Is there a better way to do this?

Ruby
1

As part of my learning Ruby I am trying to learn how to format strings.
The following is an example for formatting a U.S. phone number:

a = “1234567890”
b = “(000) 000-0000”
ai = 0

for i in 0…(b.length) -1
if b[i,1] == “0”
b[i,1] = a[ai,1]
ai += 1
end
end
puts b

Is there a better (more Rubyish) way to do this?
Obviously I will want to add more code in the future to handle
exceptions like phone numbers that are not 10 characters long or are
longer, but this will get me started.

2

On 5/4/07, Michael W. Ryder [email protected] wrote:

 ai += 1

end
end
puts b

Is there a better (more Rubyish) way to do this?
Obviously I will want to add more code in the future to handle
exceptions like phone numbers that are not 10 characters long or are
longer, but this will get me started.

I’m sure there are better ways but here is something to look at.

a = “1234567890”
area = a.slice(0…2)
exc = a.slice(3…5)
num = a.slice(6…9)
tel = “(#{area}) #{exc}-#{num}”

p tel

Harry

3

Harry K. wrote:

 b[i,1] = a[ai,1]

Harry

I am trying to come up with a “generic” formatting routine so that I
could feed it something like sform(“123456789”, “000-00-0000”) or
sform(“1234567890”, “(000) 000-0000”) or sform(“123456”, “00/00/00”) and
it would work. I could easily do something like you suggest, and for
some cases it might be better, but I want something I can for any number
of formats. Thanks for the input.

4

I am trying to come up with a “generic” formatting routine so that I
could feed it something like sform(“123456789”, “000-00-0000”) or
sform(“1234567890”, “(000) 000-0000”) or sform(“123456”, “00/00/00”) and
it would work. I could easily do something like you suggest, and for
some cases it might be better, but I want something I can for any number
of formats. Thanks for the input.

welcome.

require ‘enumerator’

def sform(num, fmt)

convert num to array (of one digit strings):

num = num.to_enum(:each_byte).map { |code| code.chr }

for each zero, replace it with a digit popped off the

front of the array of numbers (or characters):

fmt.gsub(/0/) { num.shift }
end

dan

5

Dan Z. wrote:

end

dan

Your method is much better than my C style one. With a little work to
handle exceptions it should generally work. The only problem I have
found so far is that it doesn’t handle periods in the number string
properly – i.e. sform(“12345.67”, “$00,000.00”) returns $12,345…6"
instead of $12,345.67". Something for me to work on. Thanks for the
code.

6

I am trying to come up with a “generic” formatting routine so that I
could feed it something like sform(“123456789”, “000-00-0000”) or
sform(“1234567890”, “(000) 000-0000”) or sform(“123456”, “00/00/00”) and
it would work. I could easily do something like you suggest, and for
some cases it might be better, but I want something I can for any number
of formats. Thanks for the input.

Well, that’s different :slight_smile:
So, you want to grab the first 10 digits and ignore beyond that?
Look into regular expressions.
Harry

7

On 5/4/07, Harry K. [email protected] wrote:

So, you want to grab the first 10 digits and ignore beyond that?
Look into regular expressions.
Harry

–
Kakueki.com is for sale | HugeDomains
A Look into Japanese Ruby List in English

This will grab digits and then you can do what you want with them.

arr = [“1234567890”, “(123) 867-5309”,“12/34/56/78/90444”]

arr.each do |x|
num = x.scan(/\d/).join
p num
puts
end

Harry

8

On May 3, 8:38 pm, Dan Z. [email protected] wrote:

def sform(num, fmt)

convert num to array (of one digit strings):

num = num.to_enum(:each_byte).map { |code| code.chr }

for each zero, replace it with a digit popped off the

front of the array of numbers (or characters):

fmt.gsub(/0/) { num.shift }
end

dan

Without ‘enumerator’:

def sform( str, fmt )
ary = str.split(//)
fmt.gsub( /0/ ){ ary.shift }
end

9

def sform( str, fmt )
ary = str.split(//)
fmt.gsub( /0/ ){ ary.shift }
end

Yeah, that’s much simpler. I like it.

Dan

10

On May 3, 9:03 pm, “Michael W. Ryder” [email protected]
wrote:

require ‘enumerator’
dan

Your method is much better than my C style one. With a little work to
handle exceptions it should generally work. The only problem I have
found so far is that it doesn’t handle periods in the number string
properly – i.e. sform(“12345.67”, “$00,000.00”) returns $12,345…6"
instead of $12,345.67". Something for me to work on. Thanks for the code.

I made some changes to handle the decimal point
and the case when there are fewer digits in the
string than in the format.

The part on the left of the decimal point.

def sform_left( str, fmt )
result = ‘’
fmt = fmt.split(//)
str.split(//).reverse_each{|d|
while fmt.last != ‘0’ do
result = fmt.pop + result
end
fmt.pop
result = d + result
}
result = fmt.first + result if fmt.first != ‘0’
result
end

The part on the right of the decimal point.

def sform_right( str, fmt )
ary = str.split(//)
fmt.gsub( /0/ ){ ary.shift || ‘0’ }
end

def sform( str, fmt )
str = str.split(‘.’)
fmt = fmt.split(‘.’)
result = sform_left( str[0], fmt[0])
if fmt[1]
result += “.” + sform_right( str.last, fmt.last)
end
result
end

puts sform(“12345”, “$00,000”)
puts sform(“1234”, “$00,000”)
puts sform(“123”, “$00,000”)
puts sform(“12345.6”, “$00,000.00”)
puts sform(“12345.678”, “$00,000.00”)
puts sform(“12345.678”, “$00,000”)

— output —
$12,345
$1,234
$123
$12,345.60
$12,345.67
$12,345

11

On 5/4/07, Michael W. Ryder [email protected] wrote:

I am trying to come up with a “generic” formatting routine so that I
could feed it something like sform(“123456789”, “000-00-0000”) or
sform(“1234567890”, “(000) 000-0000”) or sform(“123456”, “00/00/00”) and
it would work. I could easily do something like you suggest, and for
some cases it might be better, but I want something I can for any number
of formats. Thanks for the input.

