How to enumerate from the middle of tuple [duplicate]

Question

By default, enumerate() starts counting at 0 but if you give it a second integer argument, it'll start from that number instead but how can we enumerate from a particular value of count and elem. for example, we want to pass count as '2' and elem as 'bar'. both values have to pass to enumerate function.

elements = ('foo', 'bar', 'baz')
for count, elem in enumerate(elements):
    print (count, elem)

Answer 1

You can iterate a slice of your data and start the enumeration at a specific integer:

start_at = 1
elements = ('foo', 'bar', 'baz')
for count, elem in enumerate(elements[start_at:], start_at):
    print (count, elem)

Output:

1 bar
2 baz

Documentation:

• enumerate(iterable, start=0) (specify whatever for start to start numbering the 0th element of iterable with that number onwards)

---

Edit:

If you are working with non-sliceable iterators you have to work around it:

start_at = 3   # first one included (0 based)
stop_at = 5    # last one (not included)

elements = ('foo', 'bar', 'baz', 'wuzz', 'fizz', 'fizzbuzz', 'wuzzfizzbuzz', 'foobarfuzzbizzwuzz')

# skip first few
it = iter(elements)
for _ in range(start_at):
    print("skipping: ", next(it))      # dont print, simply: next(it)

# enumerate the ones you want
i = start_at
for _ in range(stop_at-start_at):
    print(i,next(it))                 # you might need a try: except StopIteration:
    i+=1 

to get:

skipping:  foo
skipping:  bar
skipping:  baz
3 wuzz
4 fizz
5 fizzbuzz

Answer 2

the itertools module is where you should look when you want to iterate in a non standard way

in this case use islice

to get the start index, use the .index function of the container(or hardcode it to 1 if you want to iterate from the second position regardless of the contents), the end index is the len of the container

for count, elem in enumerate(itertools.islice(elements, elements.index('bar'), len(elements))
    print(count, elem)