Re: Test for structure

2005-02-22 Thread Martin Miller
Nope, that isn't right either, in the sense that it handles all the
cases properly, including "single string" vs "list of strings'. Guess
this overly simplistic aslist() does not work after. I should have been
more suspicious and cautious before posting. Sorry.

Martin

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Re: Test for structure

2005-02-22 Thread Martin Miller
Ooops. I left out an "*" on a statement in the new aslist() function. I
should have written:

def aslist(*args):
   return list(*args) # corrected

def f(arg):
args = aslist(arg)
...

Sorry,
Martin

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Re: Test for structure

2005-02-22 Thread Martin Miller
At the end of his last post, Steve Bethard wrote:
> That said, I find that in most cases, the better option is to use
*args
> in the original function though.  For example:
>
>  def f(arg):
>  args = aslist(arg)
>  ...
>  f(42)
>  f(['spam', 'eggs', 'ham'])
>
> could probably be more easily written as:
>
>  def f(*args):
>  ...
>  f(42)
>  f('spam', 'eggs', 'ham')
>
> Of course this won't work if you have multiple list arguments.

Very interesting, but it also doesn't let you specify a default
argument value...however this gave me the idea that it would be
possible to use the *args idea to greatly simplify the proposed
aslist() function -- when one was needed to allow default argument
values and/or for handling multiple list arguments. Namely:

def aslist(*args):
return list(args)

def f(arg=None):
args = aslist(arg)
...

f()
f(42)
f('tanstaafl')
f(['spam', 'eggs', 'ham'])

This seems fairly lean and mean, with no if, isinstance, hasattr, or
try/excepts required -- although aslist() might need a check for the
single argument of None case, depending on whether it should return []
or something besides [None] in that situation.

Best,
Martin

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Re: Test for structure

2005-02-21 Thread Terry Reedy

"Steven Bethard" <[EMAIL PROTECTED]> wrote in message 
news:[EMAIL PROTECTED]
> I don't like this idea much because it depends on str and unicode _not_ 
> having a particular function.  I haven't seen any guarantee anywhere that 
> str or unicode won't ever grow an __iter__ method.  So this code seems 
> dangerous as far as future compatibility goes.

When CPython's support for the old iteration protocol goes away, which I 
expects it will someday, strings will have to grow an __iter__ method. 
Even now, I think this difference between strings and lists is more of an 
accident than a logical design.  So the test is opaque and specific to 
current CPython.  The validity of something like a = a+'', however, is 
inherent to the nature of strings.

Terry J. Reedy



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Re: Test for structure

2005-02-21 Thread Steven Bethard
Martin Miller broke the order of reading again by top-posting:
However, to handle the more general problem of allow *any* argument to
be either a single item or a list seems to require a combination of
both EAPF and LBYL. This is the best solution I've been able to come up
with so far:
def asList(arg):
[snip]
if arg is None:
return []
elif isinstance(arg, basestring): # special case strings (to
  # avoid list())
return [arg]
else:
try:
return list(arg)
except TypeError:
return [arg]
[snip]
Can this be improved or is there anything wrong or overly limiting
about it?
I don't think you're going to do a whole lot better than that, though 
you can try something like the following if you're really afraid of the 
isinstance:

def aslist(arg):
# you don't need to test None; it will be caught by the list branch
try:
arg + ''
except TypeError:
return [arg]
try:
return list(arg)
except TypeError:
return [arg]
That said, I find that in most cases, the better option is to use *args 
in the original function though.  For example:

def f(arg):
args = aslist(arg)
...
f(42)
f(['spam', 'eggs', 'ham'])
could probably be more easily written as:
def f(*args):
...
f(42)
f('spam', 'eggs', 'ham')
Of course this won't work if you have multiple list arguments.
STeVe
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Re: Test for structure

2005-02-21 Thread Martin Miller
Testing for the '__iter__' (or even '__getitem__') attribute doesn't
really address the problem, nor does trying to execute the statement
'itr = iter(a)'.

To use EAPF and answer the OP's original question, which was

> So how can I test if a variable 'a' is either a single character
> string or a list?

