String formatting: % vs. .format vs. f-string literal


There are various string formatting methods:

  • Python <2.6: "Hello %s" % name
  • Python 2.6+: "Hello {}".format(name)   (uses str.format)
  • Python 3.6+: f"{name}"   (uses f-strings)

Which is better, and for what situations?

  1. The following methods have the same outcome, so what is the difference?

    name = "Alice"
    "Hello %s" % name
    "Hello {0}".format(name)
    f"Hello {name}"
    # Using named arguments:
    "Hello %(kwarg)s" % {'kwarg': name}
    "Hello {kwarg}".format(kwarg=name)
    f"Hello {name}"
  2. When does string formatting run, and how do I avoid a runtime performance penalty?

Asked By: NorthIsUp



To answer your first question… .format just seems more sophisticated in many ways. An annoying thing about % is also how it can either take a variable or a tuple. You’d think the following would always work:

"Hello %s" % name

yet, if name happens to be (1, 2, 3), it will throw a TypeError. To guarantee that it always prints, you’d need to do

"Hello %s" % (name,)   # supply the single argument as a single-item tuple

which is just ugly. .format doesn’t have those issues. Also in the second example you gave, the .format example is much cleaner looking.

Only use it for backwards compatibility with Python 2.5.

To answer your second question, string formatting happens at the same time as any other operation – when the string formatting expression is evaluated. And Python, not being a lazy language, evaluates expressions before calling functions, so the expression log.debug("some debug info: %s" % some_info) will first evaluate the string to, e.g. "some debug info: roflcopters are active", then that string will be passed to log.debug().

Answered By: Claudiu

Assuming you’re using Python’s logging module, you can pass the string formatting arguments as arguments to the .debug() method rather than doing the formatting yourself:

log.debug("some debug info: %s", some_info)

which avoids doing the formatting unless the logger actually logs something.

Answered By: Wooble

% gives better performance than format from my test.

Test code:

Python 2.7.2:

import timeit
print 'format:', timeit.timeit("'{}{}{}'.format(1, 1.23, 'hello')")
print '%:', timeit.timeit("'%s%s%s' % (1, 1.23, 'hello')")


> format: 0.470329046249
> %: 0.357107877731

Python 3.5.2

import timeit
print('format:', timeit.timeit("'{}{}{}'.format(1, 1.23, 'hello')"))
print('%:', timeit.timeit("'%s%s%s' % (1, 1.23, 'hello')"))


> format: 0.5864730989560485
> %: 0.013593495357781649

It looks in Python2, the difference is small whereas in Python3, % is much faster than format.

Thanks @Chris Cogdon for the sample code.

Edit 1:

Tested again in Python 3.7.2 in July 2019.


> format: 0.86600608
> %: 0.630180146

There is not much difference. I guess Python is improving gradually.

Edit 2:

After someone mentioned python 3’s f-string in comment, I did a test for the following code under python 3.7.2 :

import timeit
print('format:', timeit.timeit("'{}{}{}'.format(1, 1.23, 'hello')"))
print('%:', timeit.timeit("'%s%s%s' % (1, 1.23, 'hello')"))
print('f-string:', timeit.timeit("f'{1}{1.23}{"hello"}'"))


format: 0.8331376779999999
%: 0.6314778750000001
f-string: 0.766649943

It seems f-string is still slower than % but better than format.

Answered By: lcltj

Something that the modulo operator ( % ) can’t do, afaik:

tu = (12,45,22222,103,6)
print '{0} {2} {1} {2} {3} {2} {4} {2}'.format(*tu)


12 22222 45 22222 103 22222 6 22222

Very useful.

Another point: format(), being a function, can be used as an argument in other functions:

li = [12,45,78,784,2,69,1254,4785,984]
print map('the number is {}'.format,li)   


from datetime import datetime,timedelta

once_upon_a_time = datetime(2010, 7, 1, 12, 0, 0)
delta = timedelta(days=13, hours=8,  minutes=20)

gen =(once_upon_a_time +x*delta for x in xrange(20))

print 'n'.join(map('{:%Y-%m-%d %H:%M:%S}'.format, gen))

Results in:

['the number is 12', 'the number is 45', 'the number is 78', 'the number is 784', 'the number is 2', 'the number is 69', 'the number is 1254', 'the number is 4785', 'the number is 984']

