How do I append one string to another in Python? How do I append one string to another in Python? python python

How do I append one string to another in Python?


If you only have one reference to a string and you concatenate another string to the end, CPython now special cases this and tries to extend the string in place.

The end result is that the operation is amortized O(n).

e.g.

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

used to be O(n^2), but now it is O(n).

From the source (bytesobject.c):

voidPyBytes_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;}

It's easy enough to verify empirically.

$ 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

It's important however to note that this optimisation isn't part of the Python spec. It's only in the cPython implementation as far as I know. The same empirical testing on pypy or jython for example might show the older O(n**2) performance .

$ 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

So far so good, but then,

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

ouch even worse than quadratic. So pypy is doing something that works well with short strings, but performs poorly for larger strings.


Don't prematurely optimize. If you have no reason to believe there's a speed bottleneck caused by string concatenations then just stick with + and +=:

s  = 'foo's += 'bar's += 'baz'

That said, if you're aiming for something like Java's StringBuilder, the canonical Python idiom is to add items to a list and then use str.join to concatenate them all at the end:

l = []l.append('foo')l.append('bar')l.append('baz')s = ''.join(l)


str1 = "Hello"str2 = "World"newstr = " ".join((str1, str2))

That joins str1 and str2 with a space as separators. You can also do "".join(str1, str2, ...). str.join() takes an iterable, so you'd have to put the strings in a list or a tuple.

That's about as efficient as it gets for a builtin method.


matomo