Redo for loop iteration in Python
Solution 1
No, Python doesn't have direct support for redo
. One option would something faintly terrible involving nested loops like:
for x in mylist:
while True:
...
if shouldredo:
continue # continue becomes equivalent to redo
...
if shouldcontinue:
break # break now equivalent to continue on outer "real" loop
...
break # Terminate inner loop any time we don't redo
but this mean that break
ing the outer loop is impossible within the "redo
-able" block without resorting to exceptions, flag variables, or packaging the whole thing up as a function.
Alternatively, you use a straight while
loop that replicates what for
loops do for you, explicitly creating and advancing the iterator. It has its own issues (continue
is effectively redo
by default, you have to explicitly advance the iterator for a "real" continue
), but they're not terrible (as long as you comment uses of continue
to make it clear you intend redo
vs. continue
, to avoid confusing maintainers). To allow redo
and the other loop operations, you'd do something like:
# Create guaranteed unique sentinel (can't use None since iterator might produce None)
sentinel = object()
iterobj = iter(mylist) # Explicitly get iterator from iterable (for does this implicitly)
x = next(iterobj, sentinel) # Get next object or sentinel
while x is not sentinel: # Keep going until we exhaust iterator
...
if shouldredo:
continue
...
if shouldcontinue:
x = next(iterobj, sentinel) # Explicitly advance loop for continue case
continue
...
if shouldbreak:
break
...
# Advance loop
x = next(iterobj, sentinel)
The above could also be done with a try
/except StopIteration:
instead of two-arg next
with a sentinel
, but wrapping the whole loop with it risks other sources of StopIteration
being caught, and doing it at a limited scope properly for both inner and outer next
calls would be extremely ugly (much worse than the sentinel
based approach).
Solution 2
No, it doesn't. I would suggest using a while loop and resetting your check variable to the initial value.
count = 0
reset = 0
while count < 9:
print 'The count is:', count
if not someResetCondition:
count = count + 1
Solution 3
I just meet the same question when I study perl,and I find this page.
follow the book of perl:
my @words = qw(fred barney pebbles dino wilma betty);
my $error = 0;
my @words = qw(fred barney pebbles dino wilma betty);
my $error = 0;
foreach (@words){
print "Type the word '$_':";
chomp(my $try = <STDIN>);
if ($try ne $_){
print "Sorry - That's not right.\n\n";
$error++;
redo;
}
}
and how to achieve it on Python ?? follow the code:
tape_list=['a','b','c','d','e']
def check_tape(origin_tape):
errors=0
while True:
tape=raw_input("input %s:"%origin_tape)
if tape == origin_tape:
return errors
else:
print "your tape %s,you should tape %s"%(tape,origin_tape)
errors += 1
pass
all_error=0
for char in tape_list:
all_error += check_tape(char)
print "you input wrong time is:%s"%all_error
Python has not the "redo" syntax,but we can make a 'while' loop in some function until get what we want when we iter the list.
Solution 4
This is my solution using iterators:
class redo_iter(object):
def __init__(self, iterable):
self.__iterator = iter(iterable)
self.__started = False
self.__redo = False
self.__last = None
self.__redone = 0
def __iter__(self):
return self
def redo(self):
self.__redo = True
@property
def redone(self):
return self.__redone
def __next__(self):
if not (self.__started and self.__redo):
self.__started = True
self.__redone = 0
self.__last = next(self.__iterator)
else:
self.__redone += 1
self.__redo = False
return self.__last
# Display numbers 0-9.
# Display 0,3,6,9 doubled.
# After a series of equal numbers print --
iterator = redo_iter(range(10))
for i in iterator:
print(i)
if not iterator.redone and i % 3 == 0:
iterator.redo()
continue
print('---')
- Needs explicit
continue
-
redone
is an extra feature - For Python2 use
def next(self)
instead ofdef __next__(self)
- requires
iterator
to be defined before the loop
Solution 5
Not very sophiscated but easy to read, using a while
and an increment at the end of the loop. So any continue
in between will have the effect of a redo. Sample to redo every multiple of 3:
redo = True # To ends redo condition in this sample only
i = 0
while i<10:
print(i, end='')
if redo and i % 3 == 0:
redo = False # To not loop indifinively in this sample
continue # Redo
redo = True
i += 1
Result: 00123345667899
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Matjaž Leonardis
Updated on June 04, 2022Comments
-
Matjaž Leonardis almost 2 years
Does Python have anything in the fashion of a "redo" statement that exists in some languages?
(The "redo" statement is a statement that (just like "break" or "continue") affects looping behaviour - it jumps at the beginning of innermost loop and starts executing it again.)
-
miradulo about 8 yearsThere are plenty of ways you can do this. For one, you could use a
while
loop and reset whatever your counter / condition is upon some evaluation. -
Christopher Schneider about 8 yearsNever heard of such a thing. Sounds a lot like goto
-
ShadowRanger about 8 years@ChristopherSchneider: Perl uses it (for what that's worth). Think a
continue
that doesn't perform the loop advancement step. Since it's tied to the loop itself, it's not morally distinct fromcontinue
andbreak
really; if you accept them as something other than justgoto
, thenredo
is no worse (or better).
-
-
ShadowRanger about 8 yearsUsing
reset
doesn't replicate whatredo
does in other languages.redo
iscontinue
without the loop increment/advance step, but it doesn't restart the loop from the beginning; you'd just make thecount
increment optional, not have areset
variable. -
cmaynard about 8 yearsAh, I misread your initial statement "it jumps at the beginning" to mean the initial point, not just the top of the loop. I'll modify my answer.
-
ShadowRanger about 8 yearsI'm not the OP, it was their initial statement, not mine. I'll admit I could be misreading, but the only language I know of off-hand with
redo
is Perl, and it behaves this way. Note: The edited code is fine if you're replacingfor count in range(10):
, but it's not particularly generalizable to arbitrary iterables; my second code example in my answer is the fully generalized version. -
ddelange almost 5 yearsThis should be the accepted answer. A redo would not be very pythonic, and emulating your for-loop with and flag to redo if necessary will be easy to follow for other people. Thank you