first non-null element in a list, otherwise None

[wheres pythonmonks]

This should be trivial:


I am looking to extract the first non-None element in a list, and
"None" otherwise. Here's one implementation:

x = reduce(lambda x,y: x or y, [None,None,1,None,2,None], None)
print x

1

I thought maybe a generator expression would be better, to prevent
iterating over the whole list:

x = ( x for x in [None,1,2] if x is not None ).next()
print x

1

However, the generator expression throws if the list is entirely None.

With list comprehensions, a solution is:

x = ([ x for x in [None,1,2] if x is not None ] + [ None ] )[0]

But this can be expensive memory wise. Is there a way to concatenate
generator expressions?

More importantly,

Is there a better way? (In one line?)

Thanks,

W

 

[Peter Otten]

I am looking to extract the first non-None element in a list, and
"None" otherwise. Here's one implementation:

x = reduce(lambda x,y: x or y, [None,None,1,None,2,None], None)
print x

1

I thought maybe a generator expression would be better, to prevent
iterating over the whole list:

x = ( x for x in [None,1,2] if x is not None ).next()
print x

1

However, the generator expression throws if the list is entirely None.

The next() builtin (new in Python 2.6) allows you to provide a default:

next((item for item in [None, None, "found"] if item is not None), "no

match")
'found'

next((item for item in [None, None, None] if item is not None), "no

match")
'no match'

With list comprehensions, a solution is:

x = ([ x for x in [None,1,2] if x is not None ] + [ None ] )[0]

But this can be expensive memory wise. Is there a way to concatenate
generator expressions?

itertools.chain()

Peter

 

[wheres pythonmonks]

itertools.chain()

Aha!

import itertools

x = itertools.chain( (x for x in [None,None] if x is not None), [ None ] ).next()
print x None
x = itertools.chain( (x for x in [None,7] if x is not None), [ None ] ).next()
print x 7

W

 

[Arnaud Delobelle]

This should be trivial:

I am looking to extract the first non-None element in a list, and
"None" otherwise.  Here's one implementation:

x = reduce(lambda x,y: x or y, [None,None,1,None,2,None], None)
print x

1

I thought maybe a generator expression would be better, to prevent
iterating over the whole list:

x = ( x for x in [None,1,2] if x is not None ).next()
print x

1

However, the generator expression throws if the list is entirely None.

With list comprehensions, a solution is:

x = ([ x for x in [None,1,2] if x is not None ] + [ None ] )[0]

But this can be expensive memory wise.  Is there a way to concatenate
generator expressions?

More importantly,

Is there a better way?  (In one line?)

Thanks,

W

Just for fun:

print min([None, 2, None, None, 1], key=lambda x: x is None) 2
print min([None, None, None], key=lambda x: x is None)

None

Looks clever but:
min([], key=lambda x: x is None) throws an exception.