Hi!
Is there any easiest way to get the substring starting from the 3rd
character?
my solution:
str[3…-1] which seems quite odd :-/
What if, intuitive, we can write
str[3…]?
Any smarter solution?
Gergo
–
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Hi!
Is there any easiest way to get the substring starting from the 3rd
character?
my solution:
str[3…-1] which seems quite odd :-/
Seems quite even to me 
Seriously, the ability to negative-index strings and arrays like that
is a breath of fresh air compared to traditional compiled languages.
I see no need to improve on it.
What if, intuitive, we can write
str[3…]?
I rather doubt that’s going to happen. When you execute str[3…7],
the 3…7 isn’y arbitrary syntax, it’s a Range object. If Range
objects were allowed to be unbounded, then fine. So the question
really becomes, should Range objects allow unbounded ranges?
Any smarter solution?
Write a String#substring method if you like.
Cheers,
Gavin
On Saturday, November 15, 2003, 2:48:53 AM, KONTRA wrote:
class String
def substring
/(^...)(.*)\z/ =~ self
return $2
end
end
“123456”.substring
= >“456”
Now…
I am just wondering how can I build dinamically the Regexp
I mean …
“123456”.substring(5)
= > “6”
“123456”.substring(4)
= > “56”
ideas ?
-r.
KONTRA Gergely wrote:
Hi!
Is there any easiest way to get the substring starting from the 3rd
character?my solution:
str[3…-1] which seems quite odd :-/What if, intuitive, we can write
str[3…]?Any smarter solution?
Gergo
–
General Electric - CIAT
Advanced Engineering Center
Rodrigo Bermejo
Information Technologies.
Special Applications
Dial-comm : *879-0644
Phone :(+52) 442-196-0644
Hi!
Is there any easiest way to get the substring starting from the 3rd
character?my solution:
str[3…-1] which seems quite odd :-/What if, intuitive, we can write
str[3…]?Any smarter solution?
Besids ‘get used to -1’? Well, here’s some extension to ‘String’. I
did try my best to follow POLS. The intention is that left, right,
mid emulate BASIC’s left$, right$ and mid$ while head and tail
without any arguments have the results they should have from a
functional programming point of view.
class String
def left(number = 1)
self[0…number-1]
end
def right(number = 1)
self[-number…-1]
end
def mid(from, number=1)
self[from…from+number]
end
def head(position = 0)
self[0…position]
end
def tail(position = 0)
self[position+1…-1]
end
end
Josef ‘Jupp’ Schugt
–
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“KONTRA Gergely” kgergely@mlabdial.hit.bme.hu schrieb im Newsbeitrag
news:20031114154849.GA12384@mlabdial.hit.bme.hu…
Hi!
Is there any easiest way to get the substring starting from the 3rd
character?my solution:
str[3…-1] which seems quite odd :-/What if, intuitive, we can write
str[3…]?Any smarter solution?
Not as far as I know. You can do str[3,str.length-3] but this isn’t
really an improvement unless you value the absence of negative integers in
this expression.
robert
“KONTRA Gergely” kgergely@mlabdial.hit.bme.hu wrote in message news:20031114154849.GA12384@mlabdial.hit.bme.hu…
Hi!
Is there any easiest way to get the substring starting from the 3rd
character?my solution:
str[3…-1] which seems quite odd :-/What if, intuitive, we can write
str[3…]?Any smarter solution?
Actually I find this behavior annoying:
irb(main):015:0> “hello”[3…-1]
=> “lo”
irb(main):016:0> “hello”[4…-1]
=> “o”
irb(main):017:0> “hello”[5…-1]
=> “”
irb(main):018:0> “hello”[6…-1]
=> nil
–
J. Lambert
my solution:
str[3…-1] which seems quite odd :-/
The idiom in python is str[3:]
Conversely, str[:3] gives you the 3 1st chars.
Read the explanation of slicing operations here:
http://www.python.org/doc/current/lib/typesseq.html
(read note 4 & 5)
And beware, ‘str’ is a builtin, hence a poor choice as a variable name.
Have fun,
Bernard.
I typically use str[3…str.length]… It’s probably slower, but it looks a
bit cleaner in code…
Regards,
-JD-
–On Saturday, November 15, 2003 12:48 AM +0900 KONTRA Gergely kgergely@mlabdial.hit.bme.hu wrote:
Hi!
Is there any easiest way to get the substring starting from the 3rd
character?my solution:
str[3…-1] which seems quite odd :-/
Some ways:
class String
def substring1(n)
reg = Regexp.compile("(^" + ".“n + ")(.)\z”)
reg.match(self)[2]
end
def substring2(s, e=-1)
self[s..e]
end
end
p “12345”.substring1 1
p “12345”.substring1 2
p “12345”.substring2 1
p “12345”.substring2 2
p “12345”.substring2 1, 3
class String
def substring
/(^…)(.*)\z/ =~ self
return $2
end
end
“123456”.substring
= >“456”
Now…
I am just wondering how can I build dinamically the Regexp
I mean …
“123456”.substring(5)
= > “6”
“123456”.substring(4)
= >> “56”
ideas ?
Regexen are overkill here.
class String
def substring(m, n=-1)
self[m…n]
end
end
Gavin
On Saturday, November 15, 2003, 12:24:42 PM, Rodrigo wrote:
As Gavin pointed out, regexen are overkill here. However, the
technique is worth noting:
class String
def substring(start, len = nil)
if len.nil?
