Martin v. Löwis wrote: > Nick Coghlan wrote: > > x = L[i:i+n]; del L[i:i+n]; return x > > > > By default, n = 1, so the standard behaviour of list.pop is preserved. > > This default would actually change the standard behaviour: whereas it > now returns a single element, it would then return a list containing > the single element. Too bad. I would have liked list.pop(i, n=1, step=1) to be the same as x = L[i:i+n:step]; del L[i:i+n:step]; return x Having pop()/pop(i) return an element and pop(i, n)/pop(i, n, step) return a list is a no-no, or is it? ...johahn
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