python - How to format variable number of arguments into a string?

Question

We know that formatting one argument can be done using one %s in a string:

>>> "Hello %s" % "world"
'Hello world'

for two arguments, we can use two %s (duh!):

>>> "Hello %s, %s" % ("John", "Joe")
'Hello John, Joe'

So, how can I format a variable number of arguments without having to explicitly define within the base string a number of %s equal to the number of arguments to format? it would be very cool if something like this exists:

>>> "Hello <cool_operator_here>" % ("John", "Joe", "Mary")
Hello JohnJoeMary
>>> "Hello <cool_operator_here>" % ("John", "Joe", "Mary", "Rick", "Sophie")
Hello JohnJoeMaryRickSophie

Is this even possible or the only thing I could do about it is to do something like:

>>> my_args = ["John", "Joe", "Mary"]
>>> my_str = "Hello " + ("".join(["%s"] * len(my_args)))
>>> my_str % tuple(my_args)
"Hello JohnJoeMary"

NOTE: I need to do it with the %s string formatting operator.

UPDATE:

It needs to be with the %s because a function from another library formats my string using that operator given that I pass the unformatted string and the args to format it, but it makes some checking and corrections (if needed) on the args before actually making the formatting.

So I need to call it:

>>> function_in_library("Hello <cool_operator_here>", ["John", "Joe", "Mary"])
"Hello JohnJoeMary"

Thanks for your help!

Why do you need to use the %s operator? That makes little sense; use str() instead. — Martijn Pieters
– Martijn Pieters, Commented Aug 22, 2013 at 21:23
How would the "cool operator here" specify how to connect up the values? You don't want it to just print out Hello JohnJoeMary, or Hello ("John", "Joe", "Mary"), do you? You might want Hello John, Joe, Mary or Hello John, Joe, and Mary or Hello John, Joe and Mary, or a million other things; there's no way a single operator could let you do all of those. — abarnert
– abarnert, Commented Aug 22, 2013 at 21:30
You should still be able to call it with function_in_library("Hello %s", ', '.join(["John", "Joe", "Mary"]) — Oli
– Oli, Commented Aug 22, 2013 at 21:37
Sorry, I explained the library function wrong. Anyway, I may not join the arguments before passing them to the function :\ — Gerard
– Gerard, Commented Aug 22, 2013 at 21:44
I suppose f-strings are now the best way to handle this, since they allow for function evaluation in the string literal — Mackie Messer
– Mackie Messer, Commented Oct 12, 2023 at 19:50

Martijn Pieters · Accepted Answer · 2013-08-22 22:07:33Z

29

You'd use str.join() on the list without string formatting, then interpolate the result:

"Hello %s" % ', '.join(my_args)

Demo:

>>> my_args = ["foo", "bar", "baz"]
>>> "Hello %s" % ', '.join(my_args)
'Hello foo, bar, baz'

If some of your arguments are not yet strings, use a list comprehension:

>>> my_args = ["foo", "bar", 42]
>>> "Hello %s" % ', '.join([str(e) for e in my_args])
'Hello foo, bar, 42'

or use map(str, ...):

>>> "Hello %s" % ', '.join(map(str, my_args))
'Hello foo, bar, 42'

You'd do the same with your function:

function_in_library("Hello %s", ', '.join(my_args))

If you are limited by a (rather arbitrary) restriction that you cannot use a join in the interpolation argument list, use a join to create the formatting string instead:

function_in_library("Hello %s" % ', '.join(['%s'] * len(my_args)), my_args)

edited Aug 22, 2013 at 22:07

answered Aug 22, 2013 at 21:24

Martijn Pieters

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6 Comments

Gerard Over a year ago

Though you are right, I think I didn't explain my question clear enough. I missed to state that I'm not making the formatting, an existing function is. Sorry, but thanks. I'll update my question.

Gerard Over a year ago

Sorry I may not join the arguments, each one needs to be processed by the library function before formatting. Thanks.

Martijn Pieters Over a year ago

That is a rather arbitrary restriction; in that case build the %s string elements using a separate join: "Hello %s" % ', '.join(['%s'] * len(my_args)). Note the irony here; you are creating the string using the exact same technique.

Gerard Over a year ago

Yeah, I use that example in my question, but I say I'd rather not to do so if there was another way around. Basically my question is if a somehow-magic operator even exists or I if must build the string myself? But I guess the second is the answer I'm getting :/ Sorry if this sounds rude, I don't mean to.

Martijn Pieters Over a year ago

You must build the string yourself.

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