numpy pad array with nan, getting strange float instead

Question

I'm trying to pad an array with np.nan

import numpy as np
print np.version.version
# 1.10.2
combine = lambda real, theo: np.vstack((theo, np.pad(real, (0, theo.shape[0] - real.shape[0]), 'constant', constant_values=np.nan)))
real = np.arange(20)
theoretical = np.linspace(0, 20, 100)
result = combine(real, theoretical)
np.any(np.isnan(result))
# False

Inspecting result, it seems instead of np.nan, the array is getting padded with -9.22337204e+18. What's going on here? How can I get np.nan?

Look at the result produced by np.pad(real, (0, theo.shape[0] - real.shape[0]), 'constant', constant_values=np.nan) when real is an integer array. — Warren Weckesser
– Warren Weckesser, Commented Feb 19, 2016 at 21:12
Thanks - forgot real was an integer array. Why does this behaviour happen with integers? — user2561747
– user2561747, Commented Feb 19, 2016 at 21:27

hpaulj · Accepted Answer · 2016-02-19 21:35:20Z

13

The result of pad has the same type as the input. np.nan is a float

In [874]: np.pad(np.ones(2,dtype=int),1,mode='constant',constant_values=(np.nan,))
Out[874]: array([-2147483648,           1,           1, -2147483648])

In [875]: np.pad(np.ones(2,dtype=float),1,mode='constant',constant_values=(np.nan,))
Out[875]: array([ nan,   1.,   1.,  nan])

The int pad is np.nan cast as an integer:

In [878]: np.array(np.nan).astype(int)
Out[878]: array(-2147483648)

answered Feb 19, 2016 at 21:35

hpaulj

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2 Comments

user2561747 Over a year ago

Is there an integer version of np.nan?

hpaulj Over a year ago

Masked arrays use 999999 as the default fill_value, but values like that don't propagate through calculations in the same way that np.nan does. Talk about nan often references IEEE-754 floating point standard.

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