How can I say a file is SVG without using a magic number?

Question

An SVG file is basically an XML file so I could use the string <?xml (or the hex representation: '3c 3f 78 6d 6c') as a magic number but there are a few opposing reason not to do that if for example there are extra white-spaces it could break this check.

The other images I need/expect to check are all binaries and have magic numbers. How can I fast check if the file is an SVG format without using the extension eventually using Python?

Answer 1

XML is not required to start with the <?xml preamble, so testing for that prefix is not a good detection technique — not to mention that it would identify every XML as SVG. A decent detection, and really easy to implement, is to use a real XML parser to test that the file is well-formed XML that contains the svg top-level element:

import xml.etree.cElementTree as et

def is_svg(filename):
    tag = None
    with open(filename, "r") as f:
        try:
            for event, el in et.iterparse(f, ('start',)):
                tag = el.tag
                break
        except et.ParseError:
            pass
    return tag == '{http://www.w3.org/2000/svg}svg'

Using cElementTree ensures that the detection is efficient through the use of expat; timeit shows that an SVG file was detected as such in ~200μs, and a non-SVG in 35μs. The iterparse API enables the parser to forego creating the whole element tree (module name notwithstanding) and only read the initial portion of the document, regardless of total file size.

Answer 2

You could try reading the beginning of the file as binary - if you can't find any magic numbers, you read it as a text file and match to any textual patterns you wish. Or vice-versa.