for loop in python with counter?

Question

How can I create a for loop with a counter? I have a list, and I want to read an element after each n elements. I'd initially done this

for i in enumerate(n):
    print(i)

But as expected it prints every element instead of every nth element, what would be the Python way of solving this?

Answer 1

I am not sure wich kind of value is n but usually there are this ways: (for me, n is a list)

for index, item in enumerate(n):
   if index % nth == 0: # If the output is not like you want, try doing (index + 1), or enumerate(n, start=1).
       print(item)

Other way could be:

for index in range(0, len(n), nth): # Only work with sequences
   print(n[index]) # If the output is not like you want, try doing n[index + 1]

Or:

for item in n[::nth]: # Low perfomance and hight memory consumption warning!! Only work with sequences
    print(item)

Even you can combine the first one with the last one:

for i, item in list(enumerate(n))[::nth]: # Huge low perfomance warning!!!
    print(item)

But I'm not sure if that has an advantage...

Also, if you are willing to make a function, you could do something similar to the enumerate function:

def myEnumerate(sequence, start=0, jump=1):
    n = start
    j = start // Or j = 0, that is your decision.
    for elem in sequence:
        if j % jump == 0:
            yield n, elem
            n += 1
        j += 1

for index, item in myEnumerate(n, jump=1):
    print(item)

Personally, I wouldn't do this last one. I'm not sure why but it's a feeling.

Perfomance test

n = 'a b c d e f g h i j k l m n ñ o p q r s t u v w x y z 1 2 3 4 5 6 7 8 9 ! " · $ % & / ( ) = ? ¿ Ç ç } { [ ] ; : _ ¨ ^ * ` + ´ - . , º ª \ /'.split(" ")
nth = 3    
def a():
    for i, item in enumerate(n):
       if i % nth == 0:
           item       
def b():
    for item in range(0, len(n), nth):
       n[item]           
def c():
    for item in n[::nth]:
        item    
def d():
    for i, item in list(enumerate(n))[::nth]:
       if i % nth == 0:
           item    
def enumerates(sequence, start=0, jump=1):
    n = start
    j = start
    for elem in sequence:
        if j % jump == 0:
            yield n, elem
            n += 1
        j += 1            
def e():
    for i, item in enumerates(n, jump= nth):
        item    
if __name__ == '__main__':
    import timeit
    print(timeit.timeit("a()", setup="from __main__ import a")) # 10.556324407152305
    print(timeit.timeit("b()", setup="from __main__ import b")) # 2.7166204783010137
    print(timeit.timeit("c()", setup="from __main__ import c")) # 1.0285353306076601
    print(timeit.timeit("d()", setup="from __main__ import d")) # 8.283859051918608
    print(timeit.timeit("e()", setup="from __main__ import e")) # 14.91601851631981

But if you are really looking for perfomance you should read @Martijn Pieters answer.