熊猫用空白/空字符串替换NaN

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

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

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

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.

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('')

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

Try this,

add inplace=True

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

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.