[python] Why doesn't list have safe "get" method like dictionary?

Why doesn't list have a safe "get" method like dictionary?

>>> d = {'a':'b'}
>>> d['a']
'b'
>>> d['c']
KeyError: 'c'
>>> d.get('c', 'fail')
'fail'

>>> l = [1]
>>> l[10]
IndexError: list index out of range

A reasonable thing you can do is to convert the list into a dict and then access it with the get method:

>>> my_list = ['a', 'b', 'c', 'd', 'e']
>>> my_dict = dict(enumerate(my_list))
>>> print my_dict
{0: 'a', 1: 'b', 2: 'c', 3: 'd', 4: 'e'}
>>> my_dict.get(2)
'c'
>>> my_dict.get(10, 'N/A')

Instead of using .get, using like this should be ok for lists. Just a usage difference.

>>> l = [1]
>>> l[10] if 10 < len(l) else 'fail'
'fail'

This works if you want the first element, like my_list.get(0)

>>> my_list = [1,2,3]
>>> next(iter(my_list), 'fail')
1
>>> my_list = []
>>> next(iter(my_list), 'fail')
'fail'

I know it's not exactly what you asked for but it might help others.

Try this:

>>> i = 3
>>> a = [1, 2, 3, 4]
>>> next(iter(a[i:]), 'fail')
4
>>> next(iter(a[i + 1:]), 'fail')
'fail'

Your usecase is basically only relevant for when doing arrays and matrixes of a fixed length, so that you know how long they are before hand. In that case you typically also create them before hand filling them up with None or 0, so that in fact any index you will use already exists.

You could say this: I need .get() on dictionaries quite often. After ten years as a full time programmer I don't think I have ever needed it on a list. :)

Probably because it just didn't make much sense for list semantics. However, you can easily create your own by subclassing.

class safelist(list):
    def get(self, index, default=None):
        try:
            return self.__getitem__(index)
        except IndexError:
            return default

def _test():
    l = safelist(range(10))
    print l.get(20, "oops")

if __name__ == "__main__":
    _test()

So I did some more research into this and it turns out there isn't anything specific for this. I got excited when I found list.index(value), it returns the index of a specified item, but there isn't anything for getting the value at a specific index. So if you don't want to use the safe_list_get solution which I think is pretty good. Here are some 1 liner if statements that can get the job done for you depending on the scenario:

>>> x = [1, 2, 3]
>>> el = x[4] if len(x) > 4 else 'No'
>>> el
'No'

You can also use None instead of 'No', which makes more sense.:

>>> x = [1, 2, 3]
>>> i = 2
>>> el_i = x[i] if len(x) == i+1 else None

Also if you want to just get the first or last item in the list, this works

end_el = x[-1] if x else None

You can also make these into functions but I still liked the IndexError exception solution. I experimented with a dummied down version of the safe_list_get solution and made it a bit simpler (no default):

def list_get(l, i):
    try:
        return l[i]
    except IndexError:
        return None

Haven't benchmarked to see what is fastest.

Dictionaries are for look ups. It makes sense to ask if an entry exists or not. Lists are usually iterated. It isn't common to ask if L[10] exists but rather if the length of L is 11.

For small index values you can implement

my_list.get(index, default)

as

(my_list + [default] * (index + 1))[index]

If you know in advance what index is then this can be simplified, for example if you knew it was 1 then you could do

(my_list + [default, default])[index]

Because lists are forward packed the only fail case we need to worry about is running off the end of the list. This approach pads the end of the list with enough defaults to guarantee that index is covered.

This guy worked for me:

list_get = lambda l, x: l[x:x+1] and l[x] or 0

lambdas are great for one liner helper functions like this

Credits to jose.angel.jimenez and Gus Bus.

For the "oneliner" fans…

If you want the first element of a list or if you want a default value if the list is empty try:

liste = ['a', 'b', 'c']
value = (liste[0:1] or ('default',))[0]
print(value)

returns a

and

liste = []
value = (liste[0:1] or ('default',))[0]
print(value)

returns default

Examples for other elements…

liste = ['a', 'b', 'c']
print(liste[0:1])  # returns ['a']
print(liste[1:2])  # returns ['b']
print(liste[2:3])  # returns ['c']
print(liste[3:4])  # returns []

With default fallback…

liste = ['a', 'b', 'c']
print((liste[0:1] or ('default',))[0])  # returns a
print((liste[1:2] or ('default',))[0])  # returns b
print((liste[2:3] or ('default',))[0])  # returns c
print((liste[3:4] or ('default',))[0])  # returns default

Possibly shorter:

liste = ['a', 'b', 'c']
value, = liste[:1] or ('default',)
print(value)  # returns a

It looks like you need the comma before the equal sign, the equal sign and the latter parenthesis.

More general:

liste = ['a', 'b', 'c']
f = lambda l, x, d: l[x:x+1] and l[x] or d
print(f(liste, 0, 'default'))  # returns a
print(f(liste, 1, 'default'))  # returns b
print(f(liste, 2, 'default'))  # returns c
print(f(liste, 3, 'default'))  # returns default

Tested with Python 3.6.0 (v3.6.0:41df79263a11, Dec 22 2016, 17:23:13)