2010-07-01 12:00:00
2010-07-14 20:20:00
2010-07-28 04:40:00
2010-08-10 13:00:00
2010-08-23 21:20:00
2010-09-06 05:40:00
2010-09-19 14:00:00
2010-10-02 22:20:00
2010-10-16 06:40:00
2010-10-29 15:00:00
2010-11-11 23:20:00
2010-11-25 07:40:00
2010-12-08 16:00:00
2010-12-22 00:20:00
2011-01-04 08:40:00
2011-01-17 17:00:00
2011-01-31 01:20:00
2011-02-13 09:40:00
2011-02-26 18:00:00
2011-03-12 02:20:00
Answered By: eyquem

PEP 3101 proposes the replacement of the % operator with the new, advanced string formatting in Python 3, where it would be the default.

Answered By: BrainStorm

But please be careful, just now I’ve discovered one issue when trying to replace all % with .format in existing code: '{}'.format(unicode_string) will try to encode unicode_string and will probably fail.

Just look at this Python interactive session log:

Python 2.7.2 (default, Aug 27 2012, 19:52:55) 
[GCC 4.1.2 20080704 (Red Hat 4.1.2-48)] on linux2
; s='й'
; u=u'й'
; s
; u

s is just a string (called ‘byte array’ in Python3) and u is a Unicode string (called ‘string’ in Python3):

; '%s' % s
; '%s' % u

When you give a Unicode object as a parameter to % operator it will produce a Unicode string even if the original string wasn’t Unicode:

; '{}'.format(s)
; '{}'.format(u)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
UnicodeEncodeError: 'latin-1' codec can't encode character u'u0439' in position 0: ordinal not in range(256)

but the .format function will raise “UnicodeEncodeError”:

; u'{}'.format(s)
; u'{}'.format(u)

and it will work with a Unicode argument fine only if the original string was Unicode.

; '{}'.format(u'i')

or if argument string can be converted to a string (so called ‘byte array’)

Answered By: wobmene

As I discovered today, the old way of formatting strings via % doesn’t support Decimal, Python’s module for decimal fixed point and floating point arithmetic, out of the box.

Example (using Python 3.3.5):

#!/usr/bin/env python3

from decimal import *

getcontext().prec = 50
d = Decimal('3.12375239e-24') # no magic number, I rather produced it by banging my head on my keyboard

print('%.50f' % d)



There surely might be work-arounds but you still might consider using the format() method right away.

Answered By: balu

As a side note, you don’t have to take a performance hit to use new style formatting with logging. You can pass any object to logging.debug,, etc. that implements the __str__ magic method. When the logging module has decided that it must emit your message object (whatever it is), it calls str(message_object) before doing so. So you could do something like this:

import logging

class NewStyleLogMessage(object):
    def __init__(self, message, *args, **kwargs):
        self.message = message
        self.args = args
        self.kwargs = kwargs

    def __str__(self):
        args = (i() if callable(i) else i for i in self.args)
        kwargs = dict((k, v() if callable(v) else v) for k, v in self.kwargs.items())

        return self.message.format(*args, **kwargs)

N = NewStyleLogMessage

# Neither one of these messages are formatted (or calculated) until they're
# needed

# Emits "Lazily formatted log entry: 123 foo" in log
logging.debug(N('Lazily formatted log entry: {0} {keyword}', 123, keyword='foo'))

def expensive_func():
    # Do something that takes a long time...
    return 'foo'

# Emits "Expensive log entry: foo" in log
logging.debug(N('Expensive log entry: {keyword}', keyword=expensive_func))

This is all described in the Python 3 documentation ( However, it will work with Python 2.6 as well (

One of the advantages of using this technique, other than the fact that it’s formatting-style agnostic, is that it allows for lazy values e.g. the function expensive_func above. This provides a more elegant alternative to the advice being given in the Python docs here:

Answered By: David Sanders

Yet another advantage of .format (which I don’t see in the answers): it can take object properties.

In [12]: class A(object):
   ....:     def __init__(self, x, y):
   ....:         self.x = x
   ....:         self.y = y

In [13]: a = A(2,3)

In [14]: 'x is {0.x}, y is {0.y}'.format(a)
Out[14]: 'x is 2, y is 3'

Or, as a keyword argument:

In [15]: 'x is {a.x}, y is {a.y}'.format(a=a)
Out[15]: 'x is 2, y is 3'

This is not possible with % as far as I can tell.

