How to parse a "line"?

msojka · April 26, 2009, 10:48pm

Hi there:

I have a line, the format is like this
“num_of_item item1 item2…item_n” for example

“3 Tokyo Newyork Paris”

I want ruby parse this format, extract these keywords and save them in
an array.
I googled and search many forums, but can’t find any matches.

Would anyone give me any ideas??
Thanks a lot!

msojka · April 26, 2009, 10:57pm

On Apr 26, 2009, at 3:48 PM, Martin S. wrote:

Would anyone give me any ideas??

If you just need to bust up the words via the spaces, this should
probably work for you:

items = “3 Tokyo Newyork Paris”.split[1…-1]
=> [“Tokyo”, “Newyork”, “Paris”]

James Edward G. II

msojka · April 26, 2009, 11:12pm

On Sun, Apr 26, 2009 at 10:56 PM, James G. [email protected]
wrote:

an array.
I googled and search many forums, but can’t find any matches.

Would anyone give me any ideas??

If you just need to bust up the words via the spaces, this should probably
work for you:

items = “3 Tokyo Newyork Paris”.split[1…-1]
=> [“Tokyo”, “Newyork”, “Paris”]

This is another opportunity to show Ruby’s versatile style, e.g.
functional

_, *items = “3 .a b c”.split

R.

msojka · April 27, 2009, 12:56am

On Sun, Apr 26, 2009 at 4:11 PM, Robert D. [email protected]
wrote:

I want ruby parse this format, extract these keywords and save them in

This is another opportunity to show Ruby’s versatile style, e.g. functional

_, *items = “3 .a b c”.split

I like…

(items = my_item_list).split.shift

Both you and I have examples, though, that do not return what the
statement using [1…-1] does. Mine, returns 3. Ah well, “items”
exists, though

Todd

msojka · April 27, 2009, 1:00am

On Sun, Apr 26, 2009 at 5:55 PM, Todd B. [email protected]
wrote:

(items = my_item_list).split.shift

Oops…

(items = my_item_list.split).shift

Todd

msojka · April 27, 2009, 2:25am

Thank you Todd, but the number of the keywords are dynamic.

and the array size is not fixed until program reads the first keywords
of the input line.
The line may be “3 tokyo newyork paris”, also maybe
“6 toyota bmw honda GM Ford”

So I can’t hard coded this array in the program.
How can I parse the line?

Thanks!
Todd B. wrote:

On Sun, Apr 26, 2009 at 5:55 PM, Todd B. [email protected]
wrote:

(items = my_item_list).split.shift

Oops…

(items = my_item_list.split).shift

Todd

msojka · April 27, 2009, 3:08am

Martin S. wrote:

Thank you Todd, but the number of the keywords are dynamic.

and the array size is not fixed until program reads the first keywords
of the input line.

Explain why these solutions won’t work for you:

strs = [“6 toyota bmw honda GM Ford”, “3 tokyo newyork paris”]

strs.each do |str|
puts “1) split with subscripts:”
p str.split()[1…-1]

puts “2) split with parallel assignment:”
first, *therest = str.split()
p therest

puts
end

–output:–

split with subscripts:
[“toyota”, “bmw”, “honda”, “GM”, “Ford”]
split with parallel assignment:
[“toyota”, “bmw”, “honda”, “GM”, “Ford”]
split with subscripts:
[“tokyo”, “newyork”, “paris”]
split with parallel assignment:
[“tokyo”, “newyork”, “paris”]

msojka · April 27, 2009, 4:11am

On Sun, Apr 26, 2009 at 7:25 PM, Martin S. [email protected]
wrote:

Thank you Todd, but the number of the keywords are dynamic.

and the array size is not fixed until program reads the first keywords
of the input line.
The line may be “3 tokyo newyork paris”, also maybe
“6 toyota bmw honda GM Ford”

So I can’t hard coded this array in the program.
How can I parse the line?

If you look closely, you’ll see that James’ method grabs everything
from 1 to -1 (the end) of the array, omitting the zeroth element.

Robert’s method assigns the garbage – in this case, the whole array
– to a dummy variable, and the important stuff to the var that you
care about (items).

My way was only slightly different. I opted to create the whole
array, and then drop the first without creating an extra object. I
had to use (items = my_item_list.split) inside parens like that
because #shift returns the object you popped off the front of the
list, and not the actual remaining stuff.

In all examples given, the size doesn’t really matter… hah! Or as
they say, it depends

There are probably several ways to do this; I just like that particular
one.

hth,
Todd

msojka · April 27, 2009, 4:06am

On Apr 26, 2009, at 7:25 PM, Martin S. wrote:

Thank you Todd, but the number of the keywords are dynamic.

and the array size is not fixed until program reads the first keywords
of the input line.
The line may be “3 tokyo newyork paris”, also maybe
“6 toyota bmw honda GM Ford”

So I can’t hard coded this array in the program.
How can I parse the line?

Did you try our code? All of our solutions work for all examples
you’ve posted so far…

James Edward G. II

msojka · April 27, 2009, 4:48am

On Sun, Apr 26, 2009 at 9:11 PM, Todd B. [email protected]
wrote:

because #shift returns the object you popped off the front of the
list, and not the actual remaining stuff.

In all examples given, the size doesn’t really matter… hah! Or as
they say, it depends

There are probably several ways to do this; I just like that particular one.

