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Python Python Collections (Retired) Slices Slice Functions

Sergei Miroshnikov
Sergei Miroshnikov
3,313 Points

Bummer! Try again! This is amazing error message ! Please explain why my solution is not valid ...

def first_4(iterable):
    result = []
    for i in (range(0, iterable)):
        result.append(i)
        if i == 3:
            break
    return result

I have tried another function I wrote

def first_4(iterable):
    result = []
    for i in range(0, iterable):
        result.append(i)
    return result[:4]

2 Answers

Brendan Whiting
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
Brendan Whiting
Front End Web Development Techdegree Graduate 84,735 Points

I made a few tweaks to your code to make it work.

  • We can't put an iterable into the range function, I changed it to len(iterable) which is the number of items in it

  • You want to append the item in the iterable at the index of i, not just i itself.

def first_4(iterable): 
  result = [] 
  for i in (range(0, len(iterable))): 
    result.append(iterable[i]) 
    if i == 3: break 
  return result

Alternatively, we could have just put 3 as the 2nd item in the range, and it would have stopped on its own, rather than having to break the loop.

And there is also a much much easier way. We can use list slicing like this:

def first_4(iterable): 
  return iterable[:4]