I think the best answer would be to use Robin Munn's suggestion (see
http://groups-beta.google.com/group/comp.lang.python/msg/c8befd4bed517bbc)
as mentioned in the link in Simon Brunning's post) namely:

try:
a + ''
except TypeError:
pass
else:
a = [a]

However, to handle the more general problem of allow *any* argument to
be either a single item or a list seems to require a combination of
both EAPF and LBYL. This is the best solution I've been able to come up
with so far:

def asList(arg):
"""Makes sure the argument it is passed is a Python list.
If it is, it is just returned, otherwise a (possibly empty)
list is created and returned with the single item in it.

asList() can used to create flexible interfaces which allow
arguments to be passed to them that are either single items or
lists of items. By applying this function before using the
values in arguments, single and multi-valued cases can be
handled by general list-handling code in the function or
method.

As a special case, a single argument with the value None is
converted into an empty list (instead of converted into the
list [None]).

asList(arg) ==> list
"""

if arg is None:
return []
elif isinstance(arg, basestring): # special case strings (to
  # avoid list())
return [arg]
else:
try:
return list(arg)
except TypeError:
return [arg]


if __name__ == "__main__":

def example(items=None):
"""Sample function that can be called with a single argument
that can be a single or list of items.
"""
itemList = asList(items)
if not itemList:
print "example() called with empty list or None argument"
else:
print "example() called with argument containing %d " \
  "thing%s" % \
  (len(itemList), ('','s')[len(itemList)>1])
for i, item in enumerate(itemList):
print "  items[%d] = %s" % (i, repr(item))

example(42)
example((1,2,3))
example([4,5,6,7])
example('abc')
example(u'def')
example(["aaa", 111, (4,5), 2.01])
example(None) #  Note that this will become an empty list
example() #  same in this case

Which produces the following output:

example() called with argument containing 1 thing
  items[0] = 42
example() called with argument containing 3 things
  items[0] = 1
  items[1] = 2
  items[2] = 3
example() called with argument containing 4 things
  items[0] = 4
  items[1] = 5
  items[2] = 6
  items[3] = 7
example() called with argument containing 1 thing
  items[0] = 'abc'
example() called with argument containing 1 thing
  items[0] = u'def'
example() called with argument containing 4 things
  items[0] = 'aaa'
  items[1] = 111
  items[2] = (4, 5)
  items[3] = 2.0098
example() called with empty list or None argument
example() called with empty list or None argument

Can this be improved or is there anything wrong or overly limiting
about it?

TIA,
Martin


=
Steven Bethard wrote:
> Terry Hancock wrote:
>  > But you probably shouldn't do that. You should probably just test
to
>  > see if the object is iterable --- does it have an __iter__ method?
>  >
>  > Which might look like this:
>  >
>  > if hasattr(a, '__iter__'):
>  > print "'a' quacks like a duck"
>
> Martin Miller top-posted:
> > I don't believe you can use the test for a __iter__ attribute in
this
> > case, for the following reason:
> >
> c1 = 'abc'
> c2 = ['de', 'fgh', 'ijkl']
> hasattr(c1, '__iter__')
> > False
> >
> hasattr(c2, '__iter__')
> > True
>
> Right.  str and unicode objects support iteration through the old
> __getitem__ protocol, not the __iter__ protocol.  If you want to use
> something as an iterable, just use it and catch the exception:
>
> try:
>  itr = iter(a)
> except TypeError:
>  # 'a' is not iterable
> else:
>  # 'a' is iterable
>
> Another lesson in why EAPF is often better than LBYL in Python[1].
> 
> STeVe
> 
> [1] http://www.python.org/moin/PythonGlossary

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Re: Test for structure

2005-02-21 Thread Steven Bethard
> Steven Bethard wrote:
>>
>>Right.  str and unicode objects support iteration through the old
>>__getitem__ protocol, not the __iter__ protocol.  If you want to use
>>something as an iterable, just use it and catch the exception:
>>
>>try:
>> itr = iter(a)
>>except TypeError:
>> # 'a' is not iterable
>>else:
>> # 'a' is iterable
Martin Miller broke the order of reading by top-posting:
In either case, you can't tell a string and list apart, which is what
the OP wanted to know, namely how to differentiate the two.
Yes, sorry, I should have marked my post OT.  It was an answer to Terry 
Hancock's post suggesting hasattr(x, '__iter__'), not the OP.

