On Thu, 19 May 2016 03:00 am, MRAB wrote: > I think your results are inconsistent. > > For an odd number of characters you have "abc" + "de" + "fg", i.e. more > on the left, but for an even number of characters you have "a" + "bcd" + > "ef", i.e. more on the right.
Correct. That's intentional. I didn't start with an algorithm. I started by manually extracting the N middle characters, for various values of N, then wrote code to get the same result. For example, if I start with an odd-length string, like "inquisition", then the "middle N" cases for odd-N are no-brainers, because they have to be centered on the middle character: N=1 's' N=3 'isi' N=5 'uisit' For even-N, I had a choice: N=2 'is' or 'si' and to be perfectly frank, I didn't really care much either way and just arbitrarily picked the second, based on the fact that string.center() ends up with a slight bias to the right: py> 'ab'.center(5, '*') '**ab*' For even-length string, like "aardvark", the even-N case is the no-brainer: N=2 "dv" N=4 "rdva" N=6 "ardvar" but with odd-N I have a choice: N=3 "rdv" or "dva" In this case, I *intentionally* biased it the other way, so that (in some sense) overall the mid() function would be unbiased: - half the cases, there's no bias at all; - a quarter of the time, there's a bias to the right; - a quarter of the time, there's a bias to the left; - so on average, the bias is zero. > My own solution is: > > > def mid(string, n): > """Return middle n chars of string.""" > if n <= 0: > return '' > if n > len(string): > return string > ofs = (len(string) - n) // 2 > return string[ofs : ofs + n] > > > If there's an odd number of characters remaining, it always has more on > the right. Thanks to you and Ian (who independently posted a similar solution), that's quite good too if you don't care about the bias. -- Steven -- https://mail.python.org/mailman/listinfo/python-list
