On Oct 5, 6:05 pm, MRAB <pyt...@mrabarnett.plus.com> wrote: > Scott wrote: > > I create a list of logs called LogList. Here is a sample: > > > LogList = > > ["inbound tcp office 192.168.0.125 inside 10.1.0.91 88", > > "inbound tcp office 192.168.0.220 inside 10.1.0.31 2967", > > "inbound udp lab 172.24.0.110 inside 10.1.0.6 161", > > "inbound udp office 192.168.0.220 inside 10.1.0.13 53"] > > > I want to sort the list on index 3 of each string - the first IP > > Address. > > > I only need strings with similar, first IP's to be together. I don't > > need all of the IP's to be in order. For example: > > either: > > SortedList = > > ["inbound udp lab 172.24.0.110 inside 10.1.0.6 161", > > "inbound tcp office 192.168.0.220 inside 10.1.0.31 2967", > > "inbound udp office 192.168.0.220 inside 10.1.0.13 53", > > "inbound tcp office 192.168.0.125 inside 10.1.0.91 88"] > > -or- > > SortedList = > > ["inbound tcp office 192.168.0.220 inside 10.1.0.31 2967", > > "inbound udp office 192.168.0.220 inside 10.1.0.13 53", > > "inbound udp lab 172.24.0.110 inside 10.1.0.6 161", > > "inbound tcp office 192.168.0.125 inside 10.1.0.91 88"] > > -or- > > etc. > > > would be fine. > > > I'm reading a lot on sort, sorted, cmp, etc. but I'm just not getting > > how to use an element of a string as a "key" within a list of strings. > > I'm using Python 2.6.2. > > Forget about cmp, just use the 'key' argument of the list's 'sort' > method or the 'sorted' function (the latter is better if you want to > keep the original list). The 'key' argument expects a function (anything > callable, actually) that accepts a single argument (the item) and > returns a value to be used as the key, and the items will be sorted > according to that key. In this case you want the items sorted by the > fourth 'word', so split the item into words and return the one at index > 3: > > def key_word(item): > return item.split()[3] > > SortedList = sorted(LogList, key=key_word) > > If the function is short and simple enough, lambda is often used instead > of a named function: > > SortedList = sorted(LogList, key=lambda item: item.split()[3])
Ok, the lambda worked as advertised. THANK YOU!! Thanks for giving both a def and lambda example. I'll be saving them. -Scott -- http://mail.python.org/mailman/listinfo/python-list
