# Initializing numpy matrix to something other than zero or one

### Question

I have the following code:

``````r = numpy.zeros(shape = (width, height, 9))
``````

It creates a width x height x 9 matrix filled with zeros. Instead, I'd like to know if there's a function or way to initialize them instead to NaN.

Is there any? Without having to resort to manually doing loops and such?

Thanks

1
155
11/10/2009 12:17:30 AM

You rarely need loops for vector operations in numpy. You can create an uninitialized array and assign to all entries at once:

``````>>> a = numpy.empty((3,3,))
>>> a[:] = numpy.nan
>>> a
array([[ NaN,  NaN,  NaN],
[ NaN,  NaN,  NaN],
[ NaN,  NaN,  NaN]])
``````

I have timed the alternatives `a[:] = numpy.nan` here and `a.fill(numpy.nan)` as posted by Blaenk:

``````\$ python -mtimeit "import numpy as np; a = np.empty((100,100));" "a.fill(np.nan)"
10000 loops, best of 3: 54.3 usec per loop
\$ python -mtimeit "import numpy as np; a = np.empty((100,100));" "a[:] = np.nan"
10000 loops, best of 3: 88.8 usec per loop
``````

The timings show a preference for `ndarray.fill(..)` as the faster alternative. OTOH, I like numpy's convenience implementation where you can assign values to whole slices at the time, the code's intention is very clear.

223
11/28/2017 4:44:10 PM

Another option is to use `numpy.full`, an option available in NumPy 1.8+

``````a = np.full([height, width, 9], np.nan)
``````

This is pretty flexible and you can fill it with any other number that you want.