Write python generators below(yield): The group generator takes an iterable and

Question

Write python generators below(yield):

The group generator takes an iterable and an int (call it n) as parameters: it produces lists of n values: the first list contains the first n values from the iterable; the second list contains the second n values from the iterable, etc. until there are fewer than n values left to put in the returned list. For example for i in gropu('abcdefghijklm',4): print(i,end='') prints ['a','b','c','d'] ['e','f','g','h'] ['i','j','k','l']. Hint: call iter and next directly, building a list with n values,so it doesn’t violate the conditions for using iterables.

Explanation / Answer

def group(iterable,n):
count = 0
lis = []
it = iter(iterable)
while True:
try:
lis.append(next(it))
count += 1
if count == n:
yield lis
lis = []
count = 0
except:
return
  
for val in group('abcdefghijklm', 4):
print val

# code link: https://paste.ee/p/R7JBI

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