breaking out of outer loops

[noemailplease0001]

Any elegant way of breaking out of the outer for loop than below, I
seem to have come across something, but it escapes me

for i in outerLoop:
for j in innerLoop:
if condition:
break
else:
continue
break

Thanks,
K

[Paul Rubin]

no***************@gmail.com writes:

Any elegant way of breaking out of the outer for loop than below, I
seem to have come across something, but it escapes me

for i in outerLoop:
for j in innerLoop:
if condition:
break
else:
continue
break

You can do it with a try/except/raise statement but I generally prefer
to wrap both loops in a function and use a "return" statement.

[pataphor]

On Tue, 29 Jan 2008 11:51:04 -0800 (PST)
no***************@gmail.com wrote:

Any elegant way of breaking out of the outer for loop than below, I
seem to have come across something, but it escapes me

for i in outerLoop:
for j in innerLoop:
if condition:
break
else:
continue
break

Ha! Think outside the box to begin with ...

P.

def cross(args):
ans = [[]]
for arg in args:
ans = [x+[y] for x in ans for y in arg]
return ans

def test():
L = [[0,1,2]]*3
for a,b,c in cross(L):
print a,b,c

if __name__=='__main__':
test()

[Arnaud Delobelle]

On Jan 29, 8:55*pm, pataphor <[e-mail]:

On Tue, 29 Jan 2008 11:51:04 -0800 (PST)

[e-mail] wrote:

Any elegant way of breaking out of the outer for loop than below, I
seem to have come across something, but it escapes me

for i in outerLoop:
* *for j in innerLoop:
* * * *if condition:
* * * * * break
* *else:
* * * *continue
* * break

Ha! Think outside the box to begin with ...

P.

def cross(args):
* * ans = [[]]
* * for arg in args:
* * * * ans = [x+[y] for x in ans for y in arg]
* * return ans * *

While we're at it, a generator version:

def iproduct(head=None, *tail):
if head is None:
return ((),)
else:
return ((x,)+y for x in head for y in iproduct(*tail))

for a, b, c in iproduct('124', 'ab', 'AB'):
print a, b, c

;-)

--
Arnaud

[Jeremy Sanders]

no***************@gmail.com wrote:

Any elegant way of breaking out of the outer for loop than below, I
seem to have come across something, but it escapes me

for i in outerLoop:
for j in innerLoop:
if condition:
break
else:
continue
break

Perhaps Python needs a "continue N" or a "break N" statement :-)

for i in outerLoop:
for j in innerLoop:
if condition:
break 2

Seeing as we can't have a goto :-)

Jeremy

--
Jeremy Sanders
http://www.jeremysanders.net/

[Diez B. Roggisch]

no***************@gmail.com schrieb:

Any elegant way of breaking out of the outer for loop than below, I
seem to have come across something, but it escapes me

for i in outerLoop:
for j in innerLoop:
if condition:
break
else:
continue
break

It's working because for-loops else statements are are executed only if
the loop hasn't been terminated unexpectedly. Which is what happens
here: if the inner loop is breaked, it's else is not executed. So the
outer loop's break is called.

Diez

[Steven Bethard]

Jeremy Sanders wrote:

no***************@gmail.com wrote:

>Any elegant way of breaking out of the outer for loop than below, I
seem to have come across something, but it escapes me

for i in outerLoop:
for j in innerLoop:
if condition:
break
else:
continue
break

Perhaps Python needs a "continue N" or a "break N" statement :-)

for i in outerLoop:
for j in innerLoop:
if condition:
break 2

Seeing as we can't have a goto :-)

Not true! You just have to import it:

http://entrian.com/goto/

STeVe