Performance difference between scipy and numpy norm

Question

I have always assumed `scipy.linalg.norm()` and `numpy.linalg.norm()` to be equivalent (scipy version used to not accept an axis argument, but now it does). However the following simple examples yields significantly different performances: what is the reason behind that? ``` In [1]: from scipy.linalg import norm as normsp In [2]: from numpy.linalg import norm as normnp In [3]: import numpy as np In [4]: a = np.random.random(size=(1000, 2000)) In [5]: %timeit normsp(a) The slowest run took 5.69 times longer than the fastest. This could mean that an intermediate result is being cached. 100 loops, best of 3: 2.85 ms per loop In [6]: %timeit normnp(a) The slowest run took 6.39 times longer than the fastest. This could mean that an intermediate result is being cached. 1000 loops, best of 3: 558 µs per loop ``` scipy version is 0.18.1, numpy is 1.11.1

Accepted Answer

Looking the source code reveals that `scipy` has its own `norm` function, which wraps around the `numpy.linalg.norm` or a BLAS function that is slower but handles floating point overflows better (see discussion on this PR).

However, in the example that you give it doesn't look like SciPy uses a BLAS function, so I do not think it's responsible for the time difference you see. But scipy does do some other checks before calling the numpy version of norm. In particular, that infinite check `a = np.asarray_chkfinite(a)` is a suspect for causing the performance difference:

```
In [103]: %timeit normsp(a)
100 loops, best of 3: 5.1 ms per loop

In [104]: %timeit normnp(a)
1000 loops, best of 3: 744 µs per loop

In [105]: %timeit np.asarray_chkfinite(a)
100 loops, best of 3: 4.13 ms per loop
```

So it looks like `np.asarray_chkfinite` roughly accounts for the difference in time taken to evaluate the norms.

Performance difference between scipy and numpy norm

Solution

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