JoVe - 28 days ago 9x

Python Question

In which case using objects like numpy.r_ or numpy.c_ is better (more efficient, more suitable) than using fonctions like concatenate or vstack for example ?

I am trying to understand a code where the programmer wrote something like:

`return np.r_[0.0, 1d_array, 0.0] == 2`

where

`1d_array`

Why not using np.concatenate (for example) instead ? Like :

`return np.concatenate([[0.0], 1d_array, [0.0]]) == 2`

It is more readable and apparently it does the same thing.

Answer

`np.r_`

is implemented in the `numpy/lib/index_tricks.py`

file. This is pure Python code, with no special compiled stuff. So it is not going to be any faster than the equivalent written with `concatenate`

, `arange`

and `linspace`

. It's useful only if the notation fits your way of thinking and your needs.

In your example it just saves converting the scalars to lists or arrays:

```
In [452]: np.r_[0.0, np.array([1,2,3,4]), 0.0]
Out[452]: array([ 0., 1., 2., 3., 4., 0.])
```

error with the same arguments:

```
In [453]: np.concatenate([0.0, np.array([1,2,3,4]), 0.0])
...
ValueError: zero-dimensional arrays cannot be concatenated
```

correct with the added []

```
In [454]: np.concatenate([[0.0], np.array([1,2,3,4]), [0.0]])
Out[454]: array([ 0., 1., 2., 3., 4., 0.])
```

`hstack`

takes care of that by passing all arguments through `[atleast_1d(_m) for _m in tup]`

:

```
In [455]: np.hstack([0.0, np.array([1,2,3,4]), 0.0])
Out[455]: array([ 0., 1., 2., 3., 4., 0.])
```

So at least in simple cases it is most similar to `hstack`

.

But the real usefulness of `r_`

comes when you want to use ranges

```
np.r_[0.0, 1:5, 0.0]
np.hstack([0.0, np.arange(1,5), 0.0])
np.r_[0.0, slice(1,5), 0.0]
```

`r_`

lets you use the `:`

syntax that is used in indexing. That's because it is actually an instance of a class that has a `__getitem__`

method. `index_tricks`

uses this programming trick several times.

They've thrown in other bells-n-whistles

Using an `imaginary`

step, uses `np.linspace`

to expand the slice rather than `np.arange`

.

```
np.r_[-1:1:6j, [0]*3, 5, 6]
```

produces:

```
array([-1. , -0.6, -0.2, 0.2, 0.6, 1. , 0. , 0. , 0. , 5. , 6. ])
```

There are more details in the documentation.

I did some time tests for many slices in http://stackoverflow.com/a/37625115/901925

Source (Stackoverflow)

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