我測驗了 200 萬個字串s.replace('a', '')在哪里，并且在開始、中間或結束時可能是一個例外值。那個單一的例外值使它慢得多：s'a''b'
在 TIO，他們的 Python 3.8 預發布版本：
a = 'a' * 10**6; s = f'{a}{a}' 5 ms 5 ms 5 ms
a = 'a' * 10**6; s = f'b{a}{a}' 24 ms 24 ms 24 ms
a = 'a' * 10**6; s = f'{a}b{a}' 25 ms 25 ms 25 ms
a = 'a' * 10**6; s = f'{a}{a}b' 25 ms 25 ms 25 ms
a = 'a' * 10**6; s = f'b{a}b{a}b' 25 ms 25 ms 25 ms
在裝有 Python 3.10 的舊筆記本電腦上：
a = 'a' * 10**6; s = f'{a}{a}' 4 ms 4 ms 4 ms
a = 'a' * 10**6; s = f'b{a}{a}' 93 ms 94 ms 95 ms
a = 'a' * 10**6; s = f'{a}b{a}' 94 ms 95 ms 95 ms
a = 'a' * 10**6; s = f'{a}{a}b' 94 ms 94 ms 95 ms
a = 'a' * 10**6; s = f'b{a}b{a}b' 95 ms 95 ms 96 ms
那個單一的例外值是如何'b讓它變得這么慢的？
完整的基準代碼：
from timeit import repeat
setups = [
"a = 'a' * 10**6; s = f'{a}{a}'",
"a = 'a' * 10**6; s = f'b{a}{a}'",
"a = 'a' * 10**6; s = f'{a}b{a}'",
"a = 'a' * 10**6; s = f'{a}{a}b'",
"a = 'a' * 10**6; s = f'b{a}b{a}b'",
]
for _ in range(3):
for setup in setups:
times = repeat("s.replace('a', '')", setup, number=1)
print(f'{setup:{max(map(len, setups))}}',
*('= ms ' % (t * 1e3) for t in sorted(times)[:3]))
print()
uj5u.com熱心網友回復：
str.replace()在 C 函式中的https://github.com/python/cpython/blob/main/Objects/unicodeobject.c#L10604中實作replace()。這是該函式的摘錄。它表明如果回傳的字串 ( new_size) 的大小為 0，我們會提前停止并回傳一個空字串。否則，我們將執行遍歷輸入字串并逐個執行替換的冗長任務。
new_size = slen n * (len2 - len1);
if (new_size == 0) {
u = unicode_new_empty();
goto done;
}
if (new_size > (PY_SSIZE_T_MAX / rkind)) {
PyErr_SetString(PyExc_OverflowError,
"replace string is too long");
goto error;
}
u = PyUnicode_New(new_size, maxchar);
if (!u)
goto error;
assert(PyUnicode_KIND(u) == rkind);
res = PyUnicode_DATA(u);
ires = i = 0;
if (len1 > 0) {
while (n-- > 0) {
/* look for next match */
j = anylib_find(rkind, self,
sbuf rkind * i, slen-i,
str1, buf1, len1, i);
if (j == -1)
break;
else if (j > i) {
/* copy unchanged part [i:j] */
memcpy(res rkind * ires,
sbuf rkind * i,
rkind * (j-i));
ires = j - i;
}
/* copy substitution string */
if (len2 > 0) {
memcpy(res rkind * ires,
buf2,
rkind * len2);
ires = len2;
}
i = j len1;
}
if (i < slen)
/* copy tail [i:] */
memcpy(res rkind * ires,
sbuf rkind * i,
rkind * (slen-i));
}
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