search tags | f strings, thousands separator, format integers with commas

In [1]:
value = 12345

For python >= 2.7

In [2]:
'{:,}'.format(value)
Out[2]:
'12,345'

For python >= 3.6

In [3]:
f'{value:,}'
Out[3]:
'12,345'

Examples using a print statement

In [1]:
a = 123456

In [2]:
print('The magic number is = {alpha:,}'.format(alpha=a))
The magic number is = 123,456

In [3]:
print('The magic number is = {:,}'.format(a))
The magic number is = 123,456

In [4]:
print('The magic number is = {:}'.format(a))
The magic number is = 123456

Ref:-

• https://stackoverflow.com/questions/1823058/how-to-print-number-with-commas-as-thousands-separators
• https://mail.python.org/pipermail/python-list/2020-May/896878.html

Given a data frame such as

       X           Y
0  12345  67891011.0
1  54321         NaN
2  67890   1234567.0

how can I get

        X             Y
0  12,345  67,891,011.0
1  54,321           nan
2  67,890   1,234,567.0

Use .apply(np.vectorize(lambda x: f“{x:,}”)) or .applymap(lambda x: f“{x:,}”). I prefer the .apply() approach as it can be applied on a single column or multiple columns or entire dataframe. The .applymap() works only on the entire dataframe but not on a series.

In both cases, the original dataframe is unaltered.

If a column is not “convertible” (ex:- a column with strings in it), an exception will be thrown.

Example:

Prepare the input

$ ipython
Python 3.8.5 (default, Sep  3 2020, 21:29:08) [MSC v.1916 64 bit (AMD64)]
Type 'copyright', 'credits' or 'license' for more information
IPython 7.19.0 -- An enhanced Interactive Python. Type '?' for help.

In [1]:
import pandas as pd
df = pd.DataFrame([{'X': 12345, 'Y': 67891011},
                   {'X': 54321, 'Y': None},
                   {'X': 67890, 'Y': 1234567}])
df
Out[1]:
       X           Y
0  12345  67891011.0
1  54321         NaN
2  67890   1234567.0

Use .apply() on a single column

In [2]:
import numpy as np
df['X'].apply(np.vectorize(lambda x: f"{x:,}"))
Out[2]:
0    12,345
1    54,321
2    67,890
Name: X, dtype: object

In [3]:
df['Y'].apply(np.vectorize(lambda x: f"{x:,}"))
Out[3]:
0    67,891,011.0
1             nan
2     1,234,567.0
Name: Y, dtype: object

Use .apply() on multiple columns

In [4]:
df[['X', 'Y']].apply(np.vectorize(lambda x: f"{x:,}"))
Out[4]:
        X             Y
0  12,345  67,891,011.0
1  54,321           nan
2  67,890   1,234,567.0

Use .apply() on the entire dataframe

In [5]:
df.apply(np.vectorize(lambda x: f"{x:,}"))
Out[5]:
        X             Y
0  12,345  67,891,011.0
1  54,321           nan
2  67,890   1,234,567.0

Use .applymap() on the entire dataframe

In [6]:
df.applymap(lambda x: f"{x:,}")
Out[6]:
        X             Y
0  12,345  67,891,011.0
1  54,321           nan
2  67,890   1,234,567.0

The original dataframe remains intact.

In [7]:
df
Out[7]:
       X           Y
0  12345  67891011.0
1  54321         NaN
2  67890   1234567.0