Suppose you open a file, and do an seek() somewhere in the file, how do you know the current file line ?
(I personally solved with an ad-hoc file class that maps the seek position to the line after scanning the file, but I wanted to s开发者_开发问答ee other hints and to add this question to stackoverflow, as I was not able to find the problem anywhere on google)
When you use seek(), python gets to use pointer offsets to jump to the desired position in the file. But in order to know the current line number, you have to examine each character up to that position. So you might as well abandon seek() in favor of read():
Replace
f = open(filename, "r")
f.seek(55)
with
f = open(filename, "r")
line=f.read(55).count('\n')+1
print(line)
Perhaps you do not wish to use f.read(num) since this may require a lot of memory if num is very large. In that case, you could use a generator like this:
import itertools
import operator
line_number=reduce(operator.add,( f.read(1)=='\n' for _ in itertools.repeat(None,num)))
pos=f.tell()
This is equivalent to f.seek(num)
with the added benefit of giving you line_number
.
Here's how I would approach the problem, using as much laziness as possible:
from random import randint
from itertools import takewhile, islice
file = "/etc/passwd"
f = open(file, "r")
f.seek(randint(10,250))
pos = f.tell()
print "pos=%d" % pos
def countbytes(iterable):
bytes = 0
for item in iterable:
bytes += len(item)
yield bytes
print 1+len(list(takewhile(lambda x: x <= pos, countbytes(open(file, "r")))))
For a slightly less readable but much more lazy approach, use enumerate
and dropwhile
:
from random import randint
from itertools import islice, dropwhile
file = "/etc/passwd"
f = open(file, "r")
f.seek(randint(10,250))
pos = f.tell()
print "pos=%d" % pos
def countbytes(iterable):
bytes = 0
for item in iterable:
bytes += len(item)
yield bytes
print list(
islice(
dropwhile(lambda x: x[1] <= pos, enumerate(countbytes(open(file, "r"))))
, 1))[0][0]+1
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