Pandas Replace NaN with blank/empty string

I have a Pandas Dataframe as shown below:

 1 2 3 0 a NaN read 1 b l unread 2 c NaN read

I want to remove the NaN values with an empty string so that it looks like so:

 1 2 3 0 a "" read 1 b l unread 2 c "" read

0

8 Answers

df = df.fillna('')

or just

df.fillna('', inplace=True)

This will fill na's (e.g. NaN's) with ''.

If you want to fill a single column, you can use:

df.column1 = df.column1.fillna('')

One can use df['column1'] instead of df.column1.

4

import numpy as np
df1 = df.replace(np.nan, '', regex=True)

This might help. It will replace all NaNs with an empty string.

8

If you are reading the dataframe from a file (say CSV or Excel) then use :

df.read_csv(path , na_filter=False)

df.read_excel(path , na_filter=False)

This will automatically consider the empty fields as empty strings ''

---

If you already have the dataframe

df = df.replace(np.nan, '', regex=True)

df = df.fillna('')

4

Use a formatter, if you only want to format it so that it renders nicely when printed. Just use the df.to_string(... formatters to define custom string-formatting, without needlessly modifying your DataFrame or wasting memory:

df = pd.DataFrame({ 'A': ['a', 'b', 'c'], 'B': [np.nan, 1, np.nan], 'C': ['read', 'unread', 'read']})
print df.to_string( formatters={'B': lambda x: '' if pd.isnull(x) else '{:.0f}'.format(x)})

To get:

 A B C
0 a read
1 b 1 unread
2 c read

3

Try this,

add inplace=True

import numpy as np
df.replace(np.NaN, '', inplace=True)

1

using keep_default_na=False should help you:

df = pd.read_csv(filename, keep_default_na=False)

If you are converting DataFrame to JSON, NaN will give error so best solution is in this use case is to replace NaN with None.
Here is how:

df1 = df.where((pd.notnull(df)), None)

I tried with one column of string values with nan.

To remove the nan and fill the empty string:

df.columnname.replace(np.nan,'',regex = True)

To remove the nan and fill some values:

df.columnname.replace(np.nan,'value',regex = True)

I tried df.iloc also. but it needs the index of the column. so you need to look into the table again. simply the above method reduced one step.