Am 13.03.16 um 20:39 schrieb BartC:
switch(c)
if case(ord("A"),ord("B"),ord("C"),ord("D"),ord("E"),ord("F"),
ord("G"),ord("H"),ord("I"),ord("J"),ord("K"),ord("L"),
ord("M"),ord("N"),ord("O"),ord("P"),ord("Q"),ord("R"),
ord("S"),ord("T"),ord("U"),ord("V"),ord("W"),ord("X"),
ord("Y"),ord("Z")):
upper+=1
elif case(ord("a"),ord("b"),ord("c"),ord("d"),ord("e"),ord("f"),
ord("g"),ord("h"),ord("i"),ord("j"),ord("k"),ord("l"),
ord("m"),ord("n"),ord("o"),ord("p"),ord("q"),ord("r"),
ord("s"),ord("t"),ord("u"),ord("v"),ord("w"),ord("x"),
ord("y"),ord("z")):
lower+=1
elif case(ord("0"),ord("1"),ord("2"),ord("3"),ord("4"),ord("5"),
ord("6"),ord("7"),ord("8"),ord("9")):
digits+=1
else:
other+=1
It worked, but took 110 seconds; 80 seconds without the ord's and
comparing strings (but I still think it's perverse that integer ops are
slower than string ops).
I assume you run this in a big loop. What about a single hash-table lookup?
from collections import Counter
counts=Counter()
for c in whatever:
counts[c]+=1
upper=sum(counts[x] for x in range(ord('A'), ord('Z')+1)
lower=sum(counts[x] for x in range(ord('a'), ord('z')+1)
....
I think this is equally readable as the switch version, and should be
much faster.
Christian
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