>-----Original Message-----
>From: Tutor [mailto:tutor-bounces+crk=godblessthe...@python.org] On
>Behalf Of Dave Angel
>Sent: Wednesday, October 29, 2014 5:30 AM
>To: tutor@python.org
>Subject: Re: [Tutor] Would somebody kindly...
>
>"Clayton Kirkwood" <c...@godblessthe.us> Wrote in message:
>>
>>
>> !-----Original Message-----
>> !From: Tutor [mailto:tutor-bounces+crk=godblessthe...@python.org] On
>> !Behalf Of Dave Angel
>> !Sent: Tuesday, October 28, 2014 6:34 PM
>> !To: tutor@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. 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? 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 [].
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.
Clayton
>
>
>
>--
>DaveA
>
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