Is there a Python equivalent to Ruby's string interpolation?

Question:

Ruby example:

name = "Spongebob Squarepants"
puts "Who lives in a Pineapple under the sea? n#{name}."

The successful Python string concatenation is seemingly verbose to me.

Asked By: Caste

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Source

Answers:

Python 3.6 will add literal string interpolation similar to Ruby’s string interpolation. Starting with that version of Python (which is scheduled to be released by the end of 2016), you will be able to include expressions in “f-strings”, e.g.

name = "Spongebob Squarepants"
print(f"Who lives in a Pineapple under the sea? {name}.")

Prior to 3.6, the closest you can get to this is

name = "Spongebob Squarepants"
print("Who lives in a Pineapple under the sea? %(name)s." % locals())

The % operator can be used for string interpolation in Python. The first operand is the string to be interpolated, the second can have different types including a “mapping”, mapping field names to the values to be interpolated. Here I used the dictionary of local variables locals() to map the field name name to its value as a local variable.

The same code using the .format() method of recent Python versions would look like this:

name = "Spongebob Squarepants"
print("Who lives in a Pineapple under the sea? {name!s}.".format(**locals()))

There is also the string.Template class:

tmpl = string.Template("Who lives in a Pineapple under the sea? $name.")
print(tmpl.substitute(name="Spongebob Squarepants"))
Answered By: Sven Marnach

Python’s string interpolation is similar to C’s printf()

If you try:

name = "SpongeBob Squarepants"
print "Who lives in a Pineapple under the sea? %s" % name

The tag %s will be replaced with the name variable. You should take a look to the print function tags: http://docs.python.org/library/functions.html

Answered By: Oleiade

Since Python 2.6.X you might want to use:

"my {0} string: {1}".format("cool", "Hello there!")
Answered By: EinLama
import inspect
def s(template, **kwargs):
    "Usage: s(string, **locals())"
    if not kwargs:
        frame = inspect.currentframe()
        try:
            kwargs = frame.f_back.f_locals
        finally:
            del frame
        if not kwargs:
            kwargs = globals()
    return template.format(**kwargs)

Usage:

a = 123
s('{a}', locals()) # print '123'
s('{a}') # it is equal to the above statement: print '123'
s('{b}') # raise an KeyError: b variable not found

PS: performance may be a problem. This is useful for local scripts, not for production logs.

Duplicated:

Answered By: Paulo Cheque

You can also have this

name = "Spongebob Squarepants"
print "Who lives in a Pineapple under the sea? n{name}.".format(name=name)

http://docs.python.org/2/library/string.html#formatstrings

Answered By: Quan To

I’ve developed the interpy package, that enables string interpolation in Python.

Just install it via pip install interpy.
And then, add the line # coding: interpy at the beginning of your files!

Example:

#!/usr/bin/env python
# coding: interpy

name = "Spongebob Squarepants"
print "Who lives in a Pineapple under the sea? n#{name}."
Answered By: Syrus Akbary Nieto

String interpolation is going to be included with Python 3.6 as specified in PEP 498. You will be able to do this:

name = 'Spongebob Squarepants'
print(f'Who lives in a Pineapple under the sea? n{name}')

Note that I hate Spongebob, so writing this was slightly painful. 🙂

Answered By: kirbyfan64sos

For old Python (tested on 2.4) the top solution points the way. You can do this:

import string

def try_interp():
    d = 1
    f = 1.1
    s = "s"
    print string.Template("d: $d f: $f s: $s").substitute(**locals())

try_interp()

And you get

d: 1 f: 1.1 s: s
Answered By: Michael Fox

Python 3.6 and newer have literal string interpolation using f-strings:

name='world'
print(f"Hello {name}!")
Answered By: Alejandro Silva