Getting a default value on index out of range in Python [duplicate]

Answer 1

In the non-Python spirit of "ask for permission, not forgiveness", here's another way:

b = a[4] if len(a) > 4 else 'sss'

Answer 2

You could also define a little helper function for these cases:

def default(x, e, y):
    try:
        return x()
    except e:
        return y

It returns the return value of the function x, unless it raised an exception of type e; in that case, it returns the value y. Usage:

b = default(lambda: a[4], IndexError, 'sss')

Edit: Made it catch only one specified type of exception.

Suggestions for improvement are still welcome!

Answer 3

try:
    b = a[4]
except IndexError:
    b = 'sss'

A cleaner way (only works if you're using a dict):

b = a.get(4,"sss") # exact same thing as above

Here's another way you might like (again, only for dicts):

b = a.setdefault(4,"sss") # if a[4] exists, returns that, otherwise sets a[4] to "sss" and returns "sss"

Answer 4

For a common case where you want the first element, you can do

next(iter([1, 2, 3]), None)

I use this to "unwrap" a list, possibly after filtering it.

next((x for x in [1, 3, 5] if x % 2 == 0), None)

or

cur.execute("SELECT field FROM table")
next(cur.fetchone(), None)

Answer 5

In the Python spirit of "ask for forgiveness, not permission", here's one way:

try:
    b = a[4]
except IndexError:
    b = 'sss'

Answer 6

I’m all for asking permission (i.e. I don’t like the try…except method). However, the code gets a lot cleaner when it’s encapsulated in a method:

def get_at(array, index, default):
    if index < 0: index += len(array)
    if index < 0: raise IndexError('list index out of range')
    return array[index] if index < len(a) else default

b = get_at(a, 4, 'sss')