What is the '@=' symbol for in Python? What is the '@=' symbol for in Python? python python

What is the '@=' symbol for in Python?


From the documentation:

The @ (at) operator is intended to be used for matrix multiplication. No builtin Python types implement this operator.

The @ operator was introduced in Python 3.5. @= is matrix multiplication followed by assignment, as you would expect. They map to __matmul__, __rmatmul__ or __imatmul__ similar to how + and += map to __add__, __radd__ or __iadd__.

The operator and the rationale behind it are discussed in detail in PEP 465.


@= and @ are new operators introduced in Python 3.5 performing matrix multiplication. They are meant to clarify the confusion which existed so far with the operator * which was used either for element-wise multiplication or matrix multiplication depending on the convention employed in that particular library/code. As a result, in the future, the operator * is meant to be used for element-wise multiplication only.

As explained in PEP0465, two operators were introduced:

  • A new binary operator A @ B, used similarly as A * B
  • An in-place version A @= B, used similarly as A *= B

Matrix Multiplication vs Element-wise Multiplication

To quickly highlight the difference, for two matrices:

A = [[1, 2],    B = [[11, 12],     [3, 4]]         [13, 14]]
  • Element-wise multiplication will yield:

    A * B = [[1 * 11,   2 * 12],          [3 * 13,   4 * 14]]
  • Matrix multiplication will yield:

    A @ B  =  [[1 * 11 + 2 * 13,   1 * 12 + 2 * 14],           [3 * 11 + 4 * 13,   3 * 12 + 4 * 14]]

Usage in Numpy

So far, Numpy used the following convention:

Introduction of the @ operator makes the code involving matrix multiplications much easier to read. PEP0465 gives us an example:

# Current implementation of matrix multiplications using dot functionS = np.dot((np.dot(H, beta) - r).T,            np.dot(inv(np.dot(np.dot(H, V), H.T)), np.dot(H, beta) - r))# Current implementation of matrix multiplications using dot methodS = (H.dot(beta) - r).T.dot(inv(H.dot(V).dot(H.T))).dot(H.dot(beta) - r)# Using the @ operator insteadS = (H @ beta - r).T @ inv(H @ V @ H.T) @ (H @ beta - r)

Clearly, the last implementation is much easier to read and interpret as an equation.