How do I remove NaN values from a NumPy array?

Question

How do I remove NaN values from a NumPy array?

[1, 2, NaN, 4, NaN, 8]   ⟶   [1, 2, 4, 8]

Answer 1

To remove NaN values from a NumPy array x:

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

Explanation

The inner function numpy.isnan returns a boolean/logical array which has the value True everywhere that x is not-a-number. Since 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, in order to retrieve just the non-NaN values.

Answer 2

pandas introduces an option to convert all data types to missing values.

• https://pandas.pydata.org/docs/user_guide/missing_data.html

The np.isnan() function is not compatible with all data types, e.g.

>>> import numpy as np
>>> values = [np.nan, "x", "y"]
>>> np.isnan(values)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: ufunc 'isnan' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''

The pd.isna() and pd.notna() functions are compatible with many data types and pandas introduces a pd.NA value:

>>> import numpy as np
>>> import pandas as pd

>>> values = pd.Series([np.nan, "x", "y"])
>>> values
0    NaN
1      x
2      y
dtype: object
>>> values.loc[pd.isna(values)]
0    NaN
dtype: object
>>> values.loc[pd.isna(values)] = pd.NA
>>> values.loc[pd.isna(values)]
0    <NA>
dtype: object
>>> values
0    <NA>
1       x
2       y
dtype: object

#
# using map with lambda, or a list comprehension
#

>>> values = [np.nan, "x", "y"]
>>> list(map(lambda x: pd.NA if pd.isna(x) else x, values))
[<NA>, 'x', 'y']
>>> [pd.NA if pd.isna(x) else x for x in values]
[<NA>, 'x', 'y']

Answer 3

In case it helps, for simple 1d arrays:

x = np.array([np.nan, 1, 2, 3, 4])

x[~np.isnan(x)]
>>> array([1., 2., 3., 4.])

but if you wish to expand to matrices and preserve the shape:

x = np.array([
    [np.nan, np.nan],
    [np.nan, 0],
    [1, 2],
    [3, 4]
])

x[~np.isnan(x).any(axis=1)]
>>> array([[1., 2.],
           [3., 4.]])

I encountered this issue when dealing with pandas .shift() functionality, and I wanted to avoid using .apply(..., axis=1) at all cost due to its inefficiency.

Answer 4

Simply fill with

 x = numpy.array([
 [0.99929941, 0.84724713, -0.1500044],
 [-0.79709026, numpy.NaN, -0.4406645],
 [-0.3599013, -0.63565744, -0.70251352]])

x[numpy.isnan(x)] = .555

print(x)

# [[ 0.99929941  0.84724713 -0.1500044 ]
#  [-0.79709026  0.555      -0.4406645 ]
#  [-0.3599013  -0.63565744 -0.70251352]]

Answer 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.

Answer 6

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)]

Answer 7

The accepted answer changes shape for 2d arrays. I present a solution here, using the Pandas dropna() functionality. It works for 1D and 2D arrays. In the 2D case you can choose weather to drop the row or column containing np.nan.

import pandas as pd
import numpy as np

def dropna(arr, *args, **kwarg):
    assert isinstance(arr, np.ndarray)
    dropped=pd.DataFrame(arr).dropna(*args, **kwarg).values
    if arr.ndim==1:
        dropped=dropped.flatten()
    return dropped

x = np.array([1400, 1500, 1600, np.nan, np.nan, np.nan ,1700])
y = np.array([[1400, 1500, 1600], [np.nan, 0, np.nan] ,[1700,1800,np.nan]] )


print('='*20+' 1D Case: ' +'='*20+'\nInput:\n',x,sep='')
print('\ndropna:\n',dropna(x),sep='')

print('\n\n'+'='*20+' 2D Case: ' +'='*20+'\nInput:\n',y,sep='')
print('\ndropna (rows):\n',dropna(y),sep='')
print('\ndropna (columns):\n',dropna(y,axis=1),sep='')

print('\n\n'+'='*20+' x[np.logical_not(np.isnan(x))] for 2D: ' +'='*20+'\nInput:\n',y,sep='')
print('\ndropna:\n',x[np.logical_not(np.isnan(x))],sep='')

Result:

==================== 1D Case: ====================
Input:
[1400. 1500. 1600.   nan   nan   nan 1700.]

dropna:
[1400. 1500. 1600. 1700.]


==================== 2D Case: ====================
Input:
[[1400. 1500. 1600.]
 [  nan    0.   nan]
 [1700. 1800.   nan]]

dropna (rows):
[[1400. 1500. 1600.]]

dropna (columns):
[[1500.]
 [   0.]
 [1800.]]


==================== x[np.logical_not(np.isnan(x))] for 2D: ====================
Input:
[[1400. 1500. 1600.]
 [  nan    0.   nan]
 [1700. 1800.   nan]]

dropna:
[1400. 1500. 1600. 1700.]

Answer 8

If you're using numpy

# first get the indices where the values are finite
ii = np.isfinite(x)

# second get the values
x = x[ii]

Answer 9

A simplest way is:

numpy.nan_to_num(x)

Documentation: https://docs.scipy.org/doc/numpy/reference/generated/numpy.nan_to_num.html

Answer 10

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

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

Answer 11

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)]

Answer 12

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

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

Answer 13

Try this:

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

For more, read on List Comprehensions.