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    <p>If the "lines" your file is divided into are of reasonable lengths, and there are no binary sequences in it that "reading as text" would break, you can use <code>fileinput</code>'s handy "make believe I'm rewriting a file in place" functionality:</p> <pre><code> import re import fileinput tagre = re.compile(r"&lt;o:p&gt;.*?&lt;/o:p&gt;") def sub(mo): return mo.group().replace(r"'", r"\'") for line in fileinput.input('thefilename', inplace=True): print tagre.sub(sub, line), </code></pre> <p>If not, you'll have to simulate the "in-place rewriting" yourself, e.g. (oversimplified...):</p> <pre><code> with open('thefilename', 'rb') as inf: with open('fixed', 'wb') as ouf: while True: b = inf.read(1024*1024) if not b: break ouf.write(tagre.sub(sub, b)) </code></pre> <p>and then move <code>'fixed'</code> to take place of <code>'thefilename'</code> (either in code, or manually) if you need that filename to remain after the fixing.</p> <p>This is oversimplified because one of the crucial <code>&lt;o:p&gt; ... &lt;/o:p&gt;</code> parts might end up getting split between two successive megabyte "blocks" and therefore not identified (in the first example, I'm assuming each such part is always fully contained within a "line" -- if that's not the case then you should not use that code, but the following, anyway). Fixing this requires, alas, more complicated code...:</p> <pre><code> with open('thefilename', 'rb') as inf: with open('fixed', 'wb') as ouf: while True: b = getblock(inf) if not b: break ouf.write(tagre.sub(sub, b)) </code></pre> <p>with e.g.</p> <pre><code> partsofastartag = '&lt;', '&lt;o', '&lt;o:', '&lt;o:p' def getblock(inf): b = '' while True: newb = inf.read(1024 * 1024) if not newb: return b b += newb if any(b.endswith(p) for p in partsofastartag): continue if b.count('&lt;o:p&gt;') != b.count('&lt;/o:p&gt;'): continue return b </code></pre> <p>As you see, this is pretty delicate code, and therefore, what with it being untested, I can't <em>know</em> that it is correct for your problem. In particular, can there be cases of <code>&lt;o:p&gt;</code> that are NOT matched by a closing <code>&lt;/o:p&gt;</code> or vice versa? If so, then a call to <code>getblock</code> could end up returning the whole file in quite a costly way, and even the RE matching and substitution might backfire (the latter would also occur if SOME of the single-quotes in such tags are already properly escaped, but not all).</p> <p>If you have at least a GB or so, avoiding the delicate issues with block division, at least, IS feasible, since everything should fit in memory, making the code much simpler:</p> <pre><code> with open('thefilename', 'rb') as inf: with open('fixed', 'wb') as ouf: b = inf.read() ouf.write(tagre.sub(sub, b)) </code></pre> <p>However, the other issues mentioned above (possible unbalanced opening/closing tags, etc) might remain -- only you can study your existing defective data and see if it affords such a reasonably simple approach at fixing!</p>
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    1. COThanks, that worked great, all I need to do now is find all the other queries with unescaped quotes that don't have such a handy way of finding them...
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