How can I do this

ruby-talk
1

Hello,

I have a exercise where I must count differences between two strings.

Now I can do

for 0 .. string1.length

But is there a better mor rubisch way to do this ?

Roelof

2

In general I would prefer using #each over 'for', like

(0..string1.length).each {...}

If you only want to produce e.g. a single number out of a string comparison maybe #reduce/#inject would fit.

(0..string1.length).reduce(...) {...}

Regards, Paul

···

Am 29.05.2014 11:52, schrieb Roelof Wobben:

Hello,

I have a exercise where I must count differences between two strings.

Now I can do

for 0 .. string1.length

But is there a better mor rubisch way to do this ?

Roelof

3

irb:0> s="foo"
irb:0> s.chars.each {|c| p c}
"f"
"o"
"o"
=> ["f", "o", "o"]
irb:0> s.length.times {|i| p i}
0
1
2
=> 3
irb:0> s.chars.zip(t.chars)
=> [["f", "b"], ["o", "a"], ["o", "r"]]

- --
All the best, Sandor Szücs

···

On 29/05/14 11:52, Roelof Wobben wrote:

Hello,

I have a exercise where I must count differences between two
strings.

Now I can do

for 0 .. string1.length

But is there a better mor rubisch way to do this ?

4

Paul Götze schreef op 29-5-2014 12:11:

In general I would prefer using #each over 'for', like

(0..string1.length).each {...}

If you only want to produce e.g. a single number out of a string comparison maybe #reduce/#inject would fit.

(0..string1.length).reduce(...) {...}

Regards, Paul

Thanks for the tip.
I will read the documentation and try to figure out how reduce can help me.

Roelof

5

You might consider one of the routines which computes a distance between two strings such as the following:

···

#===================================================
#
# LEVENSHTEIN DISTANCE
#
#==================================================
def dameraulevenshtein(seq1, seq2)
     oneago = nil
     thisrow = (1..seq2.size).to_a + [0]
     seq1.size.times do |x|
         twoago, oneago, thisrow = oneago, thisrow, [0] * seq2.size + [x + 1]
         seq2.size.times do |y|
             delcost = oneago[y] + 1
             addcost = thisrow[y - 1] + 1
             subcost = oneago[y - 1] + ((seq1 != seq2[y]) ? 1 : 0)
             thisrow[y] = [delcost, addcost, subcost].min
             if (x > 0 and y > 0 and seq1 == seq2[y-1] and seq1[x-1] == seq2[y] and seq1 != seq2[y])
                 thisrow[y] = [thisrow[y], twoago[y-2] + 1].min
             end
         end
     end
     return thisrow[seq2.size - 1]
end

I'm not the author of this. I copied it from somewhere back in June 11;
Tom Reilly

Paul Götze wrote:

In general I would prefer using #each over 'for', like

(0..string1.length).each {...}

If you only want to produce e.g. a single number out of a string comparison maybe #reduce/#inject would fit.

(0..string1.length).reduce(...) {...}

Regards, Paul

Am 29.05.2014 11:52, schrieb Roelof Wobben:

Hello,

I have a exercise where I must count differences between two strings.

Now I can do

for 0 .. string1.length

But is there a better mor rubisch way to do this ?

Roelof