Why do I require an "elif" statement here?

[Jim]

Could somebody tell me why I need the "elif char == '\n'" in the
following code?
This is required in order the pick up lines with just spaces in them.
Why doesn't
the "else:" statement pick this up?

OLD_INDENT = 5 # spaces
NEW_INDENT = 4 # spaces

print 'Reindent.py:'
print '\nFrom file %s' % infile
print 'Change %i space indentation to %i space indentation.' % (
OLD_INDENT, NEW_INDENT)
print 'And place revised file into %s' % outfile

whitespace = ' '
n = 0
nline = 0

for line in input.readlines ():
nline += 1
# Only look at lines that start with a space.
if line[0] == whitespace:
i = 0
for char in line:
i += 1
if char == whitespace:
pass
elif char == '\n': # Why do I need this for a
blank line with only spaces?
output.write(li ne)
break
else: # Why doesn't the blank line
get picked up here?
x = line.count(whit espace*OLD_INDE NT,0,i)
# Reindent lines that have exactly a multiple of
OLD_INDENT.
if x 0 and (i-1)%OLD_INDENT == 0:
output.write(wh itespace*NEW_IN DENT*x+line.lst rip())
n += 1
break
else:
output.write(li ne)
break
else:
output.write(li ne)

input.close()
output.close()
print 'Total number of %i lines reindented out of %i lines.' % (n,
nline)

[danielx]

No offense. I didn't mean there was anything wrong with your way, just
that it wasn't "neat". By that, I meant, you were still using lots of
for loops and if blocks.

Justin Azoff wrote:

danielx wrote:

I'm surprised no one has mentioned neat-er, more pythonic ways of doing
this. I'm also surprised no one mentioned regular expressions. Regular
expressions are really powerful for searching and manipulating text.

I suppose I put those things too close together. I meant I was
surprised for each of the two Separate (non)occurances .

[snip]

I'm surprised you don't count my post as a neat and pythonic way of
doing this. I'm also surprised that you mention regular expressions
after neat and pythonic. While regular expressions often serve a
purpose, they are rarely neat.

Anyway, here's my solution, which does Not use regular expressions:

def reindent(line):
## we use slicing, because we don't know how long line is
head = line[:OLD_INDENT]
tail = line[OLD_INDENT:]
## if line starts with Exactly so many spaces...
if head == whitespace*OLD_ INDENT and not tail.startswith (' '):
return whitespace*NEW_ INDENT + tail
else: return line # our default

[snip]

This function is broken. Not only does it still rely on global
variables to work, it does not actually reindent lines correctly. Your

It's just an illustration.

function only changes lines that start with exactly OLD_INDENT spaces,
ignoring any lines that start with a multiple of OLD_INDENT.

Maybe I misread, but that's what I thought he wanted. Did you not see
my code comments expressing that very intent? Anyway, it wouldn't be
that hard to modify what I gave to do multiple replacement: just add a
recursive call.

>
--
- Justin

[John Machin]

danielx wrote:

No offense. I didn't mean there was anything wrong with your way, just
that it wasn't "neat". By that, I meant, you were still using lots of
for loops and if blocks.

Justin Azoff wrote:

danielx wrote:

I'm surprised no one has mentioned neat-er, more pythonic ways of doing
this. I'm also surprised no one mentioned regular expressions. Regular
expressions are really powerful for searching and manipulating text.

I suppose I put those things too close together. I meant I was
surprised for each of the two Separate (non)occurances .

And what would regexes give you? A Pythonic way of calculating the
number of leading spaces??? N.B. in your (mis)reading of the OP's
intent, you don't need/use the number of leading spaces.

Anyway, here's my solution, which does Not use regular expressions:
>
def reindent(line):
## we use slicing, because we don't know how long line is
head = line[:OLD_INDENT]
tail = line[OLD_INDENT:]
## if line starts with Exactly so many spaces...
if head == whitespace*OLD_ INDENT and not tail.startswith (' '):
return whitespace*NEW_ INDENT + tail
else: return line # our default

[snip]

This function is broken. Not only does it still rely on global
variables to work, it does not actually reindent lines correctly. Your

It's just an illustration.

An illustration of ... what?

>

function only changes lines that start with exactly OLD_INDENT spaces,
ignoring any lines that start with a multiple of OLD_INDENT.

Maybe I misread, but that's what I thought he wanted.

(a) Such a requirement is rather implausible.
(b) No "maybe":

x = line.count(whit espace*OLD_INDE NT,0,i)
# Reindent lines that have exactly a multiple of
OLD_INDENT.
if x 0 and (i-1)%OLD_INDENT == 0:
output.write(wh itespace*NEW_IN DENT*x+line.lst rip())

Did you not see
my code comments expressing that very intent?

Those comments merely documented the brokenness.

Anyway, it wouldn't be
that hard to modify what I gave to do multiple replacement: just add a
recursive call.

.... which would make it uglier.

[John Savage]

"sp*****@gmail. com" <sp*****@gmail. comwrites:

>-------------------------------
whitespace = " "
old_indent = 3
new_indent = 5

x = " starts with 3 spaces"

x = x.replace(white space*old_inden t, whitespace*new_ indent)
-------------------------------

In this example though, it will replace the 3 spaces no matter where
they are at, not just in the beginning... still, it's probably more
practical for most use cases.

You'd corner it with:

if x.startswith(' '*3): x=x.replace(' '*3,' '*5,1)
--
John Savage (my news address is not valid for email)

[sp1d3rx]

John Savage wrote:

"sp*****@gmail. com" <sp*****@gmail. comwrites:

-------------------------------
whitespace = " "
old_indent = 3
new_indent = 5

x = " starts with 3 spaces"

x = x.replace(white space*old_inden t, whitespace*new_ indent)
-------------------------------

In this example though, it will replace the 3 spaces no matter where
they are at, not just in the beginning... still, it's probably more
practical for most use cases.

You'd corner it with:

if x.startswith(' '*3): x=x.replace(' '*3,' '*5,1)

As others have stated, this will only get lines that start with only 1
set of "old_indent " and not multiples of "old_indent ".

--
John Savage (my news address is not valid for email)