Replace Nan Values in a List with Zero (0)

replace the NaN value zero after an operation with arrays

If you have Python 2.6 you have the math.isnan() function to find NaN values.

With this we can use a list comprehension to replace the NaN values in a list as follows:

import math
mylist = [0 if math.isnan(x) else x for x in mylist]

If you have Python 2.5 we can use the NaN != NaN trick from this question so you do this:

mylist = [0 if x != x else x for x in mylist]

Function to replace all NaN values with zero:

Use boolean mask.

Suppose the following dataframe:

>>> df
     A  B    C
0  0.0  1  2.0
1  NaN  4  5.0  # <- NaN should be replace by 0.1
2  6.0  7  NaN  # <- NaN should be replace by 0

m1 = df.isna().any()  # Is there a NaN in columns (not mandatory)
m2 = df.eq(0).any()   # Is there a 0 in columns

# Replace by 0
df.update(df.loc[:, m1 & ~m2].fillna(0))

# Replace by 0.1
df.update(df.loc[:, m1 & m2].fillna(0.1))

Only the second mask is useful

Output result:

>>> df
     A  B    C
0  0.0  1  2.0
1  0.1  4  5.0
2  6.0  7  0.0

convert nan value to zero

This should work:

from numpy import *

a = array([[1, 2, 3], [0, 3, NaN]])
where_are_NaNs = isnan(a)
a[where_are_NaNs] = 0

In the above case where_are_NaNs is:

In [12]: where_are_NaNs
Out[12]: 
array([[False, False, False],
       [False, False,  True]], dtype=bool)

Replacing NaN/Inf/NA to 0 in a list object

We can replace the non finite values to 0 by looping over the list elements with lapply

lapply(dat, function(x) replace(x, !is.finite(x), 0))
#     [,1] [,2]
#[1,]    2    0
#[2,]    0    0

#$b
#     [,1] [,2]
#[1,]   -3    0
#[2,]    0    1

I want to replace NaN values with 0 but not able to with the below code

In your code you passed to_replace="NaN".

Note that you actually passed here a string containing just these 3 letters.

In Pandas you can pass np.nan, but only as the value to be assigned
to a cell in a DataFrame. The same pertains to a Numpy array.

You can not pass to_replace=np.nan, because the comparison rules are
that one np.nan is NOT equal to another np.nan.

One of possible solutions is to run:

df2 = df2.where(~df2.isna(), 0)

Other, simpler solution, as richardec suggested, is to use fillna,
but the argument should be 0 (zero) not "o" (a char):

df2 = df2.fillna(0)

Replacing nan values in a Pandas data frame with lists

You have to handle the three cases (empty string, NaN, NaN in list) separately.

For the NaN in list you need to loop over each occurrence and replace the elements one by one.

NB. applymap is slow, so if you know in advance the columns to use you can subset them

For the empty string, replace them to NaN, then fillna.

sub = 'X'
(df.applymap(lambda x: [sub if (pd.isna(e) or e=='')
                        else e for e in x]
                       if isinstance(x, list) else x)
   .replace('', float('nan'))
   .fillna(sub)
 )

Output:

  col1  col2       col3    col4
0    X  Jhon  [X, 1, 2]  [k, j]
1  1.0     X  [1, 1, 5]       3
2  2.0     X          X       X
3  3.0  Samy  [1, 1, X]  [b, X]

Used input:

from numpy import nan
df = pd.DataFrame({'col1': {0: nan, 1: 1.0, 2: 2.0, 3: 3.0},
                   'col2': {0: 'Jhon', 1: nan, 2: '', 3: 'Samy'},
                   'col3': {0: [nan, 1, 2], 1: [1, 1, 5], 2: nan, 3: [1, 1, nan]},
                   'col4': {0: ['k', 'j'], 1: '3', 2: nan, 3: ['b', '']}})

Replacing NaN with 0 in Python

This should work:

import math
[0 if math.isnan(i) else i for i in ls] 

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