[EMAIL PROTECTED]:
> Len(rows) recalculates each time the while loop begins. Now that I
> think of it, "rows != []" is faster than "len(rows) > 0."
the difference is very small, and "len(rows)" is faster than "rows != []"
(the latter creates a new list for each test).
and as usual, using the co
Em Ter, 2006-02-28 às 09:10 -0800, [EMAIL PROTECTED] escreveu:
> Although I don't know if this is faster or more efficient than your
> current solution, it does look cooler:
[snip]
> print [x for x in grouper]
This is not cool. Do
print list(grouper)
--
"Quem excele em empregar a força militar
You don't need to copy the list; but if you don't, your original list
will be emptied.
Len(rows) recalculates each time the while loop begins. Now that I
think of it, "rows != []" is faster than "len(rows) > 0."
By the way, you can also do this using (gasp) a control index:
def grouprows(rows):
gmail.com> writes:
> Although I don't know if this is faster or more efficient than your
> current solution, it does look cooler:
>
> def grouprows(inrows):
> rows = []
> rows[:] = inrows # makes a copy because we're going to be
> deleting
> while len(rows) > 0:
> rowspan
Although I don't know if this is faster or more efficient than your
current solution, it does look cooler:
def grouprows(inrows):
rows = []
rows[:] = inrows # makes a copy because we're going to be
deleting
while len(rows) > 0:
rowspan = rows[0]["rowspan"]
yield rows[
gmail.com> writes:
>
> Python lets you iterate through a list using an integer index, too,
> although if you do so we will make fun of you. You can accomplish it
> with a while loop, as in:
>
> i = 0
> while i < len(rows):
>if rows[i] == "This code looks like BASIC without the WEND, doesn
gmail.com> writes:
>
> A couple questions:
>
> 1- what is j?
> 2- what does the rows[x][y] object look like? I assume it's a dict
> that has a "rowspan" key. Can rows[x][y]["rowspan"] sometimes be 0?
>
> Perhaps you're looking for something like this:
> rowgroups = []
> rowspan = 0
> for i
Python lets you iterate through a list using an integer index, too,
although if you do so we will make fun of you. You can accomplish it
with a while loop, as in:
i = 0
while i < len(rows):
if rows[i] == "This code looks like BASIC without the WEND, doesn't
it?":
rowgroups.append("Pretty
A couple questions:
1- what is j?
2- what does the rows[x][y] object look like? I assume it's a dict
that has a "rowspan" key. Can rows[x][y]["rowspan"] sometimes be 0?
Perhaps you're looking for something like this:
rowgroups = []
rowspan = 0
for i in range( len(rows) ):
if rowspan <= 0:
Hi,
I have a list of rows which contains a list of cells (from a html table), and I
want to create an array of logical row groups (ie group rows by the rowspan). I
am only concerned with checking the rowspan of specific columns, so that makes
it easier, but I am having trouble implementing it in p
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