Re: Looping over lists
En Sat, 05 May 2007 05:15:32 -0300, kaens [EMAIL PROTECTED] escribió: I think the for i in range() is more readable (Maybe because I'm coming from a c-style syntax language background) - but what would the benefits of using enumerate be (other that being more . . . pythonesque?) If you want to iterate over a sequence and process each item, you could do this (as one would do in C): for i in range(len(values)): dosomethingwith(values[i]) Problems: values[i] gets translated into values.__getitem__(i), and that requires a lookup for the __getitem__ method and a function call; and you don't actually need the index i, but it's created anyway. In this case the best (and fastest) way in Python is: for item in values: dosomethingwith(item) No spurious variable, no method lookup, no aditional function call. But what if you need *also* the index? On earlier Python versions one had to go back to the first method; enumerate() was made just for this case: for i, item in enumerate(values): dosomethingwith(item, i) You get all the advantages of the iterator approach plus the index available. -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Re: Looping over lists
Tommy Grav [EMAIL PROTECTED] writes:
In C this would be equivalent to:
for(i = 0; i n; i++) {
for(j=i+1; j n; j++) {
print a[i], a[j]
for i in xrange(n):
for j in xrange(i+1, n):
print a[i], a[j]
--
http://mail.python.org/mailman/listinfo/python-list
Re: Looping over lists
prad wrote: On Friday 04 May 2007 18:40:53 Tommy Grav wrote: Can anyone help me with the right approach for this in python? for each in a: for item in a[a.index(each)+1:]: print each,item will produce 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5 a.index(each) gives the index of the each value in the a list. then you just add 1 to it so you start at the index value beside each's index. If there are equal items you may not get what you expect: items = [1, 2, 1.0] for a in items: ... for b in items[items.index(a)+1:]: ... print a, b ... 1 2 1 1.0 2 1.0 1.0 2 1.0 1.0 Peter -- http://mail.python.org/mailman/listinfo/python-list
Re: Looping over lists
In [EMAIL PROTECTED], prad wrote: On Friday 04 May 2007 18:40:53 Tommy Grav wrote: Can anyone help me with the right approach for this in python? for each in a: for item in a[a.index(each)+1:]: print each,item will produce 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5 But only if the elements in the list are unique. And the runtime is suboptimal because `index()` is doing a linear search -- the outer loop becomes slower and slower with each iteration. Ciao, Marc 'BlackJack' Rintsch -- http://mail.python.org/mailman/listinfo/python-list
Re: Looping over lists
I think the for i in range() is more readable (Maybe because I'm coming from a c-style syntax language background) - but what would the benefits of using enumerate be (other that being more . . . pythonesque?) On 5/5/07, Dennis Lee Bieber [EMAIL PROTECTED] wrote: On Fri, 4 May 2007 19:26:17 -0700, [EMAIL PROTECTED] (Alex Martelli) declaimed the following in comp.lang.python: for i in range(n): for j in range(i+1, n): print a[i], a[j] Ah, but wouldn't the cleaner Python be something like a = [1, 2, 3, 4, 5, 3, 6] #just to confuse matters for pos, val in enumerate(a): ... for v2 in a[pos+1:]: ... print val, v2 ... 1 2 1 3 1 4 1 5 1 3 1 6 2 3 2 4 2 5 2 3 2 6 3 4 3 5 3 3 3 6 4 5 4 3 4 6 5 3 5 6 3 6 -- WulfraedDennis Lee Bieber KD6MOG [EMAIL PROTECTED] [EMAIL PROTECTED] HTTP://wlfraed.home.netcom.com/ (Bestiaria Support Staff: [EMAIL PROTECTED]) HTTP://www.bestiaria.com/ -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
Re: Looping over lists
On May 5, 3:15 am, kaens [EMAIL PROTECTED] wrote: I think the for i in range() is more readable (Maybe because I'm coming from a c-style syntax language background) - but what would the benefits of using enumerate be (other that being more . . . pythonesque?) It doesn't create a whole new list just for iterating. On 5/5/07, Dennis Lee Bieber [EMAIL PROTECTED] wrote: On Fri, 4 May 2007 19:26:17 -0700, [EMAIL PROTECTED] (Alex Martelli) declaimed the following in comp.lang.python: for i in range(n): for j in range(i+1, n): print a[i], a[j] Ah, but wouldn't the cleaner Python be something like a = [1, 2, 3, 4, 5, 3, 6] #just to confuse matters for pos, val in enumerate(a): ... for v2 in a[pos+1:]: ... print val, v2 ... 1 2 1 3 1 4 1 5 1 3 1 6 2 3 2 4 2 5 2 3 2 6 3 4 3 5 3 3 3 6 4 5 4 3 4 6 5 3 5 6 3 6 -- WulfraedDennis Lee Bieber KD6MOG [EMAIL PROTECTED] [EMAIL PROTECTED] HTTP://wlfraed.home.netcom.com/ (Bestiaria Support Staff: [EMAIL PROTECTED]) HTTP://www.bestiaria.com/ -- http://mail.python.org/mailman/listinfo/python-list -- http://mail.python.org/mailman/listinfo/python-list
