In Python, how to check if a string only contains certain characters?
Final(?) edit
Answer, wrapped up in a function, with annotated interactive session:
>>> import re
>>> def special_match(strg, search=re.compile(r'[^a-z0-9.]').search):
... return not bool(search(strg))
...
>>> special_match("")
True
>>> special_match("az09.")
True
>>> special_match("az09.\n")
False
# The above test case is to catch out any attempt to use re.match()
# with a `$` instead of `\Z` -- see point (6) below.
>>> special_match("az09.#")
False
>>> special_match("az09.X")
False
>>>
Note: There is a comparison with using re.match() further down in this answer. Further timings show that match() would win with much longer strings; match() seems to have a much larger overhead than search() when the final answer is True; this is puzzling (perhaps it's the cost of returning a MatchObject instead of None) and may warrant further rummaging.
==== Earlier text ====
The [previously] accepted answer could use a few improvements:
(1) Presentation gives the appearance of being the result of an interactive Python session:
reg=re.compile('^[a-z0-9\.]+$')
>>>reg.match('jsdlfjdsf12324..3432jsdflsdf')
True
but match() doesn't return True
(2) For use with match(), the ^ at the start of the pattern is redundant, and appears to be slightly slower than the same pattern without the ^
(3) Should foster the use of raw string automatically unthinkingly for any re pattern
(4) The backslash in front of the dot/period is redundant
(5) Slower than the OP's code!
prompt>rem OP's version -- NOTE: OP used raw string!
prompt>\python26\python -mtimeit -s"t='jsdlfjdsf12324..3432jsdflsdf';import
re;reg=re.compile(r'[^a-z0-9\.]')" "not bool(reg.search(t))"
1000000 loops, best of 3: 1.43 usec per loop
prompt>rem OP's version w/o backslash
prompt>\python26\python -mtimeit -s"t='jsdlfjdsf12324..3432jsdflsdf';import
re;reg=re.compile(r'[^a-z0-9.]')" "not bool(reg.search(t))"
1000000 loops, best of 3: 1.44 usec per loop
prompt>rem cleaned-up version of accepted answer
prompt>\python26\python -mtimeit -s"t='jsdlfjdsf12324..3432jsdflsdf';import
re;reg=re.compile(r'[a-z0-9.]+\Z')" "bool(reg.match(t))"
100000 loops, best of 3: 2.07 usec per loop
prompt>rem accepted answer
prompt>\python26\python -mtimeit -s"t='jsdlfjdsf12324..3432jsdflsdf';import
re;reg=re.compile('^[a-z0-9\.]+$')" "bool(reg.match(t))"
100000 loops, best of 3: 2.08 usec per loop
(6) Can produce the wrong answer!!
>>> import re
>>> bool(re.compile('^[a-z0-9\.]+$').match('1234\n'))
True # uh-oh
>>> bool(re.compile('^[a-z0-9\.]+\Z').match('1234\n'))
False
Here's a simple, pure-Python implementation. It should be used when performance is not critical (included for future Googlers).
import string
allowed = set(string.ascii_lowercase + string.digits + '.')
def check(test_str):
set(test_str) <= allowed
Regarding performance, iteration will probably be the fastest method. Regexes have to iterate through a state machine, and the set equality solution has to build a temporary set. However, the difference is unlikely to matter much. If performance of this function is very important, write it as a C extension module with a switch statement (which will be compiled to a jump table).
Here's a C implementation, which uses if statements due to space constraints. If you absolutely need the tiny bit of extra speed, write out the switch-case. In my tests, it performs very well (2 seconds vs 9 seconds in benchmarks against the regex).
#define PY_SSIZE_T_CLEAN
#include <Python.h>
static PyObject *check(PyObject *self, PyObject *args)
{
const char *s;
Py_ssize_t count, ii;
char c;
if (0 == PyArg_ParseTuple (args, "s#", &s, &count)) {
return NULL;
}
for (ii = 0; ii < count; ii++) {
c = s[ii];
if ((c < '0' && c != '.') || c > 'z') {
Py_RETURN_FALSE;
}
if (c > '9' && c < 'a') {
Py_RETURN_FALSE;
}
}
Py_RETURN_TRUE;
}
PyDoc_STRVAR (DOC, "Fast stringcheck");
static PyMethodDef PROCEDURES[] = {
{"check", (PyCFunction) (check), METH_VARARGS, NULL},
{NULL, NULL}
};
PyMODINIT_FUNC
initstringcheck (void) {
Py_InitModule3 ("stringcheck", PROCEDURES, DOC);
}
Include it in your setup.py:
from distutils.core import setup, Extension
ext_modules = [
Extension ('stringcheck', ['stringcheck.c']),
],
Use as:
>>> from stringcheck import check
>>> check("abc")
True
>>> check("ABC")
False
EDIT: Changed the regular expression to exclude A-Z
Regular expression solution is the fastest pure python solution so far
reg=re.compile('^[a-z0-9\.]+$')
>>>reg.match('jsdlfjdsf12324..3432jsdflsdf')
True
>>> timeit.Timer("reg.match('jsdlfjdsf12324..3432jsdflsdf')", "import re; reg=re.compile('^[a-z0-9\.]+$')").timeit()
0.70509696006774902
Compared to other solutions:
>>> timeit.Timer("set('jsdlfjdsf12324..3432jsdflsdf') <= allowed", "import string; allowed = set(string.ascii_lowercase + string.digits + '.')").timeit()
3.2119350433349609
>>> timeit.Timer("all(c in allowed for c in 'jsdlfjdsf12324..3432jsdflsdf')", "import string; allowed = set(string.ascii_lowercase + string.digits + '.')").timeit()
6.7066690921783447
If you want to allow empty strings then change it to:
reg=re.compile('^[a-z0-9\.]*$')
>>>reg.match('')
False
Under request I'm going to return the other part of the answer. But please note that the following accept A-Z range.
You can use isalnum
test_str.replace('.', '').isalnum()
>>> 'test123.3'.replace('.', '').isalnum()
True
>>> 'test123-3'.replace('.', '').isalnum()
False
EDIT Using isalnum is much more efficient than the set solution
>>> timeit.Timer("'jsdlfjdsf12324..3432jsdflsdf'.replace('.', '').isalnum()").timeit()
0.63245487213134766
EDIT2 John gave an example where the above doesn't work. I changed the solution to overcome this special case by using encode
test_str.replace('.', '').encode('ascii', 'replace').isalnum()
And it is still almost 3 times faster than the set solution
timeit.Timer("u'ABC\u0131\u0661'.encode('ascii', 'replace').replace('.','').isalnum()", "import string; allowed = set(string.ascii_lowercase + string.digits + '.')").timeit()
1.5719811916351318
In my opinion using regular expressions is the best to solve this problem
Simpler approach? A little more Pythonic?
>>> ok = "0123456789abcdef"
>>> all(c in ok for c in "123456abc")
True
>>> all(c in ok for c in "hello world")
False
It certainly isn't the most efficient, but it's sure readable.