The Python Oracle

How do I format a string using a dictionary in python-3.x?

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Chapters
00:00 How Do I Format A String Using A Dictionary In Python-3.X?
00:37 Answer 1 Score 43
00:43 Answer 2 Score 89
01:02 Answer 3 Score 512
01:11 Answer 4 Score 89
02:48 Thank you

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Full question
https://stackoverflow.com/questions/5952...

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Tags
#python #string #dictionary #python3x

#avk47



ANSWER 1

Score 512


Is this good for you?

geopoint = {'latitude':41.123,'longitude':71.091}
print('{latitude} {longitude}'.format(**geopoint))



ANSWER 2

Score 89


To unpack a dictionary into keyword arguments, use **. Also,, new-style formatting supports referring to attributes of objects and items of mappings:

'{0[latitude]} {0[longitude]}'.format(geopoint)
'The title is {0.title}s'.format(a) # the a from your first example



ANSWER 3

Score 89


As Python 3.0 and 3.1 are EOL'ed and no one uses them, you can and should use str.format_map(mapping) (Python 3.2+):

Similar to str.format(**mapping), except that mapping is used directly and not copied to a dict. This is useful if for example mapping is a dict subclass.

What this means is that you can use for example a defaultdict that would set (and return) a default value for keys that are missing:

>>> from collections import defaultdict
>>> vals = defaultdict(lambda: '<unset>', {'bar': 'baz'})
>>> 'foo is {foo} and bar is {bar}'.format_map(vals)
'foo is <unset> and bar is baz'

Even if the mapping provided is a dict, not a subclass, this would probably still be slightly faster.

The difference is not big though, given

>>> d = dict(foo='x', bar='y', baz='z')

then

>>> 'foo is {foo}, bar is {bar} and baz is {baz}'.format_map(d)

is about 10 ns (2 %) faster than

>>> 'foo is {foo}, bar is {bar} and baz is {baz}'.format(**d)

on my Python 3.4.3. The difference would probably be larger as more keys are in the dictionary, and


Note that the format language is much more flexible than that though; they can contain indexed expressions, attribute accesses and so on, so you can format a whole object, or 2 of them:

>>> p1 = {'latitude':41.123,'longitude':71.091}
>>> p2 = {'latitude':56.456,'longitude':23.456}
>>> '{0[latitude]} {0[longitude]} - {1[latitude]} {1[longitude]}'.format(p1, p2)
'41.123 71.091 - 56.456 23.456'

Starting from 3.6 you can use the interpolated strings too:

>>> f'lat:{p1["latitude"]} lng:{p1["longitude"]}'
'lat:41.123 lng:71.091'

You just need to remember to use the other quote characters within the nested quotes. Another upside of this approach is that it is much faster than calling a formatting method.




ANSWER 4

Score 43


print("{latitude} {longitude}".format(**geopoint))