Terry J. Reedy <[email protected]> added the comment:
The mapping passed to str.translate must map ints representing codepoints to
either either ints or strings. Translate can extract binary codepoints for the
new string from either. Ints are slightly faster, so I am inclined not to
switch.
import timeit
class ParseMapN(dict):
def __missing__(self, key): return 120
class ParseMapS(dict):
def __missing__(self, key): return 'x'
trans1 = ParseMapN.fromkeys(range(128), 120)
trans1.update((ord(c), ord('(')) for c in "({[")
trans1.update((ord(c), ord(')')) for c in ")}]")
trans1.update((ord(c), ord(c)) for c in "\"'\\\n#")
trans2 = ParseMapN.fromkeys(range(128), 'x')
trans2.update((ord(c), '(') for c in "({[")
trans2.update((ord(c), ')') for c in ")}]")
trans2.update((ord(c), c) for c in "\"'\\\n#")
for i in (1, 10, 100, 1000):
code = '\t a([{b}])b"c\'d\n' * i
print('N ', i)
print(timeit.repeat(
'code.translate(trans1)',
number=10000, globals = globals()))
print('S ', i)
print(timeit.repeat(
'code.translate(trans2)',
number=10000, globals = globals()))
N 1 [0.056562504, 0.056747570, 0.05654651, 0.056460749, 0.056428776]
S 1 [0.060795346, 0.062304155, 0.061043432, 0.062349345, 0.061191301]
N 10 [0.076474600, 0.076463227, 0.076560984, 0.076581582, 0.076010091]
S 10 [0.080739106, 0.080798745, 0.08034192, 0.080987501, 0.080617369]
N 100 [0.28529922, 0.28383868, 0.283949046, 0.284461512, 0.284291203]
S 100 [0.289629521, 0.288535418, 0.289154560, 0.28811548, 0.28862180]
N1000 [2.23882157, 2.2383192, 2.2384120, 2.2377972, 2.23854982]
S1000 [2.24242237, 2.2426845, 2.2424623, 2.2420030, 2.24254871]
The pattern of all S repeats being greater than all corresponding N repeats was
true for 2 other runs.
----------
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Python tracker <[email protected]>
<https://bugs.python.org/issue32940>
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