I would like to convert either 0x123, 012 or 36#foo into an integer.
To do so, I wrote this:
def str2int(s):
if re.findall('(?i)^0x',s):
return int(s, 16)
if re.findall('^(?i)0',s):
return int(s, 8)
m = re.findall('(?i)^(\d+)\#([0-9a-z]+)',s)
if m:
return int(m[0][1], m[0][0])
raise AssertionError, 'Unknown value'
I feel it is a bit complicated. Is there any builtin method?
Yes, you can use ast.literal_eval.
>>> import ast
>>> ast.literal_eval("0x123")
291
>>> ast.literal_eval("012")
10
>>> ast.literal_eval("36#foo")
36
However, note that literal_eval("012") will only work in 2.7 and lower because 3.x no longer supports that style of octal literal. But this will work:
>>> ast.literal_eval("0o12")
10
echo $(( 36#foo )) gives 20328 not 36
'echo $(( 36#foo ))' give you 20328? Also how would you expect that to be parsed? It’s not even following any pattern (and it’s so very different than your examples in the question).
int('foo', 36) -> 20328.Solution without regular expressions:
def convert (s):
if s.lower().startswith('0x'):
s = '16#' + s[2:]
elif s.startswith('0'):
s = '8#' + s[1:]
elif '#' not in s:
s = '10#' + s
base, num = s.split('#', 1)
return int(num, int(base))
>>> testcases = [('0x123', 291), ('012', 10), ('36#foo', 20328)]
>>> for s, n in testcases:
print(s, n, n == convert(s))
0x123 291 True
012 10 True
36#foo 20328 True
int will do, when you pass 0 as second argument:
int('0x123', 0)
=> 291
int('0o12', 0)
=> 10
If you want to support comments, str.partition is the simplest way I can think about:
int('36#foo'.partition('#')[0], 0)
=> 36
base#numberis not a built-in supported expression format.