小编典典

如何在 Python 中将一个字符串附加到另一个字符串?

all

除了以下之外,我想要一种有效的方法在 Python 中将一个字符串附加到另一个字符串。

var1 = "foo"
var2 = "bar"
var3 = var1 + var2

有什么好的内置方法可以使用吗?


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2022-03-03

共1个答案

小编典典

如果您只有一个对字符串的引用,并且您将另一个字符串连接到末尾,CPython 现在是特殊情况,并尝试在适当的位置扩展字符串。

最终结果是操作摊销 O(n)。

例如

s = ""
for i in range(n):
    s+=str(i)

以前是 O(n^2),现在是 O(n)。

从源(bytesobject.c):

void
PyBytes_ConcatAndDel(register PyObject **pv, register PyObject *w)
{
    PyBytes_Concat(pv, w);
    Py_XDECREF(w);
}


/* The following function breaks the notion that strings are immutable:
   it changes the size of a string.  We get away with this only if there
   is only one module referencing the object.  You can also think of it
   as creating a new string object and destroying the old one, only
   more efficiently.  In any case, don't use this if the string may
   already be known to some other part of the code...
   Note that if there's not enough memory to resize the string, the original
   string object at *pv is deallocated, *pv is set to NULL, an "out of
   memory" exception is set, and -1 is returned.  Else (on success) 0 is
   returned, and the value in *pv may or may not be the same as on input.
   As always, an extra byte is allocated for a trailing \0 byte (newsize
   does *not* include that), and a trailing \0 byte is stored.
*/

int
_PyBytes_Resize(PyObject **pv, Py_ssize_t newsize)
{
    register PyObject *v;
    register PyBytesObject *sv;
    v = *pv;
    if (!PyBytes_Check(v) || Py_REFCNT(v) != 1 || newsize < 0) {
        *pv = 0;
        Py_DECREF(v);
        PyErr_BadInternalCall();
        return -1;
    }
    /* XXX UNREF/NEWREF interface should be more symmetrical */
    _Py_DEC_REFTOTAL;
    _Py_ForgetReference(v);
    *pv = (PyObject *)
        PyObject_REALLOC((char *)v, PyBytesObject_SIZE + newsize);
    if (*pv == NULL) {
        PyObject_Del(v);
        PyErr_NoMemory();
        return -1;
    }
    _Py_NewReference(*pv);
    sv = (PyBytesObject *) *pv;
    Py_SIZE(sv) = newsize;
    sv->ob_sval[newsize] = '\0';
    sv->ob_shash = -1;          /* invalidate cached hash value */
    return 0;
}

凭经验验证很容易。

$ python -m timeit -s"s=''" "for i in xrange(10):s+='a'"
1000000 loops, best of 3: 1.85 usec per loop
$ python -m timeit -s"s=''" "for i in xrange(100):s+='a'"
10000 loops, best of 3: 16.8 usec per loop
$ python -m timeit -s"s=''" "for i in xrange(1000):s+='a'"
10000 loops, best of 3: 158 usec per loop
$ python -m timeit -s"s=''" "for i in xrange(10000):s+='a'"
1000 loops, best of 3: 1.71 msec per loop
$ python -m timeit -s"s=''" "for i in xrange(100000):s+='a'"
10 loops, best of 3: 14.6 msec per loop
$ python -m timeit -s"s=''" "for i in xrange(1000000):s+='a'"
10 loops, best of 3: 173 msec per loop

但重要的 是要注意这种优化不是 Python 规范的一部分。据我所知,它仅在 cPython 实现中。例如,对 pypy 或 jython
的相同经验测试可能会显示较旧的 O(n**2) 性能。

$ pypy -m timeit -s"s=''" "for i in xrange(10):s+='a'"
10000 loops, best of 3: 90.8 usec per loop
$ pypy -m timeit -s"s=''" "for i in xrange(100):s+='a'"
1000 loops, best of 3: 896 usec per loop
$ pypy -m timeit -s"s=''" "for i in xrange(1000):s+='a'"
100 loops, best of 3: 9.03 msec per loop
$ pypy -m timeit -s"s=''" "for i in xrange(10000):s+='a'"
10 loops, best of 3: 89.5 msec per loop

到目前为止一切顺利,但是,

$ pypy -m timeit -s"s=''" "for i in xrange(100000):s+='a'"
10 loops, best of 3: 12.8 sec per loop

哎哟甚至比二次更糟糕。所以 pypy 对短字符串效果很好,但对大字符串效果很差。

2022-03-03