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

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

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

Is this good for you?

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

The Python 2 syntax works in Python 3 as well:

>>> class MyClass:
...     def __init__(self):
...         self.title = 'Title'
... 
>>> a = MyClass()
>>> print('The title is %(title)s' % a.__dict__)
The title is Title
>>> 
>>> path = '/path/to/a/file'
>>> print('You put your file here: %(path)s' % locals())
You put your file here: /path/to/a/file

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.

Most answers formatted only the values of the dict.

If you want to also format the key into the string you can use dict.items():

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

Output:

(‘latitude’, 41.123) (‘longitude’, 71.091)

If you want to format in an arbitry way, that is, not showing the key-values like tuples:

from functools import reduce
print("{} is {} and {} is {}".format(*reduce((lambda x, y: x + y), [list(item) for item in geopoint.items()])))

Output:

latitude is 41.123 and longitude is 71.091

geopoint = {'latitude':41.123,'longitude':71.091}

# working examples.
print(f'{geopoint["latitude"]} {geopoint["longitude"]}') # from above answer
print('{geopoint[latitude]} {geopoint[longitude]}'.format(geopoint=geopoint)) # alternate for format method  (including dict name in string).
print('%(latitude)s %(longitude)s'%geopoint) # thanks @tcll

Use format_map to do what you want

print('{latitude} {longitude}'.format_map(geopoint))

This has the advantage that

• the dictionary does not have to be blown up into parameters (compared to **geopoint) and that
• the format string only has access to the provided map and not the entire scope of variables (compared to F-strings).