"TypeError: not enough arguments for format string " using %s in Python 3.6

Question

I'm trying to create and play with classes in Python 3.6 but I'm getting the following error when I try to call a method that prints info about my class:

'TypeError: not enough arguments for format string'

And here's the code I'm trying to run:

class Restaurant(object):

    def __init__(self, name, type):
        self.name = name
        self.type = type

    def describe(self):
        print("Restaurant name: %s , Cuisine type: %s" % self.name, self.type)

    def open_restaurant(self):
        print("%s is now open!" % self.name)


my_restaurant = Restaurant("Domino", "Pizza")

my_restaurant.describe()
my_restaurant.open_restaurant()

The open_restaurant() method works fine but with my_restaurant.describe() I receive the error message I mentioned.

The syntax for having multiple arguments after % is % (self.name, self.type) — UnholySheep
– UnholySheep, Commented Feb 2, 2018 at 13:09
good answers already given. But please do not use type as a variable name. — Edwin van Mierlo
– Edwin van Mierlo, Commented Feb 2, 2018 at 13:26

James Schinner · Accepted Answer · 2018-02-02 13:08:58Z

2

Seeing as you are using python3.6. f-strings are much nicer.

class Restaurant(object):


    def __init__(self, name, type):
        self.name = name
        self.type = type

    def describe(self):
        print(f"Restaurant name: {self.name}, Cuisine type: {self.type}")

    def open_restaurant(self):
        print(f"{self.name} is now open!")

my_restaurant = Restaurant("Domino", "Pizza")

my_restaurant.describe()
my_restaurant.open_restaurant()

Restaurant name: Domino, Cuisine type: Pizza
Domino is now open!

answered Feb 2, 2018 at 13:08

James Schinner

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1 Comment

J. Devez Over a year ago

Indeed much more elegant! Thanks for the quick answer.

ColonelFazackerley · Accepted Answer · 2018-02-02 13:22:06Z

2

This works (note the tuple to the right of "%")

print("Restaurant name: %s , Cuisine type: %s" % (self.name, self.type))

You might consider the newer form

print("Restaurant name: {} , Cuisine type: {}".format(self.name, self.type))

answered Feb 2, 2018 at 13:22

ColonelFazackerley

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Comments

John Zwinck · Accepted Answer · 2018-02-02 13:09:06Z

1

You need to pass a tuple:

print("Restaurant name: %s , Cuisine type: %s" % (self.name, self.type))

But really, the % type of string formatting has been mostly obsolete in Python for many years, and you probably should not be using it in Python 3.6. Instead: try this:

print("Restaurant name: {} , Cuisine type: {}".format(self.name, self.type))

Or of course:

print("Restaurant name:", self.name, ", Cuisine type:", self.type)

answered Feb 2, 2018 at 13:09

John Zwinck

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6 Comments

FHTMitchell Over a year ago

This is bad advice. Don't do this: print("Restaurant name:", self.name, ", Cuisine type:", self.type), it's so easy to make mistakes (like you have, there is an extra space before the comma). In python 3.6 you should be using f-strings (or .format for setting up a string template).

J. Devez Over a year ago

Worked! Thank you very much

J. Devez Over a year ago

@FHTMitchell I think I will use print(f"Restaurant name: {self.name} Cuisine type: {self.type}")

chepner Over a year ago

Upvoting despite the bad last suggestion, if only because it addresses the problem in the question in addition to suggesting the newer format method.

John Zwinck Over a year ago

@FHTMitchell: You claim I have added an extra space before the comma. But you will see that "extra" space is not extra at all, it is present in the original code from the question, and is not an accident (at least, not on my part, I was only replicating what the question writer did).

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