Python equivalent to C strtod

Question

I am working on converting parts of a C++ program to Python, but I have some trouble replacing the C function strtod. The strings I'm working on consists of simple mathmatical-ish equations, such as "KM/1000.0". The problem is that the both constants and numbers are mixed and I'm therefore unable to use float().

How can a Python function be written to simulate strtod which returns both the converted number and the position of the next character?

Answer 1

I'm not aware of any existing functions that would do that.

However, it's pretty easy to write one using regular expressions:

import re

# returns (float,endpos)
def strtod(s, pos):
  m = re.match(r'[+-]?\d*[.]?\d*(?:[eE][+-]?\d+)?', s[pos:])
  if m.group(0) == '': raise ValueError('bad float: %s' % s[pos:])
  return float(m.group(0)), pos + m.end()

print strtod('(a+2.0)/1e-1', 3)
print strtod('(a+2.0)/1e-1', 8)

A better overall approach might be to build a lexical scanner that would tokenize the expression first, and then work with a sequence of tokens rather than directly with the string (or indeed go the whole hog and build a yacc-style parser).

Answer 2

You can create a simple C strtod wrapper:

#include <stdlib.h>

double strtod_wrap(const char *nptr, char **endptr)
{
   return strtod(nptr, endptr);
}

compile with:

gcc -fPIC -shared -o libstrtod.dll strtod.c

(if you're using Python 64 bit, the compiler must be 64-bit as well)

and call it using ctypes from python (linux: change .dll to .so in the lib target and in the code below, this was tested on Windows):

import ctypes

_strtod = ctypes.CDLL('libstrtod.dll')
_strtod.strtod_wrap.argtypes = (ctypes.c_char_p, ctypes.POINTER(ctypes.c_char_p))
_strtod.strtod_wrap.restype = ctypes.c_double

def strtod(s):
    p = ctypes.c_char_p(0)
    s = ctypes.create_string_buffer(s.encode('utf-8'))
    result = _strtod.strtod_wrap(s, ctypes.byref(p))
    return result,ctypes.string_at(p)

print(strtod("12.5hello"))

prints:

(12.5, b'hello')

(It's not as hard as it seems, since I learned how to do that just 10 minutes ago)

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Answer 3

parse the number yourself.

a recursive-descent parser is very easy for this kind of input. first write a grammar:

float ::= ipart ('.' fpart)* ('e' exp)*
ipart ::= digit+
fpart ::= digit+
exp   ::= ('+'|'-') digit+
digit = ['0'|'1'|'2'|'3'|'4'|'5'|'6'|'7'|'8'|'9']

now converting this grammar to a function should be straightforward...

Answer 4

I'd use a regular expression for this:

import re
mystring = "1.3 times 456.789 equals 593.8257 (or 5.93E2)"
def findfloats(s):
    regex = re.compile(r"[+-]?\b\d+(?:\.\d+)?(?:e[+-]?\d+)?\b", re.I)
    for match in regex.finditer(mystring):
        yield (match.group(), match.start(), match.end())

This finds all floating point numbers in the string and returns them together with their positions.

>>> for item in findfloats(mystring):
...     print(item)
...
('1.3', 0, 3)
('456.789', 10, 17)
('593.8257', 25, 33)
('5.93E2', 38, 44)