testing for valid reference: obj vs. None!=obs vs. obj is not None

[alf]

Hi,

I have a reference to certain objects. What is the most pythonic way to
test for valid reference:

if obj:

if None!=obs:

if obj is not None:

--
alfz1

[Carl Banks]

alf wrote:

Hi,

I have a reference to certain objects. What is the most pythonic way to
test for valid reference:

if obj:

if None!=obs:

if obj is not None:

If you're checking whether an object is None or not, the third is the
best way.

Some people might say you should use the first: this works sometimes,
but sometimes an object that is not None can be false, in which case
this test will fail. (I'll give you an example of one place where it
bit me: in ElementTree. An Element is false if there are no
subelements, thus to test whether a certain element exist, you can't
just say "if aaa.find('bbb') ", because a bbb tag exists, it is false
unless it has subelements of its own. You must instead say "if
aaa.find('bbb') is not None".)

The second test is slower than the first and adds nothing. Whether
something is None is an identity test; identity tests should use is.
Carl Banks

[Bruno Desthuilliers]

alf a écrit :

Hi,

I have a reference to certain objects. What is the most pythonic way to
test for valid reference:

if obj:

Don't do this:

for o in [0, '', [], {}, ()]:
print obj, bool(obj), obj is None

if None!=obs:

In python, assignement is a statement, not an expression, so there's no
way you could write 'if obj = None' by mistake (-syntax error). So
this style is unpythonic. Also, None is a singleton, and identity test
is way faster than equality test.

if obj is not None:

That's the good one.

[BJörn Lindqvist]

I have a reference to certain objects. What is the most pythonic way to

test for valid reference:

if obj:

if None!=obs:

if obj is not None:

The third way is the most precise way. It is often used in combination
with default arguments.

def __init__(self, amount = None):
if amount is not None:
self.amount = amount
else:
self.amount = self.calc_amoun t()

However, the first way is shorter and more concise. It really depends
on what obj is supposed to be and where you get obj from. If it is a
database and obj is an instance:

obj = db.get("sometab le", id = 33)

(assuming db.get returns None if the object isn't found) Then "if
obj:" clearly is the right test. In general, I think "If obj:" is the
right answer, except when dealing with default arguments. And you must
be aware that some objects, like 0, [] or {} evaluate to False -- it
is possible that that could create some subtle bugs.

--
mvh Björn

[Sandra-24]

alf wrote:

Hi,

I have a reference to certain objects. What is the most pythonic way to
test for valid reference:

if obj:

if None!=obs:

if obj is not None:

I like this way the most. I used timeit to benchmark this against the
first one, expecting it to be faster (the first is a general false
test, the last should just be comparing pointers) but it's slower.
Still I don't expect that to ever matter, so I use it wherever I wish
to test for None, it reads the best of all of them.

-Sandra

[Carl Banks]

Bruno Desthuilliers wrote:

In python, assignement is a statement, not an expression, so there's no
way you could write 'if obj = None' by mistake (-syntax error). So
this style is unpythonic. Also, None is a singleton, and identity test
is way faster than equality test.

Playing Devil's advocate here: if you were to write "x!=None", then x's
__eq__ method is invoked, which might not account for the possibility
that the other operand is None.

However, if you write "None!=x", then None's __eq__ method is invoked,
which accounts for any given type.
Carl Banks

[Paul Rubin]

alf <ask@mewrites :

I have a reference to certain objects. What is the most pythonic way
to test for valid reference:

If you're intending to use None as a sentinel for an invalid reference,
then use

if obj is not None:

You could also make a unique sentinel:

Sentinel = object()
...
if obj is not Sentinel: ...

"if obj" isn't necessarily what you want; obj might be a valid but
empty container.

"if obj != None" is also wrong, for reasons someone else explained.

[Bruno Desthuilliers]

Carl Banks wrote:

Bruno Desthuilliers wrote:

>In python, assignement is a statement, not an expression, so there's no
way you could write 'if obj = None' by mistake (-syntax error). So
this style is unpythonic. Also, None is a singleton, and identity test
is way faster than equality test.

Playing Devil's advocate here: if you were to write "x!=None", then x's
__eq__ method is invoked, which might not account for the possibility
that the other operand is None.

However, if you write "None!=x", then None's __eq__ method is invoked,
which accounts for any given type.

Well... Since we all know it should be an identity test anyway... <g>

--
bruno desthuilliers
python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for
p in 'o****@xiludom. gro'.split('@')])"

[Gabriel Genellina]

At Monday 4/9/2006 17:02, alf wrote:

>I have a reference to certain objects. What is the most pythonic way to
test for valid reference:

By "valid reference" you mean, you have initially:
obj = None
and you want to detect whether obj is bound to another, different,
object, right?

if obj:

This checks whether obj is considered True, not whether it is None.
e.g. obj=[] would not pass.

if None!=obs:

Would be OK, but the next one is better:

if obj is not None:

This is what you are looking for! :)

Gabriel Genellina
Softlab SRL

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[alf]

alf wrote:

Hi,

I have a reference to certain objects. What is the most pythonic way to
test for valid reference:

if obj:

if None!=obs:

if obj is not None:

thx for all answers - now "if obj is not None:" in an obvious choice ...