You know after looking at your original post, maybe you are trying to
create a digest, i.e. a Hash or associative array, with your keys as
the first “column”.

In that case, you might try…

h = {}
f = my_string_list.each_line do |line|
key = (value = line.split).shift
h[key] = value
end

…or something like that.

Todd

msojka · April 27, 2009, 4:57am

On Sun, Apr 26, 2009 at 9:47 PM, Todd B. [email protected]
wrote:

In that case, you might try…

h = {}
f = my_string_list.each_line do |line|
key = (value = line.split).shift
h[key] = value
end

…or something like that.

It’s late and I’m golfing with myself. Apologies to the list for this
almost one-liner…

h = {}
s.each_line{|i|h[(v=i.split).shift]=v}

msojka · April 27, 2009, 5:06am

Oh my god…

You and Jaem’s codes really work!

Thanks a lot!
Todd B. wrote:

My way was only slightly different. I opted to create the whole
array, and then drop the first without creating an extra object. I
had to use (items = my_item_list.split) inside parens like that
because #shift returns the object you popped off the front of the
list, and not the actual remaining stuff.

In all examples given, the size doesn’t really matter… hah! Or as
they say, it depends

There are probably several ways to do this; I just like that particular
one.

hth,
Todd

msojka · April 27, 2009, 5:35am

On Sun, Apr 26, 2009 at 10:32 PM, Martin S. [email protected]
wrote:

it works!

s.each instead of s.each_line

Using 1.9.1. Sorry I didn’t mention that. I haven’t been following
the differences, but I seem to recall that was a change somewhere
along the line.

Todd

msojka · April 27, 2009, 5:59am

Todd B. wrote:

My way was only slightly different. I opted to create the whole
array, and then drop the first without creating an extra object.

Note that you made ruby go through the laborious chore of shifting every
element in the array over one spot to the left.

msojka · April 27, 2009, 6:05am

On Sun, Apr 26, 2009 at 10:59 PM, 7stud – [email protected]
wrote:

Todd B. wrote:

My way was only slightly different. I opted to create the whole
array, and then drop the first without creating an extra object.

Note that you made ruby go through the laborious chore of shifting every
element in the array over one spot to the left.

Is that really what happens? I thought it just reassigned a starting
point, and indexed from there. At least, that would make more sense
to me if it did.

Todd

msojka · April 27, 2009, 5:32am

it works!

s.each instead of s.each_line

thanks!

Todd B. wrote:

On Sun, Apr 26, 2009 at 9:47 PM, Todd B. [email protected]
wrote:

In that case, you might try…

h = {}
f = my_string_list.each_line do |line|
key = (value = line.split).shift
h[key] = value
end

…or something like that.

It’s late and I’m golfing with myself. Apologies to the list for this
almost one-liner…

h = {}
s.each_line{|i|h[(v=i.split).shift]=v}

msojka · April 27, 2009, 6:08am

On Sun, Apr 26, 2009 at 11:03 PM, Todd B. [email protected]
wrote:

to me if it did.
Let me rephrase. What is different on the underlying structure with
[1…-1] and #shift? I’m not sure if shift really moves everything,
but if it does, I wouldn’t mind knowing.

Todd

msojka · April 27, 2009, 6:45am

Todd B. wrote:

On Sun, Apr 26, 2009 at 11:03 PM, Todd B. [email protected]
wrote:

to me if it did.
Let me rephrase. What is different on the underlying structure with
[1…-1] and #shift? I’m not sure if shift really moves everything,
but if it does, I wouldn’t mind knowing.

Todd

shift changes the array in place:

arr = [10, 20, 30]

y = arr.shift
puts y
p arr

–output:–
10
[20, 30]

subscripts create a new array:

arr = [10, 20, 30]
p arr
print "id: ", arr.object_id
puts

x = arr[1…-1]
p x
print "id: ", x.object_id
puts
p arr

–output:–
[10, 20, 30]
id: 75950
[20, 30]
id: 75860
[10, 20, 30]

Is that really what happens? I thought it just reassigned a starting
point, and indexed from there. At least, that would make more sense
to me if it did.

Would it? What if you had an array that took up 3GB of memory and you
shifted off every element but the last one. Would you expect your array
to still occupy 3GB of memory?

msojka · April 27, 2009, 6:41am

On Mon, Apr 27, 2009 at 1:07 PM, Todd B. [email protected]
wrote:

point, and indexed from there. At least, that would make more sense
to me if it did.

Let me rephrase. What is different on the underlying structure with
[1…-1] and #shift? I’m not sure if shift really moves everything,
but if it does, I wouldn’t mind knowing.

Todd

If it shifts everything, it sure does it fast

p Time.now
arr = (1…100_000_000).to_a
p Time.now
arr.shift
p arr[0]
p Time.now

###output

Mon Apr 27 13:34:06 +0900 2009
Mon Apr 27 13:36:29 +0900 2009
2
Mon Apr 27 13:36:29 +0900 2009

Harry

msojka · April 27, 2009, 7:31am

h = {}

strs.each{|i|h[(v=i.split).shift]=v}

The array size is 1 by this sentense…

Todd B. wrote:

On Sun, Apr 26, 2009 at 11:45 PM, 7stud – [email protected]
wrote:

shift changes the array in place:
I looked at an old thread where Bob Hutchinson piped in about
potential GC problems with this, but that was back in '07

[20, 30]
x = arr[1…-1]
new object is created by x =

[10, 20, 30]