Redo for loop iteration in Python

Question

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.)

Answer 1

Here is a solution for python 3.8+ since now we have the := operator:

for key in mandatory_attributes:  # example with a dictionary
    while not (value := input(f"{key} (mandatory): ")):
        print("You must enter a value")

    mandatory_attributes[key] = value

Answer 2

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

Answer 3

There is no redo in python. A very understandable solution is as follow:

for x in mylist:
    redo = True
    while redo:
        redo = False

        If should_redo:
            redo = True

It's clear enough to do not add comments

Continue will work as if it was in the for loop

But break is not useable, this solution make break useable but the code is less clear.

Answer 4

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.

Answer 5

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 of def __next__(self)
• requires iterator to be defined before the loop

Answer 6

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 breaking 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).

Answer 7

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