bob gailer wrote:
Or even simplre
f = open('file.txt',r).readlines()
print [f[x+1] for x, line in enumerate(f) if line.rstrip() == "3"][0]
--
Bob Gailer
919-636-4239 Chapel Hill, NC
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On Fri, May 2, 2008 at 11:42 AM, Paul McGuire <[EMAIL PROTECTED]> wrote:
> f=open('file.txt',r)
> print_line = False
> for line in f:
>if print_line:
> print line
> print_line = False
>if line == "3":
Don't forget about the newline...(that makes four!)
Kent
_
Augghh! I can't stand it!!! If position is a boolean, then *why* must we
test if it is equal to True?!!! It's a boolean! Just test it! For that
matter, let's rename "position" to something a little more direct,
"print_line" perhaps?
Did you know that files are now iterators? If going through
On Fri, May 2, 2008 at 10:34 AM, Brain Stormer <[EMAIL PROTECTED]> wrote:
> f=open('file.txt',r)
> position =False
>
> for line in f.read().split():
Note that split() splits on any whitespace, not just line endings. In
your case it doesn't much matter I guess.
Kent
_
f=open('file.txt',r)
position =False
for line in f.read().split():
if position == True
print line
position = False
if line == "3":
position = True
else:
position = False
f.close()
On Fri, May 2, 2008 at 10:28 AM, Brain Stormer <[EMAIL PROTECTED]> wrot
You are correct. It is missing the ":" and it will print "3"
On Fri, May 2, 2008 at 10:18 AM, bob gailer <[EMAIL PROTECTED]> wrote:
> Brain Stormer wrote:
>
> > Well,
> > I was somewhat confused with all of the answers so I decided to go with
> > my/following method. Kent's method has 4 fewer
Brain Stormer wrote:
Well,
I was somewhat confused with all of the answers so I decided to go
with my/following method. Kent's method has 4 fewer lines of code
than mine and cleaner. Please correct me if I am fundamentally wrong.
f=open('file.txt',r)
for line in f.read().split():
if l
Well,
I was somewhat confused with all of the answers so I decided to go with
my/following method. Kent's method has 4 fewer lines of code than mine and
cleaner. Please correct me if I am fundamentally wrong.
f=open('file.txt',r)
for line in f.read().split():
if line == "3"
positi
On Fri, May 2, 2008 at 3:55 AM, Roel Schroeven
<[EMAIL PROTECTED]> wrote:
> Shouldn't it even be 'line = f.next()'?
Wow, I think Brain Stormer should get a prize. I'm not sure what the
prize is, but his short program has elicited incomplete and inaccurate
answers from three of the top posters to
Alan Gauld schreef:
if line == "3":
line.next
this then becomes f.next() # next is a method not
an attribute so needs the () to call it
Shouldn't it even be 'line = f.next()'?
--
The saddest aspect of life right now is that science gathers knowledge
faster than society gathe
"bob gailer" <[EMAIL PROTECTED]> wrote
if line == "3":
Assuming you adopt my approach, then each line will be a digit
followed by \n (newline). So no line will == "3". Instead use:
if line[:-1] == "3":
or
if line.rstrip() == "3"
or even
if line.startswith("3"):
personall
"Brain Stormer" <[EMAIL PROTECTED]> wrote
f = open('file.txt',r)
for line in f.read():
This reads the whole file inta a string then iterates
over all the characters(not lines) in that string.
Better to iterate over the file:
for line in f:
if line == "3":
line.next
thi
Brain Stormer wrote:
I have the following code:
f = open('file.txt',r)
for line in f.read():
That will read the entire file into a string, then access one character
at a time from the string.
If you want to work with lines:
for line in f:
if line == "3":
Assuming you adopt my approac
On Thu, May 1, 2008 at 5:04 PM, Brain Stormer <[EMAIL PROTECTED]> wrote:
> I have the following code:
>
> f = open('file.txt',r)
> for line in f.read():
> if line == "3":
>line.next
Try
f.next()
line is a string, it doesn't have a next() method. The file itself is
iterable and
I have the following code:
f = open('file.txt',r)
for line in f.read():
if line == "3":
line.next
print line
f.close()
The file.txt looks like
1
2
3
4
5
I would like the code to output "4"
but I don't know how to use the next. Are there any other way?
Thanks
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