default value in a list

Fredrik Lundh fredrik at
Sat Jan 22 02:37:41 EST 2005

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]


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