Perhaps the test for an __iter__ attribute *is* the way to go because
you can tell the difference between the two type.
I've seen this done before, e.g.:
try:
itr = iter(x)
except TypeError:
# is not iterable
else:
if hasattr(x, '__iter__'):
# is other iterable
else:
# is str or unicode
I don't like this idea much because it depends on str and unicode _not_ 
having a particular function.  I haven't seen any guarantee anywhere 
that str or unicode won't ever grow an __iter__ method.  So this code 
seems dangerous as far as future compatibility goes.

I think the technique suggested by Robin Munn nearly a year ago (and
referenced by the link in Simon Brunning's post):
http://groups-beta.google.com/group/comp.lang.python/msg/c8befd4bed517bbc
namely:
try:
a + ''
except TypeError:
pass
else:
a= [a]
would be a good usable solution, although it's not totally infallible.
Yup, if I had to do this kind of type-checking (which I don't think I 
ever do), I'd probably go with something along these lines.

STeVe
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Re: Test for structure

2005-02-21 Thread Martin Miller
Yes, both string and lists have a __getitem__ attribute:

>>> c1 = 'abc'
>>> c2 = ['de', 'fgh', 'ijkl']
>>> hasattr(c1, '__getitem__')
True
>>> hasattr(c2, '__getitem__')
True

In other words you could index elements of either one using [].

Likewise, both a string and list would produce a usable iterator using
the following logic:

try:
itr = iter(a)
except TypeError:
# 'a' is not iterable
else:
# 'a' is iterable

In either case, you can't tell a string and list apart, which is what
the OP wanted to know, namely how to differentiate the two. EAPF is
fine, but what operation would answer the string vs list question?

Perhaps the test for an __iter__ attribute *is* the way to go because
you can tell the difference between the two type. Again I don't know
because the OP doesn't give enough information. I suspect, but don't
know, that it could be so that either one string or a list of strings
as an argument could treated as a list of 0 or more strings and
accessed by indices by most of the rest of the code.

I think the technique suggested by Robin Munn nearly a year ago (and
referenced by the link in Simon Brunning's post):
http://groups-beta.google.com/group/comp.lang.python/msg/c8befd4bed517bbc
namely:

try:
a + ''
except TypeError:
pass
else:
a= [a]

would be a good usable solution, although it's not totally infallible.
It may not be possible to give a better answer without more real
information about the desired usage.

Martin


=
Steven Bethard wrote:
> Terry Hancock wrote:
>  > But you probably shouldn't do that. You should probably just test
to
>  > see if the object is iterable --- does it have an __iter__ method?
>  >
>  > Which might look like this:
>  >
>  > if hasattr(a, '__iter__'):
>  > print "'a' quacks like a duck"
>
> Martin Miller top-posted:
> > I don't believe you can use the test for a __iter__ attribute in
this
> > case, for the following reason:
> >
> c1 = 'abc'
> c2 = ['de', 'fgh', 'ijkl']
> hasattr(c1, '__iter__')
> > False
> >
> hasattr(c2, '__iter__')
> > True
>
> Right.  str and unicode objects support iteration through the old
> __getitem__ protocol, not the __iter__ protocol.  If you want to use
> something as an iterable, just use it and catch the exception:
>
> try:
>  itr = iter(a)
> except TypeError:
>  # 'a' is not iterable
> else:
>  # 'a' is iterable
>
> Another lesson in why EAPF is often better than LBYL in Python[1].
> 
> STeVe
> 
> [1] http://www.python.org/moin/PythonGlossary

