Forum: Ruby Array.sort problem

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Larme (Guest)
on 2007-04-21 20:21
(Received via mailing list)
Dear all, I'm a newbie to ruby and today when I'm writing a simple
script to process some data, I found something I can't understand.

The data is stored in several column seperated by tab or space.  I use
the following code to get the data (assuming the data comes from
standard input and all numbers are integer)

data=[]
counter = 0
while line = STDIN.gets
  data[counter] = line.split
  data[counter].map! {|str| str.to_i}
  counter += 1
end

hence data[i] is an array hold all the numbers in the ith line, data[i]
[j] is the number on ith line and jth column. Then what I want the
script to do is sorting lines according to a specified column.  I
thought the following code should work:

result = data.sort {|x, y| x[col] <=> y[col] }

where the col determine which column the script will sort according
to. However ruby raise a error saying:
"undefined method `<=>' for nil:NilClass (NoMethodError)
        from ana.rb:16:in `sort'
        from ana.rb:16
"

I have to write the code as

result = data.sort {|x, y| x[col].to_i <=> y[col].to_i }

to let the script run properly. I'm quite confused here. I think the
elements of array data are converted to integer when the code

  data[counter].map! {|str| str.to_i}

finished. However why ruby still requires a explicit conversion when I
use the data.sort?
C51b60fddee4caf0ae2bb18a7986f893?d=identicon&s=25 Hemant K. (gnufied)
on 2007-04-21 20:49
(Received via mailing list)
On 4/21/07, Larme <lalalarme@gmail.com> wrote:
>   data[counter] = line.split
>
>
> to let the script run properly. I'm quite confused here. I think the
> elements of array data are converted to integer when the code
>
>   data[counter].map! {|str| str.to_i}
>
> finished. However why ruby still requires a explicit conversion when I
> use the data.sort?
>

Yes, they are. You might be doing some typo or other crap.  Following
program runs verbatim:

data=[]
counter = 0
while line = STDIN.gets
 data[counter] = line.split
 data[counter].map! {|str| str.to_i}
 counter += 1
end

#p data.sort_by { |x| x[2] } #=> you can use sort_by as an alternative

p data.sort {|x,y| x[2] <=> y[2] } #=> This also work.
Sebastian Hungerecker (Guest)
on 2007-04-21 21:22
(Received via mailing list)
Larme wrote:
> use the data.sort?
If you have an array arr with n elements which you all turn into
integers,
arr[i] with i>=n will still be nil. So if in your case not all rows have
more
than col columns, that would explain the problem.


HTH,
Sebastian Hungerecker
Robert Klemme (Guest)
on 2007-04-21 21:30
(Received via mailing list)
On 21.04.2007 20:16, Larme wrote:
>   data[counter] = line.split
>
>
> to let the script run properly. I'm quite confused here. I think the
> elements of array data are converted to integer when the code
>
>   data[counter].map! {|str| str.to_i}
>
> finished. However why ruby still requires a explicit conversion when I
> use the data.sort?

You probably have lines with differing number of entries and thus
sometimes x[col] just returns nil.

You could do something like:

data.sort_by {|x| x[col].to_i}
data.sort_by {|x| x[col] || 0}

If you want to be sure that you extract only integers you could do

l.scan(/\d+/).map {|x| x.to_i}

instead of the split.

Lots of options...

Kind regards

  robert
Larme (Guest)
on 2007-04-21 21:41
(Received via mailing list)
On Apr 22, 3:25 am, Robert Klemme <shortcut...@googlemail.com> wrote:
>
> > script to do is sorting lines according to a specified column.  I
>
> > use the data.sort?
>
> l.scan(/\d+/).map {|x| x.to_i}
>
> instead of the split.
>
> Lots of options...
>
> Kind regards
>
>         robert


Thank all of you. Yes, the problem is that the column number is not
fixed -- it should be, but my friend who prepare the data made some
thing wrong.
Ari Brown (Guest)
on 2007-04-22 01:22
(Received via mailing list)
On Apr 21, 2007, at 2:20 PM, Larme wrote:
<snip>
> 1
> 2 data=[]
> 3 counter = 0
> 4 while line = STDIN.gets
> 5   data[counter] = line.split
> 6   data[counter].map! {|str| str.to_i}
> 7   counter += 1
> 8 end

Ok, so I'm a noob trying to learn something here...

Shouldn't line 4 be presented with a double = (==) since you're
trying do something while line == STDIN.gets?

Also, does STDIN.gets need to be stated in a previous line? Or will
it be prompted from within the 'while' statement.

<snip>
--------------------------------------------|
If you're not living on the edge,
then you're just wasting space.
Philip M. Gollucci (Guest)
on 2007-04-22 02:17
(Received via mailing list)
>> 4 while line = STDIN.gets

> Ok, so I'm a noob trying to learn something here...
>
> Shouldn't line 4 be presented with a double = (==) since you're trying
> do something while line == STDIN.gets?
No, the assignment of line = STDIN.gets happens first which reads up to
a '\n' including it. While then loops while this value is true.

When there is no more inmput, line will be assigned nothing which is
false and the loop ends.

This works identically to other languages like perl/python/php


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Ari Brown (Guest)
on 2007-04-22 16:41
(Received via mailing list)
On Apr 21, 2007, at 8:16 PM, Philip M. Gollucci wrote:

>>> 4 while line = STDIN.gets

<snip>

> No, the assignment of line = STDIN.gets happens first which reads
> up to a '\n' including it. While then loops while this value is true.

Oh, so it's not checking whether line == STDIN.gets (immediately),
but rather assigning STDIN.gets to line, and then checking whether
that's true?

Nice, I learned something!

---------------------------------------------------------------|
~Ari
"I don't suffer from insanity. I enjoy every minute of it" --1337est
man alive
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