/^.{#{start}}(.*)/m =~ self
else
/^.{#{start}}(.{#{len}})/m =~ self
end
$1
end
end
“123456”.substring(3) # => “456”
“123456”.substring(3, 2) # => “45”
-austin
On Sat, 15 Nov 2003 10:24:42 +0900, Bermejo, Rodrigo wrote:
class String
def substring
/(^…)(.*)\z/ =~ self
return $2
end
end
–
austin ziegler * austin@halostatue.ca * Toronto, ON, Canada
software designer * pragmatic programmer * 2003.11.15
* 09.51.26
Maybe some destructive versions of the above methods could be useful?
input=“diamonds are forever”
p input.right!(4) #-> “ever”
p input.left!(4) #-> “diam”
p input # → “onds are for”
On Sun, 16 Nov 2003 06:04:51 +0900, Josef ‘Jupp’ SCHUGT wrote:
class String
def left(number = 1)
self[0…number-1]
end
def right(number = 1)
self[-number…-1]
end
def mid(from, number=1)
self[from…from+number]
end
def head(position = 0)
self[0…position]
end
def tail(position = 0)
self[position+1…-1]
end
end
–
Simon Strandgaard
Gavin Sinclair wrote:
On Saturday, November 15, 2003, 2:48:53 AM, KONTRA wrote:
Seriously, the ability to negative-index strings and arrays like that
is a breath of fresh air compared to traditional compiled languages.
I see no need to improve on it.
I do, but I have no good solution to the problem.
The problem being that positive indexes go from 0 to n-1, while negative
go from -1 to -n, which could introduce confusion.
But as I said, I know of no good solutions, since -0 is not a good way,
and making the first positive index a 1 is just as bad, for several reasons.
[snip class String extension]
Useful… I have added your snippet at rubygarden (StringSub):
On Sun, 16 Nov 2003 06:04:51 +0900, Josef ‘Jupp’ SCHUGT wrote:
Besids ‘get used to -1’? Well, here’s some extension to ‘String’. I
did try my best to follow POLS. The intention is that left, right,
mid emulate BASIC’s left$, right$ and mid$ while head and tail
without any arguments have the results they should have from a
functional programming point of view.
–
Simon Strandgaard
Oops - wrong NG, my bad 
This has to do with how Ruby views indices on arrays and array-like
structures (like strings).
[h] [e] [l] [l] [o]
0 1 2 3 4 5
The indices are “before” the actual element, so (5 … -1) represents
“end of the string to end of the string”, whereas (6 … -1)
reperesents (outside fo the string to the end of the string).
Some of this has to do with performance, I think, but it’s something
that makes sense, once you’re used to it. I think that it’s more
sensible than the typical implementations.
-austin
On Mon, 17 Nov 2003 11:40:13 +0900, Jon A. Lambert wrote:
Actually I find this behavior annoying:
irb(main):015:0> “hello”[3…-1]
=> “lo”
irb(main):016:0> “hello”[4…-1]
=> “o”
irb(main):017:0> “hello”[5…-1]
=> “”
irb(main):018:0> “hello”[6…-1]
=> nil
–
austin ziegler * austin@halostatue.ca * Toronto, ON, Canada
software designer * pragmatic programmer * 2003.11.16
* 23.27.28
Yes, it should What about 3…INF here?
On 1115, Gavin Sinclair wrote:
I rather doubt that’s going to happen. When you execute str[3…7],
the 3…7 isn’y arbitrary syntax, it’s a Range object. If Range
objects were allowed to be unbounded, then fine. So the question
really becomes, should Range objects allow unbounded ranges?
–
±[ Kontra, Gergelykgergely@mcl.hu PhD student Room IB113 ]---------+
http://www.mcl.hu/~kgergely “Olyan langesz vagyok, hogy |
Mobil:(+36 20) 356 9656 ICQ: 175564914 poroltoval kellene jarnom” |
±- Magyar php mirror es magyar php dokumentacio: http://hu.php.net --+
Hi,
As Gavin pointed out, regexen are overkill here. However, the
technique is worth noting:
Another syntax sugar,
class String
def substring(start, len = nil)
if len.nil?
self[/\A.{#{start}}(.*)/m, 1]
else
self[/\A.{#{start}}(.{#{len}})/m, 1]
At Sat, 15 Nov 2003 23:58:55 +0900, Austin Ziegler wrote:
end $1end
end
–
Nobu Nakada
“Linus Sellberg” linse428@student.liu.se schrieb im Newsbeitrag
news:bp53n1$4o7$1@news.island.liu.se…
Gavin Sinclair wrote:
Seriously, the ability to negative-index strings and arrays like that
is a breath of fresh air compared to traditional compiled languages.
I see no need to improve on it.I do, but I have no good solution to the problem.
The problem being that positive indexes go from 0 to n-1, while negative
go from -1 to -n, which could introduce confusion.But as I said, I know of no good solutions, since -0 is not a good way,
This reminds me of an old mathematician’s joke: there was this professor
for math that found an epsilon so small, that half of it was already
negative… 
and making the first positive index a 1 is just as bad, for several
reasons.
Yup
robert
On Saturday, November 15, 2003, 2:48:53 AM, KONTRA wrote: