Re: Problem with the sort() function
Thanks everyone!!:-) Nicks solution coupled with John's modifications worked great for 2.2!! Yipeee...!!:):) -- http://mail.python.org/mailman/listinfo/python-list
Re: Problem with the sort() function
clementine wrote: Hi, I have an array of arrays in the form of list = [[3,'fork',0.3,1],[2,'fork,0.1,2],[3,'exec',0.2,2]] The in-built sort(),list.sort() sorts on the first element, if the first elts are equal then it sorts on the second elt and so on...But i really dont want to search on the second elt if the first elts are equal...the 1-D lists shud be left in the same position i.e. i want the sorted list to be [[2,'fork',0.1,2],[3,'fork,0.3,1],[3,'exec',0.2,2]] and not [[2,'fork',0.1,2],[3,'exec',0.2,2],[3,'fork,0.3,1]]. Try this: Py from operator import itemgetter Py list = [[3,'fork',0.3,1],[2,'fork',0.1,2],[3,'exec',0.2,2]] Py list.sort(key=itemgetter(0)) Py list [[2, 'fork', 0.10001, 2], [3, 'fork', 0.2, 1], [3, ' exec', 0.20001, 2]] If the 'key' argument isn't accepted (i.e. you aren't using Python 2.4), you'll need to do the decoration manually: def mysort(iterable, cmp=None, key=None, reverse=False): return a sorted copy of its input if sys.version_info = (2,4): return sorted(iterable, cmp, key, reverse) seq = list(iterable) if reverse: seq.reverse()# preserve stability if key is not None: seq = [(key(elem), i, elem) for i, elem in enumerate(seq)] seq.sort(cmp) if key is not None: seq = [elem for (key, i, elem) in seq] if reverse: seq.reverse() return seq list = mysort([[3,'fork',0.3,1],[2,'fork',0.1,2],[3,'exec',0.2,2]], key=lambda x: x[0]) (Taken from Raymond's code in: http://mail.python.org/pipermail/python-list/2005-January/263275.html) Cheers, Nick. -- Nick Coghlan | [EMAIL PROTECTED] | Brisbane, Australia --- http://boredomandlaziness.skystorm.net -- http://mail.python.org/mailman/listinfo/python-list
Re: Problem with the sort() function
clementine [EMAIL PROTECTED] wrote: Thanx Nick...I forgot to mention im using python 2.2 and along with a host of other things it doesnt seem to have the enumarate built in function :(:(:(...is it possible to replace it by something else? I dont think simulating it will be feasible Here's one I prepared earlier: if sys.version_info (2,3): def enumerate(l): return zip(range(len(l)), l) which will suck somewhat on large lists compared to being able to do it with iterators. -- \S -- [EMAIL PROTECTED] -- http://www.chaos.org.uk/~sion/ ___ | Frankly I have no feelings towards penguins one way or the other \X/ |-- Arthur C. Clarke her nu becomeþ se bera eadward ofdun hlæddre heafdes bæce bump bump bump -- http://mail.python.org/mailman/listinfo/python-list
Re: Problem with the sort() function
Sion Arrowsmith wrote: clementine [EMAIL PROTECTED] wrote: Thanx Nick...I forgot to mention im using python 2.2 and along with a host of other things it doesnt seem to have the enumarate built in function :(:(:(...is it possible to replace it by something else? I dont think simulating it will be feasible Here's one I prepared earlier: if sys.version_info (2,3): def enumerate(l): return zip(range(len(l)), l) which will suck somewhat on large lists compared to being able to do it with iterators. Iterators are available in python 2.2 class enumerate: def __init__(self, inlist): self.inlist = inlist self.index = 0 def next(self): if self.index = len(self.inlist): raise StopIteration thisone = self.inlist[self.index] self.index += 1 return self.index-1, thisone def __iter__(self): return self Regards, Fuzzy http://www.voidspace.org.uk/python/index.shtml -- \S -- [EMAIL PROTECTED] -- http://www.chaos.org.uk/~sion/ ___ | Frankly I have no feelings towards penguins one way or the other \X/ |-- Arthur C. Clarke her nu becomeþ se bera eadward ofdun hlæddre heafdes bæce bump bump bump -- http://mail.python.org/mailman/listinfo/python-list
Re: Problem with the sort() function
