Count the number of times an element occurs in an array

Hi,

I need to count the number of times an element occurs in an array.

At the moment I’m doing it like this:

a = [“a”, “b”, “a”, “b”, “b”]
b = a.uniq
c = []

0.upto(b.length-1) do |e|
d = 0
0.upto(a.length-1) do |f|
if b[e] == a[f]
d += 1
end
end
c << d
end

print b, c

[“a”, “b”] [2, 3]

This works, but seems really clumsy.
Can anyone give me any tips on what I could improve?
Cheers.

On Oct 5, 2009, at 02:24 , Jim B. wrote:

0.upto(b.length-1) do |e|

[“a”, “b”] [2, 3]
Some suggestions:

Your indentation is wack. Make sure you’re not mixing spaces and tabs.
2 spaces per indent. You’re mixing 2 & 3 & possibly tabs at 8.

Your variable names are meaningless. Improve that and you’ll have a
lot better time debugging when it is 4am and you’re tired.

You’re not using (sensible) iterators. I suggest you read that section
in the pickaxe book.

You’re not using good data structures or algorithms. You’re passing
over the elements of ‘a’ (see? better names means better readability!)
a.uniq.size times and doing a.size * a.uniq.size string equality
comparisons.

Take a look at this:

use a hash whose values default to 0 for proper data structure use

use count as the name, since that’s what it is storing.

count = Hash.new 0

iterate over each string only once

strings.each do |string|

increment the count for each string

count[string] += 1
end

p count

If you nuke all the comments, the code makes just as much sense (if
not more). That’s what we try to achieve in ruby whenever we can.

Hello,

2009/10/5 Jim B. [email protected]:

Hi,

I need to count the number of times an element occurs in an array.

Using Enumerable.count should do the trick:

a = [“a”, “b”, “a”, “b”, “b”]

a.uniq.each do |elem|
puts “#{elem}\t#{a.count(elem)}”
end

Produces:

a 2
b 3

Cheers,

Hi –

On Mon, 5 Oct 2009, Jim B. wrote:

Hi,

I need to count the number of times an element occurs in an array.

*** DISCLAIMER ***

This answer is just for fun, and because sometimes seeing useful
techniques in strange contexts helps people learn and remember them.

With that in mind… here’s a

strings = %w{a b a a c b b b c c c d }
=> [“a”, “b”, “a”, “a”, “c”, “b”, “b”, “b”, “c”, “c”, “c”, “d”]

hash = Hash.new {|h,k| [k, strings.count(k)] }
=> {}

hash.values_at(*strings.uniq)
=> [[“a”, 3], [“b”, 4], [“c”, 4], [“d”, 1]]

It’s not really a good way to count array elements, but it’s kind of
an interesting illustration of some hash semantics.

David

a = [“a”, “b”, “a”, “b”, “b”]
a.uniq.each do |elem|
puts “#{elem}\t#{a.count(elem)}”
end

@Fleck Jean-Julien
Thanks for the quick fix. That works perfectly and is miles better than
what I wrote.

@Ryan
Thanks for the suggestions.

Your indentation is wack. Make sure you’re not mixing spaces and tabs.
2 spaces per indent. You’re mixing 2 & 3 & possibly tabs at 8.
Sorry about that. I’m writing ruby in a virtual Windows on a Linux host.
Formatting got a bit corrupted when copying the code. I’ll watch out for
that next time.

Your variable names are meaningless. Improve that and you’ll have a
lot better time debugging when it is 4am and you’re tired.
Sorry. I just did that to make the example easier. In reality they are
called things like ‘@had_lessons_all’ and ‘@had_lessons_number’.

You’re not using (sensible) iterators. I suggest you read that section
in the pickaxe book.
Good tip. I will do that now.

You’re not using good data structures or algorithms.
No kidding :slight_smile:

count = Hash.new 0

iterate over each string only once

strings.each do |string|

increment the count for each string

count[string] += 1
end

p count

Thanks for that. I will try to rewrite my method (for the sake of
learning) in a neater way, using the technique you suggest.

Cheers

On Mon, Oct 5, 2009 at 6:14 AM, David A. Black [email protected]
wrote:

This answer is just for fun, and because sometimes seeing useful
=> {}

hash.values_at(*strings.uniq)

=> [[“a”, 3], [“b”, 4], [“c”, 4], [“d”, 1]]

It’s not really a good way to count array elements, but it’s kind of
an interesting illustration of some hash semantics.

