Whittle it on down

[DFS]

On 5/5/2016 2:04 AM, Steven D'Aprano wrote:
> On Thursday 05 May 2016 14:58, DFS wrote:
>
>> Want to whittle a list like this:
> [...]
>> Want to keep all elements containing only upper case letters or upper
>> case letters and ampersand (where ampersand is surrounded by spaces)
>
>
> Start by writing a function or a regex that will distinguish strings that
> match your conditions from those that don't. A regex might be faster, but
> here's a function version.
>
> def isupperalpha(string):
>     return string.isalpha() and string.isupper()
>
> def check(string):
>     if isupperalpha(string):
>         return True
>     parts = string.split("&")
>     if len(parts) < 2:
>         return False
>     # Don't strip leading spaces from the start of the string.
>     parts[0] = parts[0].rstrip(" ")
>     # Or trailing spaces from the end of the string.
>     parts[-1] = parts[-1].lstrip(" ")
>     # But strip leading and trailing spaces from the middle parts
>     # (if any).
>     for i in range(1, len(parts)-1):
>         parts[i] = parts[i].strip(" ")
>      return all(isupperalpha(part) for part in parts)
>
>
> Now you have two ways of filtering this. The obvious way is to extract
> elements which meet the condition. Here are two ways:
>
> # List comprehension.
> newlist = [item for item in oldlist if check(item)]
>
> # Filter, Python 2 version
> newlist = filter(check, oldlist)
>
> # Filter, Python 3 version
> newlist = list(filter(check, oldlist))
>
>
> In practice, this is the best (fastest, simplest) way. But if you fear that
> you will run out of memory dealing with absolutely humongous lists with
> hundreds of millions or billions of strings, you can remove items in place:
>
>
> def remove(func, alist):
>     for i in range(len(alist)-1, -1, -1):
>         if not func(alist[i]):
>             del alist[i]
>
>
> Note the magic incantation to iterate from the end of the list towards the
> front. If you do it the other way, Bad Things happen. Note that this will
> use less memory than extracting the items, but it will be much slower.
>
> You can combine the best of both words. Here is a version that uses a
> temporary list to modify the original in place:
>
> # works in both Python 2 and 3
> def remove(func, alist):
>     # Modify list in place, the fast way.
>     alist[:] = filter(check, alist)


You are out of your mind.