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Re: Test for structure

2005-02-20 Thread Steven Bethard
Terry Hancock wrote:
> But you probably shouldn't do that. You should probably just test to
> see if the object is iterable --- does it have an __iter__ method?
>
> Which might look like this:
>
> if hasattr(a, '__iter__'):
> print "'a' quacks like a duck"
Martin Miller top-posted:
I don't believe you can use the test for a __iter__ attribute in this
case, for the following reason:
c1 = 'abc'
c2 = ['de', 'fgh', 'ijkl']
hasattr(c1, '__iter__')
False
hasattr(c2, '__iter__')
True
Right.  str and unicode objects support iteration through the old 
__getitem__ protocol, not the __iter__ protocol.  If you want to use 
something as an iterable, just use it and catch the exception:

try:
itr = iter(a)
except TypeError:
# 'a' is not iterable
else:
# 'a' is iterable
Another lesson in why EAPF is often better than LBYL in Python[1].
STeVe
[1] http://www.python.org/moin/PythonGlossary
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Re: Test for structure

2005-02-20 Thread Martin Miller
I don't believe you can use the test for a __iter__ attribute in this
case, for the following reason:
>>> c1 = 'abc'
>>> c2 = ['de', 'fgh', 'ijkl']
>>> hasattr(c1, '__iter__')
False
>>> hasattr(c2, '__iter__')
True
>>> for i in c1: print "i=%s is an element of c1" % repr(i)
...
i='a' is an element of c1
i='b' is an element of c1
i='c' is an element of c1

In other words, even though the c1 single string variable does not have
an __iter__ attribute, it can still be used in a for loop. I think the
right answer would depend on what exactly the OP intends to do with the
argument when it is a list (or is list-like in some way) -- i.e. he
didn't say specifically that he wanted use it in a for loop.

-Martin



Terry Hancock wrote:
> On Wednesday 16 February 2005 09:08 am, alex wrote:
> > how can I check if a variable is a structure (i.e. a list)? For my
> > special problem the variable is either a character string OR a list
of
> > character strings line ['word1', 'word2',...]
> >
> > So how can I test if a variable 'a' is either a single character
string
> > or a list?
>
> The literally correct but actually wrong answer is:
>
> if type(a) == type([]):
> print "'a' is a duck"
>
> But you probably shouldn't do that. You should probably just test to
> see if the object is iterable --- does it have an __iter__ method?
>
> Which might look like this:
>
> if hasattr(a, '__iter__'):
> print "'a' quacks like a duck"
>
> That way your function will also work if a happens to be a tuple,
> a dictionary, or a user-defined class instance which is happens to
> be  iterable.
>
> Being "iterable" means that code like:
>
> for i in a:
>print "i=%s is an element of a" % repr(i)
>
> works.  Which is probably why you wanted to know, right?
>
> Cheers,
> Terry
>
> --
> --
> Terry Hancock ( hancock at anansispaceworks.com )
> Anansi Spaceworks  http://www.anansispaceworks.com

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Re: Test for structure

2005-02-18 Thread Terry Hancock
On Wednesday 16 February 2005 09:08 am, alex wrote:
> how can I check if a variable is a structure (i.e. a list)? For my
> special problem the variable is either a character string OR a list of
> character strings line ['word1', 'word2',...]
> 
> So how can I test if a variable 'a' is either a single character string
> or a list?

The literally correct but actually wrong answer is:

if type(a) == type([]):
print "'a' is a duck"

But you probably shouldn't do that. You should probably just test to
see if the object is iterable --- does it have an __iter__ method?

Which might look like this:

if hasattr(a, '__iter__'):
print "'a' quacks like a duck"

That way your function will also work if a happens to be a tuple,
a dictionary, or a user-defined class instance which is happens to
be  iterable.

Being "iterable" means that code like:

for i in a:
   print "i=%s is an element of a" % repr(i)

works.  Which is probably why you wanted to know, right?

