Initializing numpy matrix to something other than zero or one

0 votes
asked Nov 10, 2009 by devoured-elysium

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?


4 Answers

0 votes
answered by hobs

You can always use multiplication if you don't immediately recall the .empty or .full methods:

>>> np.nan * np.ones(shape=(3,2))
array([[ nan,  nan],
       [ nan,  nan],
       [ nan,  nan]])

Of course it works with any other numerical value as well:

>>> 42 * np.ones(shape=(3,2))
array([[ 42,  42],
       [ 42,  42],
       [ 42, 42]])

But the @u0b34a0f6ae's accepted answer is 3x faster (CPU cycles, not brain cycles to remember numpy syntax ;):

$ python -mtimeit "import numpy as np; X = np.empty((100,100));" "X[:] = np.nan;"
100000 loops, best of 3: 8.9 usec per loop
(predict)laneh@predict:~/src/predict/predict/webapp$ master
$ python -mtimeit "import numpy as np; X = np.ones((100,100));" "X *= np.nan;"
10000 loops, best of 3: 24.9 usec per loop
0 votes
answered Nov 10, 2009 by u0b34a0f6ae

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.

0 votes
answered Nov 9, 2014 by pietro

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

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

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

0 votes
answered Nov 27, 2015 by ntg

As said, numpy.empty() is the way to go. However, for objects, fill() might not do exactly what you think it does:

In[36]: a = numpy.empty(5,dtype=object)
In[37]: a.fill([])
In[38]: a
Out[38]: array([[], [], [], [], []], dtype=object)
In[39]: a[0].append(4)
In[40]: a
Out[40]: array([[4], [4], [4], [4], [4]], dtype=object)

One way around can be e.g.:

In[41]: a = numpy.empty(5,dtype=object)
In[42]: a[:]= [ [] for x in range(5)]
In[43]: a[0].append(4)
In[44]: a
Out[44]: array([[4], [], [], [], []], dtype=object)
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