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.

Note that ndarray.fill performs its operation in-place, so numpy.empty((3,3,)).fill(numpy.nan) will instead return None.

you could use either None or np.nan to create an array of just missing values in Python like so:

np.full(shape=5, fill_value=None)
np.full(shape=5, fill_value=np.nan)

back to your example, this works just fine:

import numpy as np
import pandas as pd

unknowns = np.full(shape=5, fill_value=None)
categories = np.random.choice(['web', 'software', 'hardware', 'biotech'], size = 15, replace = True)
categories = np.concatenate([categories, unknowns])
example = pd.DataFrame(data = {'categories': categories})
example['transformed'] = [ x if pd.isna(x) == False else 'unknown' for x in example['categories']]

print(example['transformed'].value_counts())

Lastly, this line is inefficient. example['transformed'] = [ x if pd.isna(x) == False else 'unknown' for x in example['categories']]

You do want to avoid loops & list comprehensions when using pandas

on large data, this is going to run much faster: example['transformed'] = example.categories.apply(lambda s: s if s else 'unknown')

You can use np.full() to create a multi-dimensional array pre-populated with the same values:

np.full((1, 5), np.nan)

which produces:

array([[nan, nan, nan, nan, nan]])