Cheers,
Terry

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Re: Test for structure

2005-02-16 Thread Ben Finney
alex wrote On 17/02/05 02:08:
how can I check if a variable is a structure (i.e. a list)? For my 
special problem the variable is either a character string OR a list 
of character strings line ['word1', 'word2',...]
You're trying to apply the LBYL principle.  My bet is that your "special
problem" can be solved by the EAFP principle.  (These terms are
explained in the Glossary of the tutorial,
.)
If you test for a specific set of types, your code will not work with
types that you have not considered yet behave like lists.  As you 
discover more object types that you want the code to work with, it will 
sprout more cruft for checking those types.

If you want to use some list behaviour of the object, don't check first.
Use it, and catch the TypeError exception in the event that it's not.
This way, *any* object that implements the functionality you need will
work, regardless of its type.
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Re: Test for structure

2005-02-16 Thread Steven Bethard
Michael Hartl wrote:
I use a function isListLike in cases such as this one:
# def isListLike(L):
# """Return True if L is list-like, False otherwise."""
# try:
# L + []
# return True
# except:
# return False
Then you can use a standard if-else construct:
# if isListLike(myvar):
# 
# else:
# 
What kind of situations do you use this for?  I almost never have to do 
this kind of typechecking.  If it's supposed to be a list, I just use it 
as a list...

STeVe
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Re: Test for structure

2005-02-16 Thread Michael Hartl
I use a function isListLike in cases such as this one:

# def isListLike(L):
# """Return True if L is list-like, False otherwise."""
# try:
# L + []
# return True
# except:
# return False

Then you can use a standard if-else construct:

# if isListLike(myvar):
# 
# else:
# 

Michael

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Chief Technology Officer
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Re: Test for structure

2005-02-16 Thread Steven Bethard
alex wrote:
So how can I test if a variable 'a' is either a single character string
or a list?
py> def test(x):
... return (isinstance(x, list) or
... isinstance(x, basestring) and len(x) == 1)
...
py> test('a')
True
py> test('ab')
False
py> test([])
True
py> test(['a', 'b'])
True
But definitely read Simon Brunning's post - you probably don't actually 
want to do this test.  Why do you think you want to test this?  What's 
your use case?

STeVe
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Re: Test for structure

2005-02-16 Thread Diez B. Roggisch
import types

v = []
if type(v) is types.ListType:
 pass


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Re: Test for structure

2005-02-16 Thread Chris Cioffi
Perhaps you're looking for the type() built in function and the types modules?

>>> type('aaa')

>>> type([])

>>> import types
>>> if type([]) is types.ListType:
... print 'is a list'
... 
is a list

Chris

On Wed, 16 Feb 2005 07:10:56 -0800 (PST), alex
<[EMAIL PROTECTED]> wrote:
> Hi there,
> 
> how can I check if a variable is a structure (i.e. a list)? For my
> special problem the variable is either a character string OR a list of
> character strings line ['word1', 'word2',...]
> 
> So how can I test if a variable 'a' is either a single character string
> or a list? I tried:
> 
> if a is list:
> 
> but that does not work. I also looked in the tutorial and used google
> to find an answer, but I did not.
> 
> Has anyone an idea about that?
> 
> Alex
> 
> --
> http://mail.python.org/mailman/listinfo/python-list
> 


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Re: Test for structure

2005-02-16 Thread Simon Brunning
On Wed, 16 Feb 2005 07:11:08 -0800 (PST), alex
<[EMAIL PROTECTED]> wrote:
> how can I check if a variable is a structure (i.e. a list)? For my
> special problem the variable is either a character string OR a list of
> character strings line ['word1', 'word2',...]
> 
> So how can I test if a variable 'a' is either a single character string
> or a list? I tried:
> 
> if a is list:
> 
> but that does not work. I also looked in the tutorial and used google
> to find an answer, but I did not.
> 
> Has anyone an idea about that?



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[EMAIL PROTECTED],
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Test for structure

2005-02-16 Thread alex
Hi there,

how can I check if a variable is a structure (i.e. a list)? For my
special problem the variable is either a character string OR a list of
character strings line ['word1', 'word2',...]

So how can I test if a variable 'a' is either a single character string
or a list? I tried:

if a is list:


but that does not work. I also looked in the tutorial and used google
to find an answer, but I did not.

Has anyone an idea about that?

Alex

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