On Dec 29, 8:02 pm, Steven D'Aprano <st...@remove-this- cybersource.com.au> wrote: > On Mon, 29 Dec 2008 17:38:36 -0800, Ross wrote: > > On Dec 29, 8:07 pm, Scott David Daniels <scott.dani...@acm.org> wrote: > >> Ross wrote: > >> > ... Use get to write histogram more concisely. You should be able to > >> > eliminate the if statement. > > >> > def histogram(s): > >> > d = dict() > >> > for c in s: > >> > d[c]= d.get(c,0) > >> > return d > > >> > This code returns a dictionary of all the letters to any string s I > >> > give it but each corresponding value is incorrectly the default of 0. > >> > What am I doing wrong? > > >> How is this code supposed to count? > > >> --Scott David Daniels > >> scott.dani...@acm.org > > > I realize the code isn't counting, but how am I to do this without using > > an if statement as the problem instructs? snip > So what you need is: > > * set d[c] to whatever d[c] already is plus one, or 0 plus one if it > isn't already there. > > It's a two character change to one line. Let us know if you still can't > see it.
That's actually really sophisticated. Steven is being really clever here, by writing a number in an unusual way. He said: * set d[c] to whatever d[c] already is plus one, or 0 plus one if it isn't already there. It is the same as: * set d[c] to whatever d[c] already is plus one, or one if it isn't already there. Side-by-side. Perhaps you will learn his secret someday. It has to do with 0+1=1. Armed with it, he has: something something +1, something something something +1 Which brings us (to): ( something something, something something something ) +1 Three cheers for thinking. -- http://mail.python.org/mailman/listinfo/python-list
