clever exit of nested loops
John Ladasky
john_ladasky at sbcglobal.net
Thu Sep 27 04:00:36 EDT 2018
On Wednesday, September 26, 2018 at 12:50:20 AM UTC-7, vito.d... at gmail.com wrote:
> I have "abused" the "else" clause of the loops to makes a break "broke" more loops
I did this once upon a time. In recent years, when I start writing tricky nested loops, I frequently find myself reaching for itertools.product() to flatten the loops instead.
This code accomplishes the same task as yours. I'll leave it to you to decide whether you prefer it. There are things that I dislike about it, but the flow control part is clear.
from itertools import product
msgs = ("i: {}", "\tj: {}", "\t\tk: {}")
old = 3*[None]
for new in product(range(10), repeat=3):
for n, (msg, changed) in enumerate(zip(msgs, [x!=y for x, y in zip(old, new)])):
if changed:
print(msg.format(new[n]))
if condition(*new): # your condition() took three separate arguments
break
old = new
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