zeehio - 8 months ago 38

Python Question

I am looking for a nice way to

`zip`

In the case where the iterables are lists or have a

`len`

`def zip_equal(it1, it2):`

if len(it1) != len(it2):

raise ValueError("Lengths of iterables are different")

return zip(it1, it2)

However, if

`it1`

`it2`

`TypeError: object of type 'generator' has no len()`

I imagine the

`itertools`

`def zip_equal(it1, it2):`

exhausted = False

while True:

try:

el1 = next(it1)

if exhausted: # in a previous iteration it2 was exhausted but it1 still has elements

raise ValueError("it1 and it2 have different lengths")

except StopIteration:

exhausted = True

# it2 must be exhausted too.

try:

el2 = next(it2)

# here it2 is not exhausted.

if exhausted: # it1 was exhausted => raise

raise ValueError("it1 and it2 have different lengths")

except StopIteration:

# here it2 is exhausted

if not exhausted:

# but it1 was not exhausted => raise

raise ValueError("it1 and it2 have different lengths")

exhausted = True

if not exhausted:

yield (el1, el2)

else:

return

The solution can be tested with the following code:

`it1 = (x for x in ['a', 'b', 'c']) # it1 has length 3`

it2 = (x for x in [0, 1, 2, 3]) # it2 has length 4

list(zip_equal(it1, it2)) # len(it1) < len(it2) => raise

it1 = (x for x in ['a', 'b', 'c']) # it1 has length 3

it2 = (x for x in [0, 1, 2, 3]) # it2 has length 4

list(zip_equal(it2, it1)) # len(it2) > len(it1) => raise

it1 = (x for x in ['a', 'b', 'c', 'd']) # it1 has length 4

it2 = (x for x in [0, 1, 2, 3]) # it2 has length 4

list(zip_equal(it1, it2)) # like zip (or izip in python2)

Am I overlooking any alternative solution? Is there a simpler implementation of my

`zip_equal`

PS: I wrote the question thinking in Python 3, but a Python 2 solution is also welcome.

Answer

I can a simpler solution, use `itertools.zip_longest()`

and raise an exception if the sentinel value used to pad out shorter iterables is present in the tuple produced:

```
from itertools import zip_longest
def zip_equal(*iterables):
sentinel = object()
for combo in zip_longest(*iterables, fillvalue=sentinel):
if sentinel in combo:
raise ValueError('Iterables have different lengths')
yield combo
```

Unfortunately, we can't use `zip()`

with `yield from`

to avoid a Python-code loop with a test each iteration; once the shortest iterator runs out, `zip()`

would advance all preceding iterators and thus swallow the evidence if there is but one extra item in those.

Source (Stackoverflow)