On 10/01/17 03:02, Deborah Swanson wrote:
Erik wrote, on January 09, 2017 5:47 PM
IIRC, you create it using a list comprehension which creates the
records. A list comprehension always creates a list.
Well no. The list is created with:
records.extend(Record._make(row) for row in rows)
No, the list is _extended_ by that code. The list is _created_ with a
line that will say something like "records = []" or "records = list()"
(or "records = <expression that yields a list>").
It's nice to see you agree that it's a list though. Oh, hold on ... ;)
I'm not exactly
sure if this statement is a list comprehension.
No, it's not. I was remembering an old message where someone suggested
using the _make() method and that was expressed as a list comprehension.
What you have there is a call to list.extend() passing a _generator_
comprehension as its only parameter (which in this case you can consider
to be equivalent to a list comprehension as all of the data are
generated greedily). You see that I said "list.extend()". That's because
'records' is a list.
type(records)
<class 'list'>
Yes, it's an instance of the list class. A list object. A list.
>>> type(list())
<class 'list'>
>>> type([])
<class 'list'>
>>> class foo: pass
...
>>> type(foo())
<class '__main__.foo'>
>>>
... type() will tell you what class your object is an instance of.
"<class 'list'>" tells you that your object is a list.
And it behaves like a list sometimes, but many times
not.
I think that's impossible. I'm 100% sure it's a list. Please give an
example of when 'records' does not behave like a list.
The only thing I don't think you have 100% correct is your assertion
that records is a list.
It's a list.
But that's just a quibble. The important thing in this context is that
both .sort() and sorted() treat it like a list and DTRT.
That's because it's a list :)
E.
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