Nick Coghlan wrote: def mysort(iterable, cmp=None, key=None, reverse=False): return a sorted copy of its input if sys.version_info = (2,4): return sorted(iterable, cmp, key, reverse) seq = list(iterable) if reverse: seq.reverse()# preserve stability if key is not None: seq = [(key(elem), i, elem) for i, elem in enumerate(seq)] seq.sort(cmp) if key is not None: seq = [elem for (key, i, elem) in seq] if reverse: seq.reverse() return seq You'd be better off defining things once (and using the standard name) rather than doing a test every time your function is called. Something like: import sys ... if sys.version_info (2, 3): def enumerate(iterable): # iterators not yet invented return zip(range(len(iterable)), iterable) if sys.version_info (2, 4): def sorted(iterable, cmp=None, key=None, reverse=False): return a sorted copy of its input seq = list(iterable) if reverse: seq.reverse()# preserve stability if key is not None: seq = [(key(elem), i, elem) for i, elem in enumerate(seq)] seq.sort(cmp) if key is not None: seq = [elem for (key, i, elem) in seq] if reverse: seq.reverse() return seq If you like your names better, you can use: if sys.version_info = (2, 4): mysort = sorted else: def mysort(iterable, cmp=None, key=None, reverse=False): ... or even (if you can't be bothered to look up when features happened): try: test = enumerate except NameError: def enumerate(iterable): ... try: test = sorted except NameError: def sorted(iterable, cmp=None, key=None, reverse=False): ... --Scott David Daniels [EMAIL PROTECTED] -- http://mail.python.org/mailman/listinfo/python-list
Re: Problem with the sort() function
Scott David Daniels wrote: if sys.version_info (2, 4): def sorted(iterable, cmp=None, key=None, reverse=False): return a sorted copy of its input seq = list(iterable) if reverse: seq.reverse()# preserve stability if key is not None: seq = [(key(elem), i, elem) for i, elem in enumerate(seq)] seq.sort(cmp) if key is not None: seq = [elem for (key, i, elem) in seq] if reverse: seq.reverse() return seq I think you may have some unintended indentation on the 'seq.sort' line otherwise this only sorts when a key is specified. Also, you probably want: if key is not None: if reverse: seq = [(key(elem), -i, elem) for i, elem in enumerate(seq)] else: seq = [(key(elem), i, elem) for i, elem in enumerate(seq)] to handle the case where both key and reverse are given. -- http://mail.python.org/mailman/listinfo/python-list
Re: Problem with the sort() function
Fuzzyman wrote:
Iterators are available in python 2.2
class enumerate:
def __init__(self, inlist):
self.inlist = inlist
self.index = 0
def next(self):
if self.index = len(self.inlist): raise StopIteration
thisone = self.inlist[self.index]
self.index += 1
return self.index-1, thisone
def __iter__(self):
return self
A simpler version that works with any iterable (not just sequences):
py class enumerate(object):
... def __init__(self, iterable):
... self._next = iter(iterable).next
... self._index = -1
... def __iter__(self):
... return self
... def next(self):
... self._index += 1
... return self._index, self._next()
...
py enumerate('abcde')
__main__.enumerate object at 0x011627F0
py list(enumerate('abcde'))
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e')]
STeVe
--
http://mail.python.org/mailman/listinfo/python-list
Re: Problem with the sort() function
Scott David Daniels wrote: or even (if you can't be bothered to look up when features happened): try: test = enumerate except NameError: def enumerate(iterable): ... try: test = sorted except NameError: def sorted(iterable, cmp=None, key=None, reverse=False): ... Ridiculously minor nit, but there's no reason to assign to test: try: enumerate except NameError: def enumerate(iterable): ... try: sorted except NameError: def sorted(iterable, cmp=None, key=None, reverse=False): ... Just evaluating the expression 'enumerate' or 'sorted' should raise the NameError if they don't exist. STeVe -- http://mail.python.org/mailman/listinfo/python-list
Problem with the sort() function
Hi, I have an array of arrays in the form of list = [[3,'fork',0.3,1],[2,'fork,0.1,2],[3,'exec',0.2,2]] The in-built sort(),list.sort() sorts on the first element, if the first elts are equal then it sorts on the second elt and so on...But i really dont want to search on the second elt if the first elts are equal...the 1-D lists shud be left in the same position i.e. i want the sorted list to be [[2,'fork',0.1,2],[3,'fork,0.3,1],[3,'exec',0.2,2]] and not [[2,'fork',0.1,2],[3,'exec',0.2,2],[3,'fork,0.3,1]]. I tried the following code: def mysort(x,y): return x[0]-y[0] list.sort(mysort) This somehow seems to work for a small subset of my input...but not for my real input which has more than 2000 sub-lists(and so obviously i cant trace each pass..). Can someone tell me what i'm missing? -- http://mail.python.org/mailman/listinfo/python-list
