Does Python do variable interpolation similar to "string #{var}" in Ruby?

Question

In Python, it is tedious to write:

print "foo is" + bar + '.'

Can I do something like this in Python?

print "foo is #{bar}."

I think I've found a solution, would you check it out? http://stackoverflow.com/questions/16504732/how-could-i-make-my-python-string-interpolation-implementation-work-accross-impo — HarmonicaMuse, May 12 '13 at 10:25

score 66 · Accepted Answer · edited Sep 08 '18 at 00:47

66

Python 3.6+ does have variable interpolation - prepend an f to your string:

f"foo is {bar}"

For versions of Python below this (Python 2 - 3.5) you can use str.format to pass in variables:

# Rather than this:
print("foo is #{bar}")

# You would do this:
print("foo is {}".format(bar))

# Or this:
print("foo is {bar}".format(bar=bar))

# Or this:
print("foo is %s" % (bar, ))

# Or even this:
print("foo is %(bar)s" % {"bar": bar})

edited Sep 08 '18 at 00:47

smci

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answered Aug 03 '12 at 02:41

Sean Vieira

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The 3rd one looks neat, but I remember this way kinda deprecated? is that true? – mko Aug 03 '12 at 02:52
10

There is also, for the lazy, `print "foo is %(bar)s" % locals()`. – Brian M. Hunt Aug 03 '12 at 02:58
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@yozloy - correct it is deprecated in Python 3 (as I understand it). – Sean Vieira Aug 03 '12 at 02:58
@SeanVieira what's the over taker of this solution? – mko Aug 03 '12 at 03:14
Strange the fact that Python has something explicit. – Jul 07 '15 at 18:39
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the % syntax is no longer deprecated in Python 3: http://stackoverflow.com/a/23381153/770425 – Jeff Widman Dec 18 '15 at 01:48
What does "it is explicit rather than implicit" imply in this context? `print "foo is #{bar}."`, for example, looks very explicit to me – Lem Ko Dec 13 '16 at 05:40
The quote-as-operator-context is what's implicit (to my mind) - `print "foo is #{bar}"` vs. `print 'foo is #{bar}'` is implicit compared to `"foo is {bar}".format(bar=bar)` or even Python 3.6's `f"foo is {bar}"` – Sean Vieira Dec 16 '16 at 06:50
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This is incorrect since [3.6 introduced f-strings](https://docs.python.org/3.7/reference/lexical_analysis.html#index-20) – smci Sep 07 '18 at 01:01

AXO · Answer 2 · 2017-02-18T12:31:05.167

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Python 3.6 ~~will have~~ has literal string interpolation using f-strings:

print(f"foo is {bar}.")

edited Feb 18 '17 at 12:31

answered May 14 '16 at 10:37

AXO

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In case anyone was wondering: yes, you can combine this with say raw strings, like so `rf"foo is {bar}"`. – Toni Sučić Feb 20 '17 at 07:40

warvariuc · Answer 3 · 2017-09-22T11:29:48.447

11

Python 3.6 has introduced f-strings:

print(f"foo is {bar}.")

Old answer:

Since version 3.2 Python has str.format_map which together with locals() or globals() allows you to do fast:

Python 3.3.2+ (default, Feb 28 2014, 00:52:16) 
>>> bar = "something"
>>> print("foo is {bar}".format_map(locals()))
foo is something
>>>

edited Sep 22 '17 at 11:29

answered Mar 21 '14 at 08:25

warvariuc

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dkim · Answer 4 · 2012-08-06T17:49:05.730

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I have learned the following technique from Python Essential Reference:

>>> bar = "baz"
>>> print "foo is {bar}.".format(**vars())
foo is baz.

