Array : each_with_index

Ruby
1

Hi All,

I have following lines

m=[‘a’,‘b’,‘c’]
puts m.each_with_index{|v,i| i}

which output in:

ruby try.rb
a
b
c
Exit code: 0

I was expecting to see indexes, rather than values.
Am I overlooking some trivial nuance or what.
Basically i wanted to print index values while iterating an array at
some other project.
and “each_with_index” seemed very obvious for the task but for the
unexpected result.

Raja

2

2007/7/19, Vin R. [email protected]:

ruby try.rb
some other project.
and “each_with_index” seemed very obvious for the task but for the
unexpected result.

It is. But #each_with_index returns the Enumerable. You need to put
the put into the block:

irb(main):001:0> [‘a’,‘b’,‘c’].each_with_index {|a,i| puts i}
0
1
2
=> [“a”, “b”, “c”]
irb(main):002:0>

Kind regards

robert

3

On Thu, Jul 19, 2007 at 03:46:10PM +0900, Vin R. wrote:

c

Exit code: 0

irb(main):003:0> m.each_with_index{|v,i| i}
=> [“a”, “b”, “c”]
irb(main):004:0> m.each_with_index{|v,i| p i}
0
1
2
=> [“a”, “b”, “c”]

The i variable in the block is indeed the index. But the result of the
entire
each_with_index expression is not the result of each block invocation,
but the collection
of all the array elements.

marcel

4

Alle giovedì 19 luglio 2007, Vin R. ha scritto:

and “each_with_index” seemed very obvious for the task but for the
unexpected result.

Raja

You’re telling ruby to print the return value of each with index, which,
according to the ri documentation, is the array itself. If you want to
print
the different inidces, you need to put the call to puts inside the
block:

m.each_with_index{|v,i| puts( i )}

Putting i as the last statement of the block simply makes that the
return
value of the block, but this is ignored by each_with_index, so it’s
useless.

I hope this helps

Stefano

5

Thanks All,

I got the nuance.

But my problem just got a bit more messier.
have a look down here:

r = [‘KLP’,‘OGN’ ]
msg =<<MSG
Found #{r.length} orders
#{r.map.each_with_index{|v,i| puts “(#{i+1}) #{v}\n” }}
MSG

puts msg

This prints:

ruby try.rb
(1) KLP
(2) OGN
Found 2 orders
KLPOGN

Exit code: 0

And this time I also know why, thanks again.

But what I actually wanted was this sort of output:

ruby try.rb

Found 2 orders
(1) KLP
(2) OGN

Exit code: 0

Any Help!

Thanks
raja

6

[mailto:[email protected]] On Behalf Of Vin R.:

But what I actually wanted was this sort of output:

>ruby try.rb

Found 2 orders

(1) KLP

(2) OGN

just want you to show the evolution from your original scheme (spot the
difference) to that of roberts (retaining your original inlining)

C:\family\ruby>cat -n try.rb
1 puts “—scheme 1—”
2 r = [‘KLP’,‘OGN’]
3 msg =<<MSG
4 #{puts “Found #{r.length} orders”}
5 #{r.map.each_with_index{|v,i| puts “(#{i+1}) #{v}\n” }}
6 MSG
7 msg
8
9 puts “—scheme 2—”
10 r = [‘KLP’,‘OGN’]
11 <<MSG
12 #{puts “Found #{r.length} orders”}
13 #{r.map.each_with_index{|v,i| puts “(#{i+1}) #{v}\n” }}
14 MSG
15
16 puts “—scheme 3—”
17 r = [‘KLP’,‘OGN’]
18 puts “Found #{r.length} orders”
19 r.each_with_index{|v,i| puts “(#{i+1}) #{v}\n” }

C:\family\ruby>ruby try.rb
—scheme 1—
Found 2 orders
(1) KLP
(2) OGN
—scheme 2—
Found 2 orders
(1) KLP
(2) OGN
—scheme 3—
Found 2 orders
(1) KLP
(2) OGN

kind regards -botp

7

#… puts “(#{i+1}) #{v}\n” }

oops, lose “\n” pls.

