sort out duplicated fax numbers

i have a number of forms with fax numbers to come up into arrays and then
combine to string.

after that i design the flow
1. break the string to array
now the string looks like this 12345678,23456789,34567890...
2. check record-2 again record-1,
check record-3 again record-2 & record-1
check record-4 again record-3 & record-2 & record-1
and so on... (if duplicated, drop it)

problem - step 1 break the string to array
var CDEF = new Array(faxnumbers);
alert(CDEF[0],

it returns all the numbers (12345678,23456789,34567890...

i just wish to have 12345678

anything wrong with my script?

Grateful if you could kindly advise whether step 2 is the best way to sort
out duplicated numbers?

Thanks a lot

tony

re: sort out duplicated fax numbers

Tony WONG wrote:[color=blue]
> i have a number of forms with fax numbers to come up into arrays and then
> combine to string.
>
> after that i design the flow
> 1. break the string to array
> now the string looks like this 12345678,23456789,34567890...
> 2. check record-2 again record-1,
> check record-3 again record-2 & record-1
> check record-4 again record-3 & record-2 & record-1
> and so on... (if duplicated, drop it)
>
> problem - step 1 break the string to array
> var CDEF = new Array(faxnumbers);
> alert(CDEF[0],
>
> it returns all the numbers (12345678,23456789,34567890...
>
> i just wish to have 12345678
>
> anything wrong with my script?[/color]

Yes, it's very inefficient. If you start with a comma-delimited set of
numbers, you can split it into an array using split:

var faxNumbers = '1234,2345,3456,45646,...';
var faxNumArray = faxNumbers.split(',');

Now to get unique numbers, sort the array and load only new ones into a
new array (the numbers are sorted alphabetically, but since they are all
numbers that shouldn't matter here):

faxNumArray.sort();
var faxNumUnique = [faxNumArray[0]];
for (var i=1, j=0, len=faxNumArray.length; i<len; ++i){
if (faxNumArray[i] != faxNumUnique[j]){
faxNumUnique[++j] = faxNumArray[i];
}
}
// faxNumUnique now contains a sorted, unique list of fax numbers

[color=blue]
> Grateful if you could kindly advise whether step 2 is the best way to sort
> out duplicated numbers?[/color]

No, it's not. The best way would be to use a UNIX sort unique, but
since this is JavaScript ... see above.


--
Rob
Group FAQ: <URL:http://www.jibbering.com/faq/>

re: sort out duplicated fax numbers

Tony WONG a écrit :[color=blue]
> i have a number of forms with fax numbers to come up into arrays and then
> combine to string.
>
> after that i design the flow
> 1. break the string to array
> now the string looks like this 12345678,23456789,34567890...
> 2. check record-2 again record-1,
> check record-3 again record-2 & record-1
> check record-4 again record-3 & record-2 & record-1
> and so on... (if duplicated, drop it)
>
> problem - step 1 break the string to array
> var CDEF = new Array(faxnumbers);[/color]

CDEF = faxnumbers.split(',');
[color=blue]
> alert(CDEF[0],[/color]

would be 12345678



--
Stephane Moriaux et son [moins] vieux Mac

re: sort out duplicated fax numbers

Thanks a lot.


"RobG" <rgqld@iinet.net.au>
???????:YHg8g.1218$Nv4.171223@news.optus.net.au...[color=blue]
> Tony WONG wrote:[color=green]
>> i have a number of forms with fax numbers to come up into arrays and then
>> combine to string.
>>
>> after that i design the flow
>> 1. break the string to array
>> now the string looks like this 12345678,23456789,34567890...
>> 2. check record-2 again record-1,
>> check record-3 again record-2 & record-1
>> check record-4 again record-3 & record-2 & record-1
>> and so on... (if duplicated, drop it)
>>
>> problem - step 1 break the string to array
>> var CDEF = new Array(faxnumbers);
>> alert(CDEF[0],
>>
>> it returns all the numbers (12345678,23456789,34567890...
>>
>> i just wish to have 12345678
>>
>> anything wrong with my script?[/color]
>
> Yes, it's very inefficient. If you start with a comma-delimited set of
> numbers, you can split it into an array using split:
>
> var faxNumbers = '1234,2345,3456,45646,...';
> var faxNumArray = faxNumbers.split(',');
>
> Now to get unique numbers, sort the array and load only new ones into a
> new array (the numbers are sorted alphabetically, but since they are all
> numbers that shouldn't matter here):
>
> faxNumArray.sort();
> var faxNumUnique = [faxNumArray[0]];
> for (var i=1, j=0, len=faxNumArray.length; i<len; ++i){
> if (faxNumArray[i] != faxNumUnique[j]){
> faxNumUnique[++j] = faxNumArray[i];
> }
> }
> // faxNumUnique now contains a sorted, unique list of fax numbers
>
>[color=green]
>> Grateful if you could kindly advise whether step 2 is the best way to
>> sort out duplicated numbers?[/color]
>
> No, it's not. The best way would be to use a UNIX sort unique, but since
> this is JavaScript ... see above.
>
>
> --
> Rob
> Group FAQ: <URL:http://www.jibbering.com/faq/>[/color]

re: sort out duplicated fax numbers

JRS: In article <YHg8g.1218$Nv4.171223@news.optus.net.au>, dated Wed,
10 May 2006 07:34:16 remote, seen in news:comp.lang.javascript, RobG
<rgqld@iinet.net.au> posted :[color=blue]
>Yes, it's very inefficient. If you start with a comma-delimited set of
>numbers, you can split it into an array using split:
>
> var faxNumbers = '1234,2345,3456,45646,...';
> var faxNumArray = faxNumbers.split(',');
>
>Now to get unique numbers, sort the array and load only new ones into a
>new array (the numbers are sorted alphabetically, but since they are all
>numbers that shouldn't matter here):[/color]

In sorting to determine duplicates, the sort order does not matter, and
hence there is no relevance in their being numbers. The only need is
that identical values sort to adjacent and that comparison of a given
pair is consistent.

The obvious javascript way is to do the built-in sort, then to make a
single pass removing repeats or copying non-repeats.

If the number of duplicates is large, however, it may be better to code
the sort explicitly using a reasonably efficient algorithm, but to then
modify it so that when two items are found to be equal one of them is
dropped.

OTOH, perhaps the most efficient sorts are non-stable (i.e. items
testing as identical do not have their order preserved) and do not use
comparison-for-equality.

--
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