):
   ...     return reduce(lambda x, y: x*y, L)
   

9. Overall the loop over the domain looks like:

   >>> for indices in domain:
   ...     vals = {k: v for v, k in zip(indices, to_key)}
   ...     res_ind = tuple(zip([vals[key] for key in res_expr]))
   ...     inputs_ind = [tuple(zip([vals[key] for key in expr])) 
   ...                     for expr in inputs_expr]
   ...
   ...     res[res_ind] += reduce_mult([M[i] for M, i in zip(inputs, inputs_ind)])
   

---

>>> res
array([[70., 44., 65.],
       [30., 59., 68.]]) 

That’s pretty close to what np.einsum('ij,jk->ki', A, B) returns!

Once get familiar with the dummy index (the common or repeating index) and the summation along the dummy index in the Einstein Summation (einsum), the output -> shaping is easy. Hence focus on:

1. Dummy index, the common index j in np.einsum("ij,jk->ki", a, b)
2. Summation along the dummy index j

Dummy index

For einsum("...", a, b), element wise multiplication always happens in-between matrices a and b regardless there are common indices or not. We can have einsum('xy,wz', a, b) which has no common index in the subscripts 'xy,wz'.

If there is a common index, as j in "ij,jk->ki", then it is called a dummy index in the Einstein Summation.

• Einstein Summation

An index that is summed over is a summation index, in this case "i". It is also called a dummy index since any symbol can replace "i" without changing the meaning of the expression provided that it does not collide with index symbols in the same term.

Summation along the dummy index

For np.einsum("ij,j", a, b) of the green rectangle in the diagram, j is the dummy index. The element-wise multiplication a[i][j] * b[j] is summed up along the j axis as Σ ( a[i][j] * b[j] ).

It is a dot product np.inner(a[i], b) for each i. Here being specific with np.inner() and avoiding np.dot as it is not strictly a mathematical dot product operation.

• Einstein Summation Convention: an Introduction

The dummy index can appear anywhere as long as the rules (please see the youtube for details) are met.

For the dummy index i in np.einsum(“ik,il", a, b), it is a row index of the matrices a and b, hence a column from a and that from b are extracted to generate the dot products.

Output form

Because the summation occurs along the dummy index, the dummy index disappears in the result matrix, hence i from “ik,il" is dropped and form the shape (k,l). We can tell np.einsum("... -> <shape>") to specify the output form by the output subscript labels with -> identifier.

See the explicit mode in numpy.einsum for details.

In explicit mode the output can be directly controlled by specifying
output subscript labels. This requires the identifier ‘->’ as well as
the list of output subscript labels. This feature increases the
flexibility of the function since summing can be disabled or forced
when required. The call np.einsum('i->', a) is like np.sum(a, axis=-1), and np.einsum('ii->i', a) is like np.diag(a). The difference
is that einsum does not allow broadcasting by default. Additionally
np.einsum('ij,jh->ih', a, b) directly specifies the order of the
output subscript labels and therefore returns matrix multiplication,
unlike the example above in implicit mode.

Without a dummy index

An example for having no dummy index in the einsum.

1. A term (subscript Indices, e.g. "ij") selects an element in each array.
2. Each left-hand side element is applied on the element on the right-hand side for element-wise multiplication (hence multiplication always happens).

a has shape (2,3) each element of which is applied to b of shape (2,2). Hence it creates a matrix of shape (2,3,2,2) without no summation as (i,j), (k.l) are all free indices.

# --------------------------------------------------------------------------------
# For np.einsum("ij,kl", a, b)
# 1-1: Term "ij" or (i,j), two free indices, selects selects an element a[i][j].
# 1-2: Term "kl" or (k,l), two free indices, selects selects an element b[k][l].
# 2:   Each a[i][j] is applied on b[k][l] for element-wise multiplication a[i][j] * b[k,l]
# --------------------------------------------------------------------------------
# for (i,j) in a:
#    for(k,l) in b:
#        a[i][j] * b[k][l]
np.einsum("ij,kl", a, b)

array([[[[ 0,  0],
         [ 0,  0]],

        [[10, 11],
         [12, 13]],

        [[20, 22],
         [24, 26]]],


       [[[30, 33],
         [36, 39]],

        [[40, 44],
         [48, 52]],

        [[50, 55],
         [60, 65]]]])

Examples

dot products from matrix A rows and matrix B columns

A = np.matrix('0 1 2; 3 4 5')
B = np.matrix('0 -3; -1 -4; -2 -5');
np.einsum('ij,ji->i', A, B)

# Same with
np.diagonal(np.matmul(A,B))
(A*B).diagonal()
---
[ -5 -50]
[ -5 -50]
[[ -5 -50]]