I was thinking about your question again and came up with this.
Just before I posted I saw that William had taken a similar approach
and offered it in a more compact form.
Anyway , for what it’s worth.
I was thinking the dots should only be in the formatting, not in the
input so I stripped them out of the input.

#arr = [“123456”, “00/00/00”]
arr = [“12345.67”, “$00,000.00”]
#arr = [“1234567890”, “(000) 000-0000”]

inp_arr = arr[0].delete(“.”).split(//)
fmt_arr = arr[1].split(//)
str = “”
fmt_arr.each do |x|
str << inp_arr.shift if x =~ /\d/
str << x if x !~ /\d/
end
p str

Harry

12

On Fri, May 04, 2007 at 09:45:04AM +0900, Michael W. Ryder wrote:

ai += 1

end
end
puts b

Is there a better (more Rubyish) way to do this?

I have a vague recollection that Perl has a specific feature along these
lines: ah yes, see “man perlform”. But I’ve never used it, and I think
this
is one Perlism that Ruby hasn’t copied.

It sounds to me like you actually want two different types of format:

format(“000 000-0000”,“1234567890”) # => “123 456-7890”
format(“000000.00”,“1234.4”) # => " 1234.40"

People have given you several solutions for the former. The latter is
most
easily handled by sprintf (or format % [values]) if the value is
numeric.

Actually, you could bend sprintf to do the former too:

val = “1234567890”
fmt = “(000) 000-0000”
res = sprintf(fmt.gsub(/0/,"%s"), *val.split(//))

Regards,

Brian.

13

William J. wrote:

require ‘enumerator’
found so far is that it doesn’t handle periods in the number string
fmt = fmt.split(//)

if fmt[1]
puts sform(“12345.678”, “$00,000”)

— output —
$12,345
$1,234
$123
$12,345.60
$12,345.67
$12,345

Thank you for the ideas. I figured there had to be a better way to
accomplish what I wanted without doing it like I would do it in C. With
a little work I think I can make these do everything I want to do. The
last step I need to work on is to allow it to handle formats that
replace leading zeros with blanks. This will allow the display of
dollar amounts in neat columns.

14

On Fri, May 04, 2007 at 04:42:31PM +0900, Brian C. wrote:

Actually, you could bend sprintf to do the former too:

val = “1234567890”
fmt = “(000) 000-0000”
res = sprintf(fmt.gsub(/0/,"%s"), *val.split(//))

which of course can be shrunk to

val = “1234567890”
fmt = “(000) 000-0000”
res = fmt.gsub(/0/,"%s") % val.split(//)

15

On 04.05.2007 02:43, Michael W. Ryder wrote:

ai += 1

end
end
puts b

Is there a better (more Rubyish) way to do this?
Obviously I will want to add more code in the future to handle
exceptions like phone numbers that are not 10 characters long or are
longer, but this will get me started.

I’d start by having the parts of the number separate. Then you can
easily use sprintf for the formatting. For example, with

Phone = Struct.new(:country, :area, :number) do
def to_s
sprintf("(%d) %d-%d", country, area, number)
end
end

You can do

irb(main):006:0> a = Phone.new 123, 456, 7890
=> #
irb(main):007:0> puts a
(123) 456-7890
=> nil
irb(main):008:0>

You can even add exceptions and special handling if one of the parts is
missing etc. If you need leading zeros, that’s easy with sprintf as
well.

Kind regards

robert

16

On 5/4/07, Robert K. [email protected] wrote:

 sprintf("(%d) %d-%d", country, area, number)

irb(main):008:0>

You can even add exceptions and special handling if one of the parts is
missing etc. If you need leading zeros, that’s easy with sprintf as well.

Kind regards

    robert

I answered that question, too.
But then he added more requirements.
He not only wants to deal with phone number formats, but any format.

Harry

17

On 04.05.2007 10:24, Harry K. wrote:

I answered that question, too.
But then he added more requirements.
He not only wants to deal with phone number formats, but any format.

That somehow eluded me. Thanks for pointing it out!

Kind regards

robert

18

Ken B. wrote:

if b[i,1] == “0”

puts a.sub(/(…)(…)(…)/,’(\1) \2-\3’)

when you have other formats, use if a.length == whatever to decide which
pattern to apply.

–Ken

That looks very interesting! Thanks for the code. I think it will work
great for parts of the method. With very little work this I can make
this work for formatting dollar amounts, etc.

19

On Fri, 04 May 2007 00:43:34 +0000, Michael W. Ryder wrote:

 ai += 1

end
end
puts b

Is there a better (more Rubyish) way to do this? Obviously I will want
to add more code in the future to handle exceptions like phone numbers
that are not 10 characters long or are longer, but this will get me
started.

puts a.sub(/(…)(…)(…)/,’(\1) \2-\3’)

when you have other formats, use if a.length == whatever to decide which
pattern to apply.

–Ken