8

2007/7/19, Peña, Botp [email protected]:

#… puts “(#{i+1}) #{v}\n” }

oops, lose “\n” pls.

It doesn’t hurt though - output is the same.

robert

9

2007/7/19, Vin R. [email protected]:

#{r.map.each_with_index{|v,i| puts “(#{i+1}) #{v}\n” }}
KLPOGN
(1) KLP
(2) OGN

Exit code: 0

Any Help!

You are confusing string evaluation with output. You need to remove
the puts from the string interpolation or use a different approach.
I’d keep things simple and do

r = [‘KLP’,‘OGN’ ]
print “found “, r.size, " orders\n”
r.each_with_index do |e,i|
print " (”, i+1, ") ", e, “\n”
end

robert

10

On Jul 19, 3:12 am, Vin R. [email protected] wrote:

r = [‘KLP’,‘OGN’ ]
msg =<<MSG
Found #{r.length} orders
#{r.map.each_with_index{|v,i| puts “(#{i+1}) #{v}\n” }}
MSG

puts msg

It looks like others have already clarified things for you. This is a
bit of a tangent, but I just wanted to mention that ‘map’ is
superfluous in the above code i.e. the output is identical with and
without it.

irb(main):001:0> r = [‘KLP’,‘OGN’ ]
=> [“KLP”, “OGN”]
irb(main):002:0> r
=> [“KLP”, “OGN”]
irb(main):003:0> r.map
=> [“KLP”, “OGN”]

11

From: Robert K. [mailto:[email protected]]

2007/7/19, Peña, Botp [email protected]:

> #… puts “(#{i+1}) #{v}\n” }

> oops, lose “\n” pls.

It doesn’t hurt though - output is the same.

’ just little concern for the op. he might assume puts requires \n.

but you’re right, it does not hurt. he’ll learn it later anyway…

kind regards -botp

12

On 7/19/07, Marcel Molina Jr. [email protected] wrote:

b

The i variable in the block is indeed the index. But the result of the entire
each_with_index expression is not the result of each block invocation, but the collection
of all the array elements.

marcel

Marcel Molina Jr. [email protected]

Actually Enumerable#each_with_index appears to return the receiver
object itself, identity is preserved.

irb(main):001:0> a = %w{a word array}
=> [“a”, “word”, “array”]
irb(main):002:0> a.equal?(a.each_with_index {|e, i| [e, i]})
=> true
irb(main):003:0> a = (1…3)
=> 1…3
irb(main):004:0> a.equal?(a.each_with_index {|e, i| [e, i]})
=> true

Although this is not documented, and could of course be overridden in
a class which includes enumerable or one of its subclasses.

Rick DeNatale

My blog on Ruby
http://talklikeaduck.denhaven2.com/

13

On Jul 19, 3:06 am, Peña, Botp [email protected] wrote:

11 <<MSG
12 #{puts “Found #{r.length} orders”}
13 #{r.map.each_with_index{|v,i| puts “(#{i+1}) #{v}\n” }}
14 MSG
15
16 puts “—scheme 3—”
17 r = [‘KLP’,‘OGN’]
18 puts “Found #{r.length} orders”
19 r.each_with_index{|v,i| puts “(#{i+1}) #{v}\n” }

One other variation which looks a little closer to what the OP wanted:

module Enumerable
def map_with_index
idx = -1
map{ |v| yield v,idx+=1 }
end
end

puts “—scheme 4—”
r = [‘KLP’,‘OGN’]
msg =<<ENDMSG
Found #{r.length} orders
#{r.map_with_index{ |v,i| “(#{i+1}) #{v}”}.join(“\n”)}
ENDMSG
puts msg

14

Phrogz wrote:

puts “—scheme 4—”

While I have found the thread fascinating, it appears to me
that the original code and the 4 schemes are simply trying
to avoid the definition of a function.

def msg®
puts “Found #{r.length} orders”
r.each_with_index{|v,i| puts “(#{i+1}) #{v}\n” }
end

x = [‘KLP’,‘OGN’]
msg(x)

Can anyone explain what I am missing?

Thanks
Ian

15

Thanks All
It cleared up a lot of mess up at my head.

thanks again

Raja