[Tutor] Would somebody kindly...

[Dave Angel]

"Clayton Kirkwood" <crk at godblessthe.us> Wrote in message:
> 
> 
>>-----Original Message-----
>>From: Tutor [mailto:tutor-bounces+crk=godblessthe.us at python.org] On
>>Behalf Of Dave Angel
>>Sent: Wednesday, October 29, 2014 5:30 AM
>>To: tutor at python.org
>>Subject: Re: [Tutor] Would somebody kindly...
>>
>>"Clayton Kirkwood" <crk at godblessthe.us> Wrote in message:
>>>
>>>
>>> !-----Original Message-----
>>> !From: Tutor [mailto:tutor-bounces+crk=godblessthe.us at python.org] On
>>> !Behalf Of Dave Angel
>>> !Sent: Tuesday, October 28, 2014 6:34 PM
>>> !To: tutor at python.org
>>> !Subject: Re: [Tutor] Would somebody kindly...
>>> !
>>> !
>>> !>
>>> ! Explain this double speak(>:
>>> !>  [pair for pair in values if key == pair[0]] !
>>> !>  I understand the ‘for pair in values’. I assume the first  ‘pair’
>>> !> creates the namespace
>>> !
>>> !The namespace question depends on the version of Python. Python  2.x
>>> !does not do any scoping.
>>> !
>>> !But in version 3.x, the variable pair will go away.
>>> !
>>> !So please tell us the version you're asking about.
>>>
>>> I am using 3.4.1.
>>>
>>
>>Have you somehow configured your email program to use exclamation
>>points for quoting instead of the standard greater-than symbol?
>> "!" instead of ">" ? If so, do you mind changing it  back?
>>
>>In 3.4.1, let's consider the following code.
>>
>>thingie = 2
>>mylist = [(2,55), "charlie", [2, "item2", 12]] x = [78 for item in
>>mylist if item[0] == thingie]
>>
>>What will happen in the list comprehension, and what will be the  final
>>value of x ?
>>
>>First an anonymous list object will be created.  This eventually  will
>>be bound to x, but not till the comprehension is  successfully
>>completed. Next a locally scoped variable item is  created.  This goes
>>away at the end of the comprehension,  regardless of how we exit.
>>
>>Next the 0th value from mylist is bound to item. It happens to be  a
>>tuple, but not from anything the comprehension  decides.
>>Next the expression item [0] == thingie is evaluated.  If it's  true,
>>then the int 78 is appended to the anonymous  list.
>>
>>Now the previous group of actions is repeated for the 1th value of
>>mylist. So now item is a string, and the zeroth character of the  string
>>is compared with the int 2. Not equal, so 72 doesn't get  appended.
>>
>>Similarly for the 2th item. The first element of that list is  equal to
>>2, so another 72 is appended.
>>
>>Now the anonymous list is bound to x.
>>
>>print (x)
>>[72, 72]
> 
> So, in this case, the assignment to x is external. Often I don't see an external assignment, but there is an item in the first position within the comprehension. 

The thing before the 'for' keyword  is an expression. It's
 evaluated once for each piece of the result list. If it happens
 to be a simple variable, nothing changes. It is not a name to be
 bound to, just a name to be evaluated. 

> You don't have that here. When you have [item for item in [list] if item[0] == key], after the iteration completes does item equal the matched entities or does it have the original item?

'item' that was used in the comprehension is gone, and if you
 happened to have an earlier one of same name, it's visible again.
 Like happens when a function returns,  all the variables of that
 namespace are gone.

> I understand that if we had x = [dsfasdfasdf] x will be a list (never a tuple?) with the matches, but what happens to the first item?
> 
> This is from a previous email--
> When I run:
> values = [ ('a', 1), ('b', 2), ('a', 5), ('c', 7)]
> key = 'a'
> pair=[]
> [pair for pair in values if key == pair[0]]
> print(pair)
> 
> I get [].

Naturally because the two variables are in separate namespaces.

> 
> When I run:
> values = [ ('a', 1), ('b', 2), ('a', 5), ('c', 7)]
> key = 'a'
> pair=[]
> x=[pair for pair in values if key == pair[0]]
> print(x)
> 
> I get [('a', 1), ('a', 5)]
> 
> So, what does that first pair do? I see and have used the first comprehension.
> 

The first use of the name 'pair' assigned an empty list to it.
 That's totally unaffected by anything inside your comprehension.
 



-- 
DaveA