hi,
I need to get the index of an object in a list. I know that no two objects
in the list are the same, but objects might evaluate as equal. for example
list = [obj1, obj2, obj3, obj4, obj5]
for object in list:
objectIndex = list.index(object)
print objectIndex
prints 0, 1, 2, 3,
Iain King iainking at gmail.com writes:
what's wrong with:
i = 0
for object in list:
objectIndex = i
print objectIndex
i += 1
Iain
The issues with that is you might have a complex structure below the for object
in list: with lots of continues or breaks and you don't
Kent Johnson kent at kentsjohnson.com writes:
In either case enumerate() is your friend. To find an
item by identity:
def index_by_id(lst, o):
for i, item in enumerate(lst):
if item is o:
return i
raise ValueError, %s not in list % o
If you just want the index
I am having trouble with the python interactive shell. The arrow keys render as
^[[D, ^[[A, etc making line editing impossible. The arrow keys (and function
keys) work fine in bash, but in the python shell they are printed. Any ideas
what is going on?
--
Simon Brunning simon at brunningonline.net writes:
Sounds like a readline problem. Your OS? How did you install Python?
Yea, that was it. I just had to copy readline.so from another installation.
Thanks for the quick reply
--
http://mail.python.org/mailman/listinfo/python-list
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
shandy.b at 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 in
johnzenger at 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,
johnzenger at 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:
