4.10 Reading Lines with Continuation Characters


Credit: Alex Martelli

4.10.1 Problem

You have a file that includes long logical lines split over two or more physical lines, with backslashes to indicate that a continuation line follows. You want to process a sequence of logical lines, rejoining those split lines.

4.10.2 Solution

As usual, a class is the right way to wrap this functionality in Python 2.1:

class LogicalLines:     def _ _init_ _(self, fileobj):         # Ensure that we get a line-reading sequence in the best way possible:         import xreadlines         try:             # Check if the file-like object has an xreadlines method             self.seq = fileobj.xreadlines(  )         except AttributeError:             # No, so fall back to the xreadlines module's implementation             self.seq = xreadlines.xreadlines(fileobj)         self.phys_num = 0  # current index into self.seq (physical line number)         self.logi_num = 0  # current index into self (logical line number)     def _ _getitem_ _(self, index):         if index != self.logi_num:             raise TypeError, "Only sequential access supported"         self.logi_num += 1         result = []         while 1:             # Intercept IndexError, since we may have a last line to return             try:                  # Let's see if there's at least one more line in self.seq                 line = self.seq[self.phys_num]             except IndexError:                 # self.seq is finished, so break the loop if we have any                 # more data to return; else, reraise the exception, because                 # if we have no further data to return, we're finished too                 if result: break                 else: raise             self.phys_num += 1             if line.endswith('\\\n'):                 result.append(line[:-2])             else:                 result.append(line)                 break         return ''.join(result) # Here's an example function, showing off usage: def show_logicals(fileob, numlines=5):     ll = LogicalLines(fileob)     for l in ll:         print "Log#%d, phys# %d: %s" % (             ll.logi_num, ll.phys_num, repr(l))         if ll.logi_num>numlines: break if _ _name_ _=='_ _main_ _':     from cStringIO import StringIO     ff = StringIO( r"""prima \ seconda \ terza quarta \ quinta sesta settima \ ottava """)     show_logicals( ff )

4.10.3 Discussion

This is another sequence-bunching problem, like Recipe 4.9. In Python 2.1, a class wrapper is the most natural approach to getting reusable code for sequence-bunching tasks. We need to support the sequence protocol ourselves and handle the sequence protocol in the sequence we wrap. In Python 2.1 and earlier, the sequence protocol is as follows: a sequence must be indexable by successively larger integers (0, 1, 2, ...), and it must raise an IndexError as soon as an integer that is too large is used as its index. So, if we need to work with Python 2.1 and earlier, we must behave this way ourselves and be prepared for just such behavior from the sequence we are wrapping.

In Python 2.2, thanks to iterators, the sequence protocol is much simpler. A call to the next method of an iterator yields its next item, and the iterator raises a StopIteration when it's done. Combined with a simple generator function that returns an iterator, this makes sequence bunching and similar tasks far easier:

from _ _future_ _ import generators def logical_lines(fileobj):     logical_line = []     for physical_line in fileobj:         if physical_line.ends_with('\\\n'):             logical_line.append(physical_line[:-2])         else:             yield ''.join(logical_line)+physical_line             logical_line = []     if logical_line: yield ''.join(logical_line)

4.10.4 See Also

Recipe 4.9; Perl Cookbook Recipe 8.1.



Python Cookbook
Python Cookbook
ISBN: 0596007973
EAN: 2147483647
Year: 2005
Pages: 346

flylib.com © 2008-2017.
If you may any questions please contact us: flylib@qtcs.net