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List comprehension in format string? (Python)

Let's say I have a list datalist, with len(datalist) = 4. Let's say I want each of the elements in the list to be represented in a string in this way:

s = "'{0}'::'{1}'::'{2}' '{3}'\n".format(datalist[0], datalist[1], datalist[2], datalist[3])

I don't like having to type datalist[index] so many times, and feel like there should be a more effective way. I tried:

s = "'{0}'::'{1}'::'{2}' '{3}'\n".format(datalist[i] for i in range(4))

But this doesn't work:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: tuple index out of range

Does anybody know a functioning way to achieve this efficiently and concisely?

like image 888
Sahand Avatar asked Aug 23 '17 20:08

Sahand


1 Answers

Yes, use argument unpacking with the "splat" operator *:

>>> s = "'{0}'::'{1}'::'{2}' '{3}'\n"
>>> datalist = ['foo','bar','baz','fizz']
>>> s.format(*datalist)
"'foo'::'bar'::'baz' 'fizz'\n"
>>>

Edit

As pointed out by @AChampion you can also just use indexing inside the format string itself:
>>> "'{0[0]}'::'{0[1]}'::'{0[2]}' '{0[3]}'\n".format(datalist)
"'foo'::'bar'::'baz' 'fizz'\n"
like image 180
juanpa.arrivillaga Avatar answered Sep 22 '22 04:09

juanpa.arrivillaga