Removing nan values from an array

Question

I want to figure out how to remove nan values from my array. My array looks something like this:

x = [1400, 1500, 1600, nan, nan, nan ,1700] #Not in this exact configuration

How can I remove the nan values from x?

Solution 1

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.

Solution 2

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

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

Solution 3

Try this:

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

For more, read on List Comprehensions.

Solution 4

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

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

Solution 5

@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.

View more solutions

Comments

[smci]

To be clear, by "remove NaNs" you mean filter out only the subset of non-null values. Not "fill the NaNs with some value (zero, constant, mean, median, etc.)"

[Miki Tebeka]

Or x = x[numpy.isfinite(x)]

[jmetz]

If you're using numpy both my answer and that by @lazy1 are almost an order of magnitude faster than the list comprehension - lazy1's solution is slightly faster (though technically will also not return any infinity values).

[chbrown]

Or x = x[~numpy.isnan(x)], which is equivalent to mutzmatron's original answer, but shorter. In case you want to keep your infinities around, know that numpy.isfinite(numpy.inf) == False, of course, but ~numpy.isnan(numpy.inf) == True.

[jmetz]

@dax-felizv I agree with @chbrown, NaN and Infinite are not the same in numpy. @chbrown - thanks for pointing out the shorthand for logical_not, though beware that it is considerably slower - stackoverflow.com/questions/15998188/…, stackoverflow.com/questions/13600988/…

[chbrown]

Hmm, @mutzmatron -- I figured they did the same thing underneath the hood, and I'm getting very similar results with timeit (as did @unutbu at that first link): python -m timeit -s "import numpy; bools = numpy.random.uniform(size=10000) >= 0.5" "numpy.logical_not(bools)" vs. python -m timeit -s "import numpy; bools = numpy.random.uniform(size=10000) >= 0.5" "~bools" (numpy.__version__ == '1.8.0')

[jmetz]

@chbrown - you're right, any performance gain with numpy seems to have only occurred on the second posters machine - I tested numpy.invert and numpy.logical_not and got the same result for both as for ~, on numpy v1.7.1. Not sure if architecture affects comparative performance - am testing on my chromebook (armv7l).

[Austin Richardson]

A hack but an especially useful one in the case where you are filtering nans from an array of objects with mixed types, such as a strings and nans.

[jmetz]

This is strange; according to the docs, boolean array indexing (which this is), is under advanced indexing which apparently "always returns a copy of the data", so you should be over-writing x with the new value (i.e. without the NaNs...). Can you provide any more info as to why this could be happening?

[Pier Paolo]

Welcome to SO! The solution you propose does not answer the problem: your solution substitutes NaNs with a large number, while the OP asked to entirely remove the elements.

[BoltzmannBrain]

For people looking to solve this with an ndarray and maintain the dimensions, use numpy where: np.where(np.isfinite(x), x, 0)

[Moondra]

Very clean solution.

[hypers]

Don't forget the brackets :) print ([value for value in x if not math.isnan(value)])

[towry]

TypeError: only integer scalar arrays can be converted to a scalar index

[jmetz]

@towry: this is happening because your input, x is not a numpy array. If you want to use logical indexing, it must be an array - e.g. x = np.array(x)

[Chris_Rands]

This might seem clever, but if obscures the logic and theoretically other objects (such as custom classes) can also have this property

[yeliabsalohcin]

If you're using numpy like the top answer then you can use this list comprehension answer with the np package: So returns your list without the nans: [value for value in x if not np.isnan(value)]

[Dark]

Also, to completely remove the non-finite rows, use .any(axis=1). The full code will be x=x[~pd.isnull(x).any(axis=1)] for Pandas or x=x[~np.isnan(x).any(axis=1)]for Numpy. Note that these are working on different type of variables.

[jmetz]

@Dark - thanks for the useful example for 2d data, though it's beyond the scope of the OP's question which relates only to a 1d input. Perhaps it would be useful for others posted as a separate Q and A?

[Christian O'Reilly]

Also useful because it only needs x to be specified once as opposed to solutions of the type x[~numpy.isnan(x)]. This is convenient when x is defined by a long expression and you don't want to clutter the code by creating a temporary variable to store the result of this long expression.

[smm]

It might be slow compere to x[~numpy.isnan(x)]

[seralouk]

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

[Darren Weber]

Similarly, as a list comprehension, e.g. [v for v in var if v == v]

[Darren Weber]

This can avoid TypeError: ufunc 'isnan' not supported for the input types when the var contains mixtures of nan and strings, as noted by @AustinRichardson