Equivalent of Ruby nil array in Python

Question

In Ruby an empty array returns a nil if the first element is accessed:

2.1.2 :005 > arr = []
  => [] 
2.1.2 :006 > arr.first
  => nil 
2.1.2 :007 > arr[0]
  => nil

I know Python lists are not exactly the same as Ruby arrays, but if you try the equivalent action, you get an error:

>>> l = []
>>> l[0]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>

How can I instead of an error get a False or None value?

Ideally I'd like an assignment like this to resolve to None / False if the function call returns nothing, but with an inline check rather than a if block.

tp_journal = self.pool.get('account.journal').browse(cr, uid, tp_ids, context=context)[0]

((expression) or [None])[0] would be a simple hack to achieve that. — Bakuriu
– Bakuriu, Commented Sep 16, 2014 at 14:14

Bakuriu · Accepted Answer · 2014-09-16 14:29:48Z

There is no simple way to achieve what you want. The list operations raise an error, period.

However you can work around by:

Accessing something else when the list is empty
Not using indexing

In the first case you can use the fact that empty lists are false and hence (expression) or [None] will evaluate to [None] when the expression returns an empty list, and hence:

((expression) or [None])[0]

The alternative is to avoid using indexing but using iteration and next to get the first element:

next(iter(expression), None)

Unfortunately you must explicitly call iter to get a list iterator.

To extend these methods to any index you can use slicing instead of simple indexing:

((expression)[index:] or [None])[0]

Or:

next(islice(expression, index, None), None)
# or
next(iter(expression[index:]), None)

(where islice is itertools.islice).

This works because slicing doesn't raise errors for out-of-bounds indexes.

Hoever my preferred alternative would be to use a function, if you use that often in your code:

def get_with_default(seq, index):
    try:
        return seq[index]
    except LookupError:
        return None

Kevin · Accepted Answer · 2014-09-16 14:15:01Z

1

Perhaps you could make use of the short-circuiting nature of the or operator.

seq = [23,42]
print (seq or [None])[0]
#result: 23

seq = []
print (seq or [None])[0]
#result: None

In the above code, seq or [None] evaluates to seq if seq contains any elements, and it evaluates to [None] if seq is empty. Then [0] accesses the first element of whichever list is returned.

answered Sep 16, 2014 at 14:15

Kevin

76.5k13 gold badges141 silver badges168 bronze badges

2 Comments

Max Over a year ago

I think the asker wants this to work for any index that would raise an IndexError, not just for empty lists. At least, that's how it works in Ruby.

port5432 Over a year ago

arr.first or arr[0] was what I put. @Max you have more clearly expressed what I was trying to say.

Collectives™ on Stack Overflow

Equivalent of Ruby nil array in Python

2 Answers 2

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related