[EMAIL PROTECTED] wrote:
> for n,l in enumerate(open("file")):
> print n,l # this prints current line
> print next line in this current iteration of the loop.
Depends what you want to happen when you request "next". If you want to
renumber the lines, you can call .next() on the iterator::
>>> open('temp.txt', 'w').write('1\n2\n3\n4\n5\n6\n7\n')
>>> lines_iter = open('temp.txt')
>>> for i, line in enumerate(lines_iter):
... print 'LINE %i, %r %r' % (i, line, lines_iter.next())
...
LINE 0, '1\n' '2\n'
LINE 1, '3\n' '4\n'
LINE 2, '5\n' '6\n'
Traceback (most recent call last):
File "<interactive input>", line 2, in <module>
StopIteration
If you want to peek ahead without removing the line from the iterator,
check out this recipe::
http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/304373
Which allows code like::
>>> lines_iter = peekable(open('temp.txt'))
>>> for i, line in enumerate(lines_iter):
... print 'LINE %i, %r %r' % (i, line, lines_iter.peek())
...
LINE 0, '1\n' '2\n'
LINE 1, '2\n' '3\n'
LINE 2, '3\n' '4\n'
LINE 3, '4\n' '5\n'
LINE 4, '5\n' '6\n'
LINE 5, '6\n' '7\n'
Traceback (most recent call last):
...
StopIteration
(Note that the recipe doesn't try to catch the StopIteration, but if you
want that suppressed, it should be a pretty simple change.)
STeVe
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