How to use a tuple with string.format with out splitting beforehand

Question

I've this working code:

', '.join('%s=%r' % item for item in list)

where list is list of tuples ('key', 'value').

I want to refactore the code to use the new style string.format. I've tried it this way:

', '.join(['{!s}={!r}'.format(item) for item in list])

This results in an IndexError: tuple index out of range. Are there any better solutions than:

arguments = []
for key, value in list:
    arguments.append('{!s}={!r}'.format(key, value))

', '.join(arguments)

Context:

def __repr__(self):
    """
    This method returns a string representation of the object such that eval() can recreate the object.
    The class attributes will be ordered
    e.g. Class(attribute1='String', attribute2=3)
    """
    list =[]
    for item in vars(self).items():
        list.append(item)
    list.sort(key=lambda tup: tup[0])

    arguments = []
    for key, value in list:
        arguments.append('{!s}={!r}'.format(key, value))

    return '{!s}({!s})'.format(
        type(self).__name__,
        ', '.join(arguments)
    )

`', '.join(['{!s}={!r}'.format(*item) for item in list])` try unpacking the item with the `*` operator. — Gábor Fekete, Jan 02 '18 at 13:26
https://stackoverflow.com/questions/26758341/python-3-using-tuples-in-str-format — Ilja Everilä, Jan 02 '18 at 13:27

score 1 · Accepted Answer · answered Jan 02 '18 at 13:26

1

Unpack your tuple.

>>> t = ('a', 'b')
>>> '{!s}={!r}'.format(*t)
"a='b'"

answered Jan 02 '18 at 13:26

timgeb

76,762
20
123
145

score 1 · Answer 2 · answered Jan 02 '18 at 13:30

1

What about ', '.join(['{!s}={!r}'.format(item[0], item[1]) for item in list]) ?

answered Jan 02 '18 at 13:30

Giovanni Rescia

605
5
15

How to use a tuple with string.format with out splitting beforehand

2 Answers2