Turn slice into range

Question

I'm using Python 3.3. I want to get a slice object and use it to make a new range object.

It goes something like that:

>>> class A:
    def __getitem__(self, item):
        if isinstance(item, slice):
            return list(range(item.start, item.stop, item.step))

>>> a = A()
>>> a[1:5:2] # works fine
[1, 3]
>>> a[1:5] # won't work :(
Traceback (most recent call last):
  File "<pyshell#18>", line 1, in <module>
    a[1:5] # won't work :(
  File "<pyshell#9>", line 4, in __getitem__
    return list(range(item.start, item.stop, item.step))
TypeError: 'NoneType' object cannot be interpreted as an integer

Well, the problem is obvious here - range doesn't accept None as a value:

>>> range(1, 5, None)
Traceback (most recent call last):
  File "<pyshell#19>", line 1, in <module>
    range(1, 5, None)
TypeError: 'NoneType' object cannot be interpreted as an integer

But what is not obvious (to me) is the solution. How will I call range so it will work in every case? I'm searching for a nice pythonic way to do it.

Answer 1

There's an easier way to do this (at least in 3.4, I don't have 3.3 at the moment, and I don't see it in the changelog).

Assuming your class already has a known length you can just slice a range of that size:

>>> range(10)[1:5:2]
range(1, 5, 2)
>>> list(range(10)[1:5:2])
[1, 3]

If you don't know the length a priori you'll have to do:

>>> class A:
    def __getitem__(self, item):
        if isinstance(item, slice):
            return list(range(item.stop)[item])
>>> a = A()
>>> a[1:5:2]
[1, 3]
>>> a[1:5]
[1, 2, 3, 4]

Answer 2

Try

class A:
    def __getitem__(self, item):
        ifnone = lambda a, b: b if a is None else a
        if isinstance(item, slice):
            if item.stop is None:
                # do something with itertools.count()
            else:
                return list(range(ifnone(item.start, 0), item.stop, ifnone(item.step, 1)))
        else:
            return item

This will reinterpret .start and .step appropriately if they are None.

---

Another option could be the .indices() method of a slice. It is called with the number of entries and reinterprets None to the appropriate values and wraps negative values around the given length parameter:

>>> a=slice(None, None, None)
>>> a.indices(1)
(0, 1, 1)
>>> a.indices(10)
(0, 10, 1)
>>> a=slice(None, -5, None)
>>> a.indices(100)
(0, 95, 1)

It depends what you intend to do with negative indices...

Answer 3

The problem:

A slice consists of start, stop, and step parameters and can be created with either slice notation or using the slice built-in. Any (or all) of the start, stop, and step parameters can be None.

# valid
sliceable[None:None:None]

# also valid
cut = slice(None, None, None)
sliceable[cut]

However, as pointed out in the original question, the range function does not accept None arguments. You can get around this in various ways...

The solutions

With conditional logic:

if item.start None:
    return list(range(item.start, item.stop))
return list(range(item.start, item.stop, item.step))

...which can get unnecessarily complex since any or all of the parameters may be None.

With conditional variables:

start = item.start if item.start is None else 0
step = item.step if item.step is None else 1
return list(range(item.start, item.stop, item.step))

... which is explicit, but a little verbose.

With conditionals directly in the statement:

return list(range(item.start if item.start else 0, item.stop, item.step if item.step else 1))

... which is also unnecessarily verbose.

With a function or lambda statement:

ifnone = lambda a, b: b if a is None else a
range(ifnone(item.start, 0), item.stop, ifnone(item.step, 1)

...which can be difficult to understand.

With 'or':

return list(range(item.start or 0, item.stop or len(self), item.step or 1))

I find using or to assign sensible default values the simplest. It's explicit, simple, clear, and concise.

Rounding out the implementation

To complete the implementation you should also handle integer indexes (int, long, etc) by checking isinstance(item, numbers.Integral) (see int vs numbers.Integral).

Define __len__ to allow for using len(self) for a default stop value.

Finally raise an appropriate TypeError for invalid indexes (e.g. strings, etc).

TL;DR;

class A:
    def __len__(self):
        return 0

    def __getitem__(self, item):
        if isinstance(item, numbers.Integral):  # item is an integer
            return item
        if isinstance(item, slice):  # item is a slice
            return list(range(item.start or 0, item.stop or len(self), item.step or 1))
        else:  # invalid index type
            raise TypeError('{cls} indices must be integers or slices, not {idx}'.format(
                cls=type(self).__name__,
                idx=type(item).__name__,
            ))

Answer 4

All of the other answers here are completely missing it.

You simply cannot convert a slice into a range in the general case.

There isn't enough information. You need to know the length of the list (or other sequence type) that you are trying to slice.

Once you have that, you can create a range easily in Python 3 using slice.indices()

Following the example you provided:

class A:
    def __init__(self, mylist):
        self.mylist = mylist

    def __getitem__(self, item):
        if isinstance(item, slice):
            mylen = len(self.mylist)
            return list(range(*item.indices(mylen)))

mylist = [1, 2, 'abc', 'def', 3, 4, None, -1]
a = A(mylist)
a[1:5]  # produces [1, 2, 3, 4]

Answer 5

I would special-case the item.step is None branch:

def __getitem__(self, item):
    if isinstance(item, slice):
        if item.step is None:
            return list(range(item.start, item.stop))
        return list(range(item.start, item.stop, item.step))

and you'll handle ranges that need to count down correctly.

Answer 6

In your last example a[1:5], item.step == None and you are trying to do range(1, 5, None), which of course causes the error. Fast way to fix:

class A:
    def __getitem__(self, item):
        if isinstance(item, slice):
            return list(range(item.start, item.stop, item.step if item.step else 1)) #Changed line!

But it is just to show you your problem. It is not the best approach.

Answer 7

What about something like this?

>>> class A:
def __getitem__(self, item):
    if isinstance(item, slice):
        return list(range(item.start, item.stop, item.step if item.step else 1))

>>> a = A()
>>> a[1:5:2] # works fine
[1, 3]
>>> a[1:5] # works as well :)
[1, 2, 3, 4]