CODEWITHSUNDEEP
rubystring
Nick Vanderbilt
Nick Vanderbilt

Reputation: 38420

String#count options

From the documentation for String#count I understand the first example, but I do not understand the rest of the examples:

a = "hello world"
a.count "lo"            #=> 5
a.count "lo", "o"       #=> 2
a.count "hello", "^l"   #=> 4
a.count "ej-m"          #=> 4

Any explanation will be helpful.

Upvotes: 51

Views: 57053

Answers (6)

Dyaniyal Wilson
Dyaniyal Wilson

Reputation: 1060

Using gsub in string 

a = "hello world hello hello hello hello world world world"
2.1.5 :195 > a.gsub('hello').count
=> 5

Upvotes: 5

Pete
Pete

Reputation: 18075

I'll take a stab:

Second example: using the wording "The intersection of these sets defines the characters to count in str" the parameters are "lo" and "o". The intersection of these is just "o" of which there are 2 in the string being counted on. Hence the return value of 2.

Third example: This one seems to be saying: "Any of the characters in 'hello' but not the character 'l'". Getting this from the line "Any other_str that starts with a caret (^) is negated". So, you can count the set of letters contained in the string "hello" which are found in "hello world" (i.e., h,e,l,l,o,o,l) but then comparing the intersection with the set of "^l" (i.e. h,e,o,w,o,r,d) you are left with 4 (i.e. h,e,o,o).

Fourth example: This one basically says "count all the 'e' characters and any character between 'j' and 'm'. There is one 'e' and 3 characters contained between 'j' and 'm' and all 3 happen to be the letter 'l' which leaves us with an answer of 4 again.

Upvotes: 2

Jon Kern
Jon Kern

Reputation: 3235

This is one of the dorkiest ruby methods, and pretty lousy documentation to boot. Threw me for a loop. I ended up looking at it because it looked like it should give me the count of occurrences of a given string. Nope. Not remotely close. But here is how I ended up counting string occurrences:

s="this is a string with is thrice"
s.scan(/is/).count  # => 3

Makes me wonder why someone asked for this method, and why the documentation is so lousy. Almost like the person documenting the code really did not have a clue as to the human-understandable "business" reason for asking for this feature.

count([other_str]+) → fixnum

Each _other_str_ parameter defines a set of characters to count. The intersection of these sets defines the characters to count in str. Any _other_str_ that starts with a caret (^) is negated. The sequence c1–c2 means all characters between c1 and c2.

Upvotes: 78

FMc
FMc

Reputation: 42411

Each argument defines a set of characters. The intersection of those sets determines the overall set that count uses to compute a tally.

a = "hello world"

a.count "lo"            # l o       => 5
a.count "lo", "o"       # o         => 2

And ^ can be used for negation (all letters in hello, except l)

a.count "hello", "^l"   # h e o     => 4

Ranges can be defined with -:

a.count "ej-m"          # e j k l m => 4

Upvotes: 9

Russ Cam
Russ Cam

Reputation: 125488

Let's break these down

a = "hello world"

  1. to count the number of occurrences of the letters l and o

    a.count "lo" #=> 5

  2. to find the intersect of lo and o (which is counting the number of occurrences of l and o and taking only the count of o from the occurrences):

    a.count "lo", "o" #=> 2

  3. to count the number of occurrences of h, e, l, l and o, then intersect with any that are not l (which produces the same outcome to finding occurrences of h, e and o)

    a.count "hello", "^l" #=> 4

  4. to count the number of occurrences of e and any letter between j and m (j, k, l and m):

    a.count "ej-m" #=> 4

Upvotes: 10

edmz
edmz

Reputation: 3360

If you pass more than 1 parameter to count, it will use the intersection of those strings and will use that as the search target:

a = "hello world"
a.count "lo"            #=> finds 5 instances of either "l" or "o"
a.count "lo", "o"       #=> the intersection of "lo" and "o" is "o", so it finds 2 instances
a.count "hello", "^l"   #=> the intersection of "hello" and "everything that is not "l" finds 4 instances of either "h", "e" or "o"
a.count "ej-m"          #=> finds 4 instances of "e", "j", "k", "l" or "m" (the "j-m" part)

Upvotes: 37

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