Answered By: matiasg

One situation where % may help is when you are formatting regex expressions. For example,

'{type_names} [a-z]{2}'.format(type_names='triangle|square')

raises IndexError. In this situation, you can use:

'%(type_names)s [a-z]{2}' % {'type_names': 'triangle|square'}

This avoids writing the regex as '{type_names} [a-z]{{2}}'. This can be useful when you have two regexes, where one is used alone without format, but the concatenation of both is formatted.

Answered By: Jorge Leitao

As of Python 3.6 (2016) you can use f-strings to substitute variables:

>>> origin = "London"
>>> destination = "Paris"
>>> f"from {origin} to {destination}"
'from London to Paris'

Note the f" prefix. If you try this in Python 3.5 or earlier, you’ll get a SyntaxError.


Answered By: Colonel Panic

For python version >= 3.6 (see PEP 498)



'albha      beta'
Answered By: Roushan

I would add that since version 3.6, we can use fstrings like the following

foo = "john"
bar = "smith"
print(f"My name is {foo} {bar}")

Which give

My name is john smith

Everything is converted to strings

mylist = ["foo", "bar"]
print(f"mylist = {mylist}")


mylist = [‘foo’, ‘bar’]

you can pass function, like in others formats method

print(f'Hello, here is the date : {time.strftime("%d/%m/%Y")}')

Giving for example

Hello, here is the date : 16/04/2018

Answered By: Sylvan LE DEUNFF

If your python >= 3.6, F-string formatted literal is your new friend.

It’s more simple, clean, and better performance.

In [1]: params=['Hello', 'adam', 42]

In [2]: %timeit "%s %s, the answer to everything is %d."%(params[0],params[1],params[2])
448 ns ± 1.48 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [3]: %timeit "{} {}, the answer to everything is {}.".format(*params)
449 ns ± 1.42 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [4]: %timeit f"{params[0]} {params[1]}, the answer to everything is {params[2]}."
12.7 ns ± 0.0129 ns per loop (mean ± std. dev. of 7 runs, 100000000 loops each)
Answered By: zhengcao

But one thing is that also if you have nested curly-braces, won’t work for format but % will work.


>>> '{{0}, {1}}'.format(1,2)
Traceback (most recent call last):
  File "<pyshell#3>", line 1, in <module>
    '{{0}, {1}}'.format(1,2)
ValueError: Single '}' encountered in format string
>>> '{%s, %s}'%(1,2)
'{1, 2}'
Answered By: U12-Forward

Python 3.6.7 comparative:

#!/usr/bin/env python
import timeit

def time_it(fn):
    Measure time of execution of a function
    def wrapper(*args, **kwargs):
        t0 = timeit.default_timer()
        fn(*args, **kwargs)
        t1 = timeit.default_timer()
        print("{0:.10f} seconds".format(t1 - t0))
    return wrapper

def new_new_format(s):
    print("new_new_format:", f"{s[0]} {s[1]} {s[2]} {s[3]} {s[4]}")

def new_format(s):
    print("new_format:", "{0} {1} {2} {3} {4}".format(*s))

def old_format(s):
    print("old_format:", "%s %s %s %s %s" % s)

def main():
    samples = (("uno", "dos", "tres", "cuatro", "cinco"), (1,2,3,4,5), (1.1, 2.1, 3.1, 4.1, 5.1), ("uno", 2, 3.14, "cuatro", 5.5),) 
    for s in samples:

if __name__ == '__main__':


new_new_format: uno dos tres cuatro cinco
0.0000170280 seconds
new_format: uno dos tres cuatro cinco
0.0000046750 seconds
old_format: uno dos tres cuatro cinco
0.0000034820 seconds
new_new_format: 1 2 3 4 5
0.0000043980 seconds
new_format: 1 2 3 4 5
0.0000062590 seconds
old_format: 1 2 3 4 5
0.0000041730 seconds
new_new_format: 1.1 2.1 3.1 4.1 5.1
0.0000092650 seconds
new_format: 1.1 2.1 3.1 4.1 5.1
0.0000055340 seconds
old_format: 1.1 2.1 3.1 4.1 5.1
0.0000052130 seconds
new_new_format: uno 2 3.14 cuatro 5.5
0.0000053380 seconds
new_format: uno 2 3.14 cuatro 5.5
0.0000047570 seconds
old_format: uno 2 3.14 cuatro 5.5
0.0000045320 seconds
Answered By: Felix Martinez