How can I use functions that return multiple values as inputs to a format string, without getting the TypeError: not enough arguments for format string error?
>>> def foo():
... return 1, 2
...
>>> foo()
(1, 2)
>>> print "%d,%d,%d" % foo(), 3
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: not enough arguments for format string
Expected output: "1,2,3"
string.format() is more powerful and lets you access list elements or even attributes:
>>> print "{0[0]},{0[1]},{1}".format(foo(), 3)
1,2,3
The problem with foo(), 3 is that it's a tuple and an integer, two different types. Instead you can create a 3-tuple via concatenation. If you use string.format() it's a bit trickier, as you need to unpack it first using the * operator so you can use it as arguments:
>>> foo() + (3,)
(1, 2, 3)
>>> print "%d,%d,%d" % (foo() + (3,))
1,2,3
>>> print "{},{},{}".format(*foo() + (3,))
1,2,3
Of course, you can always do it this way, which is obvious but verbose:
>>> foo1, foo2 = foo()
>>> print "%d,%d,%d" % (foo1, foo2, 3)
1,2,3
(foo() + (3,)) in your answer. :/The reason print "%d,%d,%d" % foo(), 3 is not working is because Python is thinking you're trying to print two items here: "%d,%d,%d" % foo() and then 3. Due to this the first expression clearly fails because of in-sufficient number of items:
>>> def func():
... print "%d,%d,%d" % foo(), 3
...
>>> import dis
>>> dis.dis(func)
2 0 LOAD_CONST 1 ('%d,%d,%d')
3 LOAD_GLOBAL 0 (foo)
6 CALL_FUNCTION 0
9 BINARY_MODULO
10 PRINT_ITEM
11 LOAD_CONST 2 (3)
14 PRINT_ITEM
15 PRINT_NEWLINE
16 LOAD_CONST 0 (None)
19 RETURN_VALUE
In fact even something like "%d,%d,%d" % foo() + (3,) won't work because I guess operator precedence comes into play here:
>>> def func():
print "%d,%d,%d" % foo() + (3,)
...
>>> dis.dis(func)
2 0 LOAD_CONST 1 ('%d,%d,%d')
3 LOAD_GLOBAL 0 (foo)
6 CALL_FUNCTION 0
9 BINARY_MODULO
10 LOAD_CONST 3 ((3,))
13 BINARY_ADD
14 PRINT_ITEM
15 PRINT_NEWLINE
16 LOAD_CONST 0 (None)
19 RETURN_VALUE
A working version will be:
>>> print "%d,%d,%d" % (foo() + (3,))
1,2,3
A general solution to take an arbitrary number of arguments and print them with some delimiter would be to use str.join
def foo():
return 1, 2
>>> print(','.join(map(str,foo())))
1,2
Another example
def bar():
return 4,5,6,7
>>> print(','.join(map(str,bar())))
4,5,6,7
print "%d,%d,%d" % (foo() + (3,))should do it.