Re: Looping over lists
Dustan [EMAIL PROTECTED] writes: I think the for i in range() is more readable (Maybe because I'm coming from a c-style syntax language background) - but what would the benefits of using enumerate be (other that being more . . . pythonesque?) It doesn't create a whole new list just for iterating. That's what xrange is for. -- http://mail.python.org/mailman/listinfo/python-list
Re: Looping over lists
Dustan [EMAIL PROTECTED] wrote: On May 5, 3:15 am, kaens [EMAIL PROTECTED] wrote: I think the for i in range() is more readable (Maybe because I'm coming from a c-style syntax language background) - but what would the benefits of using enumerate be (other that being more . . . pythonesque?) It doesn't create a whole new list just for iterating. As the example list was of length 5, that's not all that important in this case. In cases where it _is_ crucial, you can use xrange. The performance of the various ways of looping is substantially the same: $ python -mtimeit -s'n=5; a=n*[23]' 'for i in range(n): x=a[i]' 100 loops, best of 3: 1.4 usec per loop $ python -mtimeit -s'n=5; a=n*[23]' 'for i in xrange(n): x=a[i]' 100 loops, best of 3: 1.18 usec per loop $ python -mtimeit -s'n=5; a=n*[23]' 'for i,v in enumerate(a): x=v' 100 loops, best of 3: 1.49 usec per loop $ Here, xrange is minutely faster and enumerate slower, but each speed difference in the noise. Focusing on clarity is thus well warranted. for i in range(n): for j in range(i+1, n): print a[i], a[j] Ah, but wouldn't the cleaner Python be something like a = [1, 2, 3, 4, 5, 3, 6] #just to confuse matters for pos, val in enumerate(a): ... for v2 in a[pos+1:]: ... print val, v2 This breaks symmetry, by using enumerate in the outer loop and a slice in the inner loop; the symmetrical construction of using range in both loops is a big conceptual/clarity win -- the reader of the code needs to grasp one fewer concept (locally). Using xrange in both loops would be just as good from this POV. Alex -- http://mail.python.org/mailman/listinfo/python-list
Re: Looping over lists
Tommy Grav [EMAIL PROTECTED] wrote:
I have a list:
a = [1., 2., 3., 4., 5.]
I want to loop over a and then
loop over the elements in a
that is to the right of the current
element of the first loop
In C this would be equivalent to:
for(i = 0; i n; i++) {
for(j=i+1; j n; j++) {
print a[i], a[j]
and should yield:
1. 2.
1. 3.
1. 4.
1. 5.
2. 3.
2. 4.
2. 5.
3. 4.
3. 5.
4. 5.
Can anyone help me with the right approach for this
in python?
Closes to the C++ code would be:
for i in range(n):
for j in range(i+1, n):
print a[i], a[j]
Alex
--
http://mail.python.org/mailman/listinfo/python-list
Re: Looping over lists
On Friday 04 May 2007 18:40:53 Tommy Grav wrote: Can anyone help me with the right approach for this in python? for each in a: for item in a[a.index(each)+1:]: print each,item will produce 1 2 1 3 1 4 1 5 2 3 2 4 2 5 3 4 3 5 4 5 a.index(each) gives the index of the each value in the a list. then you just add 1 to it so you start at the index value beside each's index. -- In friendship, prad ... with you on your journey Towards Freedom http://www.towardsfreedom.com (website) Information, Inspiration, Imagination - truly a site for soaring I's -- http://mail.python.org/mailman/listinfo/python-list
Re: Looping over lists
Tommy Grav wrote:
I have a list:
a = [1., 2., 3., 4., 5.]
I want to loop over a and then
loop over the elements in a
that is to the right of the current
element of the first loop
In C this would be equivalent to:
for(i = 0; i n; i++) {
for(j=i+1; j n; j++) {
print a[i], a[j]
and should yield:
1. 2.
1. 3.
1. 4.
1. 5.
2. 3.
2. 4.
2. 5.
3. 4.
3. 5.
4. 5.
Can anyone help me with the right approach for this
in python?
Two more options:
def pop_combine(items):
items = list(items)
while items:
a = items.pop(0)
for b in items:
print a, b
def enumerate_combine(items):
for i, a in enumerate(items):
for b in items[i+1:]:
print a, b
Peter
--
http://mail.python.org/mailman/listinfo/python-list