This is quite useful when we want to refer to many variables in the formatting string:

We don't have to repeat all variables in the argument list again: compare it to the explicit keyword argument-based approaches (such as "{x}{y}".format(x=x, y=y) and "%(x)%(y)" % {"x": x, "y": y}).
We don't have to check one by one if the order of variables in the argument list is consistent with their order in the formatting string: compare it to the positional argument-based approaches (such as "{}{}".format(x, y), "{0}{1}".format(x, y) and "%s%s" % (x, y)).

edited Aug 06 '12 at 17:49

answered Aug 03 '12 at 02:47

dkim

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That's an... odd way to pass in bar... Pretty neat though, makes it more closely follow the Ruby way. – Josiah Aug 03 '12 at 02:48
This seems like the best solution if you're working with an object. For example, if you are reporting an urllib2.HTTPError you can do `"HTTP error: {error.code} {error.msg}".format(**vars())` This doesn't work with `format(**locals())` – Mike T Oct 03 '14 at 08:27

score 5 · Answer 5 · answered Aug 03 '12 at 02:38

5

String formatting

>>> bar = 1
>>> print "foo is {}.".format(bar)
foo is 1.

answered Aug 03 '12 at 02:38

Josiah

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or older but still popular: print "foo is %s" % str(bar) – Edmon Aug 03 '12 at 02:39
That's supposed to be deprecated by now, though I can't find the PEP. – Josiah Aug 03 '12 at 02:41
Not deprecated, just superseded. – Sean Vieira Aug 03 '12 at 02:42
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[I guess in Python 2 it's not](http://docs.python.org/release/3.0.1/whatsnew/3.0.html#pep-3101-a-new-approach-to-string-formatting), but it was at least supposed to be deprecated by 3.1, which was released some time ago. – Josiah Aug 03 '12 at 02:45

score 2 · Answer 6 · answered Aug 03 '12 at 02:55

2

I prefer this approach because you don't have to repeat yourself by referencing the variable twice:

alpha = 123
print 'The answer is {alpha}'.format(**locals())

answered Aug 03 '12 at 02:55

Chris Johnson

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But i guess it's very slow - unpacking a possibly big dict for the params. – warvariuc Aug 03 '12 at 05:36
@warwaruk writing to std out is the limiting factor, printing a string takes 10 times longer than formatting it, furthermore locals() returns a reference so I think this method is pretty fast – jcr Mar 21 '14 at 07:55

score 2 · Answer 7 · answered Aug 03 '12 at 02:56

2

There is a big difference between this in Ruby:

print "foo is #{bar}."

And these in Python:

print "foo is {bar}".format(bar=bar)

In the Ruby example, bar is evaluated
In the Python example, bar is just a key to the dictionary

In the case that you are just using variables the behave more or less the same, but in general, converting Ruby to Python isn't quite so simple

answered Aug 03 '12 at 02:56

John La Rooy

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This is incorrect since 3.6. We now have `print(f"foo is #{bar}")` – smci Sep 07 '18 at 01:00

score 0 · Answer 8 · edited Jul 29 '13 at 21:26

0

Yes, absolutely. Python, in my opinion, has great support for string formatting, replacements and operators.

This might be helpful:
http://docs.python.org/library/stdtypes.html#string-formatting-operations

edited Jul 29 '13 at 21:26

Slater Victoroff

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answered Aug 03 '12 at 02:36

Edmon

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score 0 · Answer 9 · answered Dec 16 '22 at 18:08

You can also use Template Strings like this:

from string import Template
s = Template('$who likes $what')
s.substitute(who='tim', what='kung pao')
# 'tim likes kung pao'

This works really well for when you want to build the string template ahead of time and inject the variable elsewhere

See Also: PEP 292 – Simpler String Substitutions

score -1 · Answer 10 · answered Jul 28 '18 at 18:12

-1

Almost every other answer didn't work for me. Probably it's because I'm on Python3.5. The only thing which worked is:

 print("Foobar is %s%s" %('Foo','bar',))

answered Jul 28 '18 at 18:12

valk

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Does Python do variable interpolation similar to "string #{var}" in Ruby?

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