Or, since we don’t use a key more than once one could do:

strings = %w{a b a a c b b b c c c d }

not_a_hash = lambda {|k| [k, strings.count(k)] }

strings.uniq.map {|s| not_a_hash[s]}

=> [[“a”, 3], [“b”, 4], [“c”, 4], [“d”, 1]]


Rick DeNatale

Blog: http://talklikeaduck.denhaven2.com/
Twitter: http://twitter.com/RickDeNatale
WWR: http://www.workingwithrails.com/person/9021-rick-denatale
LinkedIn: http://www.linkedin.com/in/rickdenatale

Cheers for the replies, they are certainly very interesting and I will
work through all of them.
However, now I’m having weird problem.

When I run the following on Linux:
a = [“a”, “b”, “a”, “b”, “b”]

a.uniq.each do |elem|
puts “#{elem}\t#{a.count(elem)}”
end

I get the expected output :
a 2
b 3

But when I run exactly the same code on Windows I get:

undefined method `count’ for [“a”, “b”, “a”, “b”, “b”]:Array
(NoMethodError)

ruby -v (Linux): ruby 1.8.7 (2008-08-11 patchlevel 72) [x86_64-linux]
ruby -v (Windows): ruby 1.8.6 (2009-09-24 patchlevel 111) [i386-mswin32]

Does anyone have an idea why this could be happening.
I have tried searching for a solution with Google, but could find
nothing helpful.

Hi –

On Mon, 5 Oct 2009, Jim B. wrote:

ruby -v (Windows): ruby 1.8.6 (2009-09-24 patchlevel 111) [i386-mswin32]

Does anyone have an idea why this could be happening.
I have tried searching for a solution with Google, but could find
nothing helpful.

count is a post-1.8.6 method.

David

On Mon, Oct 5, 2009 at 5:24 PM, Jim B. [email protected]
wrote:

I need to count the number of times an element occurs in an array.

just for fun

a = [“a”, “b”, “a”, “b”, “b”]
=> [“a”, “b”, “a”, “b”, “b”]

a.group_by{|x|x}.map{|x,y| [x,y.size]}
=> [[“a”, 2], [“b”, 3]]

kind regards -botp

On Mon, Oct 5, 2009 at 9:22 PM, Jim B. [email protected]
wrote:

ruby -v (Windows): ruby 1.8.6 (2009-09-24 patchlevel 111) [i386-mswin32]

Does anyone have an idea why this could be happening.
I have tried searching for a solution with Google, but could find
nothing helpful.

The solution from Ryan D. also works on Ruby 1.8.6
Check it out.

Harry

On Mon, Oct 5, 2009 at 9:22 PM, Jim B. [email protected]
wrote:

ruby -v (Windows): ruby 1.8.6 (2009-09-24 patchlevel 111) [i386-mswin32]

Does anyone have an idea why this could be happening.
I have tried searching for a solution with Google, but could find
nothing helpful.

This works on 1.8.6

arr = [“a”, “b”, “a”, “b”, “b”]
p arr.uniq.map{|x| [x,arr.select{|y| y == x}.length]}

#> [[“a”, 2], [“b”, 3]]

Harry

Hello,
Â
I am new to ruby and the rails and it seems that it will not run or open
to let me see if I have install it. is there amy way that you can help
me.
james

— On Mon, 10/5/09, Fleck Jean-Julien [email protected]
wrote:

From: Fleck Jean-Julien [email protected]
Subject: Re: Count the number of times an element occurs in an array
To: “ruby-talk ML” [email protected]
Date: Monday, October 5, 2009, 3:43 AM

Hello,

2009/10/5 Jim B. [email protected]:

Hi,

I need to count the number of times an element occurs in an array.

Using Enumerable.count should do the trick:

a = [“a”, “b”, “a”, “b”, “b”]

a.uniq.each do |elem|
 puts “#{elem}\t#{a.count(elem)}”
end

Produces:

a     2
b     3

Cheers,

Take a look at this:

use a hash whose values default to 0 for proper data structure use

use count as the name, since that’s what it is storing.

count = Hash.new 0

iterate over each string only once

strings.each do |string|

increment the count for each string

count[string] += 1
end

p count

If you nuke all the comments, the code makes just as much sense (if
not more). That’s what we try to achieve in ruby whenever we can.

Hi Ryan,

Don’t know if you’ll be reading this, but wanted to say thanks anyway.

After reading the section you suggest from Pickaxe I came up with this,
which is not perfect, but definitely an improvement:

array = [“a”, “b”, “a”, “b”, “b”]
count =[]
for i in 0 … array.uniq.length
temp_count = 0
array.collect{|elem| temp_count +=1 if elem == array.uniq[i]}
count << temp_count
end
p array.uniq, count

Then I worked through your code and realized what a concise solution you
had proposed. I have changed my programme accordingly and it is now
considerably shorter and easier to read.

Thanks again and thanks to everyone else who made suggestions too.

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