How can I replace all the NaN values with Zero'

Question

I have a dataframe as below

      itm Date                  Amount 
67    420 2012-09-30 00:00:00   65211
68    421 2012-09-09 00:00:00   29424
69    421 2012-09-16 00:00:00   29877
70    421 2012-09-23 00:00:00   30990
71    421 2012-09-30 00:00:00   61303
72    485 2012-09-09 00:00:00   71781
73    485 2012-09-16 00:00:00     NaN
74    485 2012-09-23 00:00:00   11072
75    485 2012-09-30 00:00:00  113702
76    489 2012-09-09 00:00:00   64731
77    489 2012-09-16 00:00:00     NaN

when I try to .apply a function to the Amount column I get the following error.

ValueError: cannot convert float NaN to integer

I have tried applying a function using .isnan from the Math Module I have tried the pandas .replace attribute I tried the .sparse data attribute from pandas 0.9 I have also tried if NaN == NaN statement in a function. I have also looked at this article How do I replace NA values with zeros in an R dataframe? whilst looking at some other articles. All the methods I have tried have not worked or do not recognise NaN. Any Hints or solutions would be appreciated.

Answer 1

I just wanted to provide a bit of an update/special case since it looks like people still come here. If you're using a multi-index or otherwise using an index-slicer the inplace=True option may not be enough to update the slice you've chosen. For example in a 2x2 level multi-index this will not change any values (as of pandas 0.15):

idx = pd.IndexSlice
df.loc[idx[:,mask_1],idx[mask_2,:]].fillna(value=0,inplace=True)

The "problem" is that the chaining breaks the fillna ability to update the original dataframe. I put "problem" in quotes because there are good reasons for the design decisions that led to not interpreting through these chains in certain situations. Also, this is a complex example (though I really ran into it), but the same may apply to fewer levels of indexes depending on how you slice.

The solution is DataFrame.update:

df.update(df.loc[idx[:,mask_1],idx[[mask_2],:]].fillna(value=0))

It's one line, reads reasonably well (sort of) and eliminates any unnecessary messing with intermediate variables or loops while allowing you to apply fillna to any multi-level slice you like!

If anybody can find places this doesn't work please post in the comments, I've been messing with it and looking at the source and it seems to solve at least my multi-index slice problems.

Answer 2

Easy way to fill the missing values:-

filling string columns: when string columns have missing values and NaN values.

df['string column name'].fillna(df['string column name'].mode().values[0], inplace = True)

filling numeric columns: when the numeric columns have missing values and NaN values.

df['numeric column name'].fillna(df['numeric column name'].mean(), inplace = True)

filling NaN with zero:

df['column name'].fillna(0, inplace = True)

Answer 3

I believe DataFrame.fillna() will do this for you.

Link to Docs for a dataframe and for a Series.

Example:

In [7]: df
Out[7]: 
          0         1
0       NaN       NaN
1 -0.494375  0.570994
2       NaN       NaN
3  1.876360 -0.229738
4       NaN       NaN

In [8]: df.fillna(0)
Out[8]: 
          0         1
0  0.000000  0.000000
1 -0.494375  0.570994
2  0.000000  0.000000
3  1.876360 -0.229738
4  0.000000  0.000000

To fill the NaNs in only one column, select just that column. in this case I'm using inplace=True to actually change the contents of df.

In [12]: df[1].fillna(0, inplace=True)
Out[12]: 
0    0.000000
1    0.570994
2    0.000000
3   -0.229738
4    0.000000
Name: 1

In [13]: df
Out[13]: 
          0         1
0       NaN  0.000000
1 -0.494375  0.570994
2       NaN  0.000000
3  1.876360 -0.229738
4       NaN  0.000000

Answer 4

The below code worked for me.

import pandas

df = pandas.read_csv('somefile.txt')

df = df.fillna(0)

Answer 5

You could use replace to change NaN to 0:

import pandas as pd
import numpy as np

# for column
df['column'] = df['column'].replace(np.nan, 0)

# for whole dataframe
df = df.replace(np.nan, 0)

# inplace
df.replace(np.nan, 0, inplace=True)

Answer 6

You should use fillna() . It works for me.

df = df.fillna(value_to_replace_null)