On Dec 28, 10:05 am, [EMAIL PROTECTED] wrote: > On Dec 27, 10:59 pm, John Machin <[EMAIL PROTECTED]> wrote: > > > On Dec 28, 4:56 am, [EMAIL PROTECTED] wrote: > > > > from itertools import groupby > > > You seem to have overlooked this important sentence in the > > documentation: "Generally, the iterable needs to already be sorted on > > the same key function" > > Yes, but I imagine this shouldn't prevent me from using and > manipulating the data.
You imagine correctly (and pointlessly) in general; however in particular it prevents you using itertools.groupby simplistically to manipulate the data in the way you want to manipulate it. > It also doesn't explain why the names get > sorted correctly and the time does not. The names in your example were NOT sorted, "correctly" or otherwise. The output order is the same as the input order: Bob, Joe, Jil. >>> seq = ['Bob', 'Joe', 'Jil'] >>> sorted(seq) ['Bob', 'Jil', 'Joe'] >>> seq == sorted(seq) False >>> > > I was trying to do this: > > count_tot = [] > for k, g in groupby(data, key=lambda r: r[NAME]): > for record in g: > count_tot.append((k,record[SALARY])) > for i in count_tot: > here I want to say add all the numbers for each person, but I'm > missing something. > > If you have any ideas about how to solve this pivot table issue, which > seems to be scant on Google, I'd much appreciate it. I know I can do > this in Excel easily with the automated wizard, but I want to know how > to do it myself and format it to my needs. Watch this space. -- http://mail.python.org/mailman/listinfo/python-list
