Removing nan values from an array
PythonArraysNumpyNanPython Problem Overview
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?
Python Solutions
Solution 1 - Python
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 - Python
filter(lambda v: v==v, x)
works both for lists and numpy array since v!=v only for NaN
Solution 3 - Python
Try this:
import math
print [value for value in x if not math.isnan(value)]
For more, read on List Comprehensions.
Solution 4 - Python
For me the answer by @jmetz didn't work, however using pandas isnull() did.
x = x[~pd.isnull(x)]
Solution 5 - Python
@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.
Solution 6 - Python
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)]
Solution 7 - Python
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)]
Solution 8 - Python
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]
Solution 9 - Python
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.]
Solution 10 - Python
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.
Solution 11 - Python
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]]
Solution 12 - Python
A simplest way is:
numpy.nan_to_num(x)
Documentation: https://docs.scipy.org/doc/numpy/reference/generated/numpy.nan_to_num.html