Removing nan values from an array

Question

If you're using numpy for your arrays, you can also use

x = x[numpy.logical_not(numpy.isnan(x))]

Equivalently

x = x[~numpy.isnan(x)]

[Thanks to chbrown for the added shorthand]

Explanation

The inner function, numpy.isnan returns a boolean/logical array which has the value True everywhere that x is not-a-number. As we want the opposite, we use the logical-not operator, ~ to get an array with Trues everywhere that x is a valid number.

Lastly we use this logical array to index into the original array x, to retrieve just the non-NaN values.

filter(lambda v: v==v, x)

works both for lists and numpy array since v!=v only for NaN

Try this:

import math
print [value for value in x if not math.isnan(value)]

For more, read on List Comprehensions.

For me the answer by @jmetz didn't work, however using pandas isnull() did.

x = x[~pd.isnull(x)]

As shown by others

x[~numpy.isnan(x)]

works. But it will throw an error if the numpy dtype is not a native data type, for example if it is object. In that case you can use pandas.

x[~pandas.isna(x)] or x[~pandas.isnull(x)]

@jmetz's answer is probably the one most people need; however it yields a one-dimensional array, e.g. making it unusable to remove entire rows or columns in matrices.

To do so, one should reduce the logical array to one dimension, then index the target array. For instance, the following will remove rows which have at least one NaN value:

x = x[~numpy.isnan(x).any(axis=1)]

See more detail here.

Doing the above :

x = x[~numpy.isnan(x)]

or

x = x[numpy.logical_not(numpy.isnan(x))]

I found that resetting to the same variable (x) did not remove the actual nan values and had to use a different variable. Setting it to a different variable removed the nans. e.g.

y = x[~numpy.isnan(x)]