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Type (3) returns an integer instead of a tuple in python, why?

type(3,) returns type int, and

t = 3,
type(t)

returns the type of tuple. Why?

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4 answers

In parentheses that make up the function call operator, the comma is not intended for constructing tuples, but for separating arguments. So it is type(3, )equivalent type(3). An extra comma at the end of the argument list is allowed by the grammar. To create a tuple, you need an additional pair of pairs:

>>> def f(x):
...     print x
... 
>>> f(3)
3
>>> f(3,)
3
>>> f((3,))
(3,)
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Inline type()is a function, so the comma is parsed as an argument delimiter, not a tuple constructor.

>>> type(3,)
<type 'int'>

>>> type((3,))
<type 'tuple'>
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, Python :

def f (a):
    print a
    print type(a)

>>> f(3,)
3
<type 'int'>

- , . - :

>>> [a,a*2 for a in range(4)]
  File "<stdin>", line 1
    [a,a*2 for a in range(4)]
             ^

:

>>> [(a,a*2) for a in range(4)]
[(0, 0), (1, 2), (2, 4), (3, 6)]

, , , :

>>> [(a,b) for a, b in zip(range(4),range(4))]
[(0, 0), (1, 1), (2, 2), (3, 3)]
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As func(bunch_of_args)you are allowed to follow the last argument with a comma, as in the

alist = [1, 2, 3, ]
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