[Tutor] changing char list to int list isn't working

[Steven D'Aprano]

On 04/05/13 14:13, Jim Mooney wrote:
> I'm turning an integer into a string so I can make a list of separate
> chars, then turn those chars back into individual ints, but the
> resulting list still looks like string chars when I print it. What am
> I doing wrong?
>
> listOfNumChars = list(str(intNum))

This creates a list of characters.


> for num in listOfNumChars:
>      num = int(num)

This walks over the list, setting the variable "num" to each character in turn, then inside the loop you set the variable "num" to the converted char->int. But num isn't linked to the list in any way -- num has no memory that the value it got came from a list. Reassigning num inside the loop doesn't touch the list in any way, so naturally the list doesn't change.


> print(listOfNumChars)
>
> # result of 455 entered is ['4', '5', '5']


To split a number into digits, the shortest way is to use a list comprehension:


digits = [int(c) for c in str(num)]


We can expand that list comp into a for-loop:


digits = []
for c in str(num):
     digits.append(int(c))



Notice that there is no need to convert the string into a list. You can iterate over the characters of a string just fine.

If you prefer to create a list, then modify it in place, we can do this:


digits = list(str(num))
for position, char in enumerate(digits):
     digits[position] = int(char)



Here we use enumerate() to iterate over pairs of (position, value) instead of just value:

py> list("abcd")
['a', 'b', 'c', 'd']
py> list(enumerate("abcd"))
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd')]



Here's another way, using map:


digits = map(int, str(num))  # Python 2.x only

digits = list(map(int, str(num)))  # Python 3.x or better



Why the difference between Python 2.x and 3.x? In 2.x, map is "eager", it runs all the way through the string as soon as you call it, returning a list. In 3.x, map is "lazy", and only generates values when and as needed. By wrapping the map generator in a call to list, that forces it to run all the way through the string.



-- 
Steven