Using range() within a function in Python

Question

I am creating a function that should return a list of every third number between start and 100 (inclusive). For example, every_three_nums(91) should return the list [91, 94, 97, 100]. Initially, I tried the following:

def every_three_nums(start):
  list = []
  a = 0
  if start > 100:
    return list
  else:
    a = range(start, 101, 3)
    print(list(a))    

print(every_three_nums(91))

And I get TypeError: 'list' object is not callable.

The same happens if I use

print(list(range(91, 101, 3)))

within the function. Despite this working outside the function.

I don't quite understand the rationale. Could you help me with this?

Eventually, I solved with:

def every_three_nums(start):
  list = []
  a = 0
  if start > 100:
    return list
  else:
    for i in range(start, 101, 3):
      list.append(i)
    return list

print(every_three_nums(91))

In the first code, you declare list as a variable in the first code and then you call the list after some time. — Albin Paul
– Albin Paul, Commented Apr 16, 2020 at 17:15

Badgy · Accepted Answer · 2020-04-16 17:16:51Z

1

You are redefining the list function to a list instance inside the function with list = []. So when you call list(a), you are treating a list object as a function. Rename that variable and it should work.

answered Apr 16, 2020 at 17:16

Badgy

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