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?
Answers:
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Method 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.
Method 2
filter(lambda v: v==v, x)
works both for lists and numpy array
since v!=v only for NaN
Method 3
Try this:
import math print [value for value in x if not math.isnan(value)]
For more, read on List Comprehensions.
Method 4
For me the answer by @jmetz didn’t work, however using pandas isnull() did.
x = x[~pd.isnull(x)]
Method 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.
Method 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)]
Method 7
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)]
Method 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]
Method 9
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('nn'+'='*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('nn'+'='*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.]
Method 10
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.
Method 11
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]]
Method 12
pandas introduces an option to convert all data types to missing values.
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']
Method 13
A simplest way is:
numpy.nan_to_num(x)
Documentation: https://docs.scipy.org/doc/numpy/reference/generated/numpy.nan_to_num.html
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0