Set thousands separators in iPython without string formatting

Question

I have been looking for the answer to the following question for more than 4 hours. Most pages list string formatting methods. This is not what I want.

I want to set a parameter in IPython for thousands separators for integers and floats. This setting should only affect how numbers are displayed in my interactive session. I want to set the parameter once. All the decisions that I need to do some formatting for each new release do not cover my need at all. I am doing some analytical analytic data and don't want to worry about formatting numbers for each line of code.

The format should be used with all integers and floats, including those stored in numpy or pandas arrays.

For those who are familiar with Mathematica, I indicate how this can be done in Mathematica: go to settings => appearance => numbers => formatting. There you can "enable automatic number formatting" and select "digital character delimiter".

Example: if I find "600 + 600" in my ipython session, I need the following output: 1'200 (where "there will be my thousands separator").

I use IPython consoles on Spyder and IPython laptops. Thank.

Answer 1

If you used str.formatand numpy.set_printoptions, you can install it globally once:

import numpy as np
import IPython

frm = get_ipython().display_formatter.formatters['text/plain']


def thousands(arg, p, cycle):
    p.text("{:,}".format(arg).replace(",","'"))

frm.for_type(int, thousands)
frm.for_type(float, thousands)

np.set_printoptions(formatter={'int_kind': lambda x: '{:,}'.format(x).replace(",","'")})

 np.set_printoptions(formatter={'float_kind': lambda x: '{:,}'.format(x).replace(",","'")})

frm = get_ipython().display_formatter.formatters['text/plain']
frm.for_type(int, thousands)
frm.for_type(float, thousands)

It does not cover all bases, but you can add more logic:

In [2]: arr = np.array([12345,12345])

In [3]: arr
Out[3]: array([12'345, 12'345])

In [4]: 123456
Out[4]: 123'456

In [5]: 123456.343
Out[5]: 123'456.343

startup.py script, PYTHONSTARTUP, , , ipython:

~$ ipython2
Python 2.7.6 (default, Jun 22 2015, 17:58:13) 
Type "copyright", "credits" or "license" for more information.

IPython 4.0.1 -- An enhanced Interactive Python.
?         -> Introduction and overview of IPython features.
%quickref -> Quick reference.
help      -> Python own help system.
object?   -> Details about 'object', use 'object??' for extra details.
(.startup.py)
(imported datetime, os, pprint, re, sys, time,np,pd)

In [1]: arr = np.array([12345,12345])

In [2]: arr
Out[2]: array([12'345, 12'345])

In [3]: 12345
Out[3]: "12'345"

pandas , display.float_format set_option

In [22]: pd.set_option("display.float_format",lambda x: "{:,}".format(x).replace(",","'"))

In [23]: pd.DataFrame([[12345.3,12345.4]])
Out[23]: 
         0        1
0 12'345.3 12'345.4

pandas pandas.core.format.IntArrayFormatter:

, script :

import IPython

import numpy as np
import pandas as pd

# numpy
np.set_printoptions(formatter={'float_kind': lambda x: '{:,}'.format(x).replace(",", "'"),
                          'int_kind': lambda x: '{:,}'.format(x).replace(",", "'")})


# pandas
class IntFormatter(pd.core.format.GenericArrayFormatter):
    pd.set_option("display.float_format", lambda x: "{:,}".format(x).replace(",", "'"))

    def _format_strings(self):
        formatter = self.formatter or (lambda x: ' {:,}'.format(x).replace(",", "'"))
        fmt_values = [formatter(x) for x in self.values]
        return fmt_values


pd.core.format.IntArrayFormatter = IntFormatter


# general
def thousands(arg, p, cycle):
    p.text("{:,}".format(arg).replace(",","'"))


frm = get_ipython().display_formatter.formatters['text/plain']
frm.for_type(int, thousands)
frm.for_type(float, thousands)

, , , :

IPython 4.0.1 -- An enhanced Interactive Python.
?         -> Introduction and overview of IPython features.
%quickref -> Quick reference.
help      -> Python own help system.
object?   -> Details about 'object', use 'object??' for extra details.
(.startup.py)
(imported datetime, os, pprint, re, sys, time,np,pd)

In [1]: pd.DataFrame([[12345,12345]])
Out[1]: 
        0       1
0  12'345  12'345

In [2]: pd.DataFrame([[12345,12345.345]])
Out[2]: 
        0          1
0  12'345 12'345.345

In [3]: np.array([12345,678910])
Out[3]: array([12'345, 678'910])

In [4]: np.array([12345.321,678910.123])
Out[4]:  array([12'345.321, 678'910.123])


In [5]: 100000
Out[5]: 100'000

In [6]: 100000.123
Out[6]: 100'000.123

In [7]: 10000000
Out[7]: 10'000'000