default value in a list

[Fredrik Lundh]

Paul McGuire wrote:

> I asked a very similar question a few weeks ago, and from the various
> suggestions, I came up with this:
>
> expand = lambda lst,default,minlen : (lst + [default]*minlen)[0:minlen]

I wouldn't trust whoever suggested that.  if you want a function, use a function:

    def expand(seq, default, minlen):
        return (seq + [default]*minlen)[:minlen]

</F>