You are given a string S. Count the number of occurrences of all the digits in the string S.
Input:
First line contains string S
Output:
Count the number of occurences of 1 in the entered number and print the output.
I tried to attempt the given problem like this:
number=$stdin.gets.chomp
number.split('').map(&:to_i)
number.each do |numbers|
i==0
while numbers===1
i+=1
end
end
puts i
But, it's not executing.
Can someone please suggest how may I do it?
A clean way to do what you want in ruby
number=$stdin.gets.chomp
number.count('1')
str = "1a2b &32T2*3"
You can count the total number of digits in the string like so:
str.count('0123456789')
#=> 6
If you wish to count the number of occurrences of each digit in the string, you could use a counting hash (see Hash::new):
str.each_char.with_object(Hash.new(0)) { |c,h| h[c] += 1 if c =~ /\d/ }
#=> {"1"=>1, "2"=>3, "3"=>2}
or use the method Enumerable#group_by:
h = str.each_char.group_by(&:itself).select { |k,_| k =~ /\d/ }
#=> {"1"=>["1"], "2"=>["2", "2", "2"], "3"=>["3", "3"]}
h.update(h) { |*,v| v.count }
#=> {"1"=>1, "2"=>3, "3"=>2}
This uses the form of Hash#update (aka merge!) that employs a block to determine the values of keys that are present in both hashes being merged (here all keys). See the doc for details.
There were several problems with your code. First you didn't actually set numbers to be an array, you can do that by saying numbers = numbers.split('').map(&:to_i)
Then you should declare the i variable outside of the loop. Instead of saying i == 0, use i = 0 for the assignment operator
Instead of using a while loop, you should use an if statement.
This code is assuming your string is only all numbers.
numbers = $stdin.gets.chomp
numbers = numbers.split('').map(&:to_i)
i = 0
numbers.each do |number|
if number === 1
i += 1
end
end
If your string consists of letters, numbers or special characters, you could do this
numbers = $stdin.gets.chomp
numbers.split('').select { |c| c=~ /1/ }.length
Related
There is a method that accepts a string as an argument. The string will be formatted so every letter is followed by a number. The method should return an "uncompressed" version of the string where every letter is repeated multiple times given based on the number that appears directly after the letter.
def uncompress(str)
new_str = ""
(0 ... str.length - 1).each do |i|
new_str += str[i] * str[i + 1].to_i
end
new_str
end
Result:
uncompress('a2b4c1') # 'aabbbbc'
Why this code does not return an error if str[i] is a number and str[i+ 1] is a letter?
As you're doing substitution in a string, let's use #gsub and toss a block to it.
irb(main):028:0> "h5a2".gsub(/[a-zA-Z]\d+/) { |x| x[0] * x[1..-1].to_i }
=> "hhhhhaa"
The regex is looking for a single character followed by a (potentially) multiple digit number. It then substitutes that with the character multiplied by the number. (Converting to an int with #to_i is critical as attempting to multiply a string by a string will not work.)
Or if you wish to modify the original string, you can use #gsub!.
irb(main):032:0> s = 'a2b4c1'
irb(main):033:0> s.gsub!(/[a-zA-Z]\d+/) { |x| x[0] * x[1..-1].to_i }
=> "aabbbbc"
irb(main):034:0> s
=> "aabbbbc"
If the format is fixed to a single character followed by a single digit multiplier, then this is easily accommodated by a minor modification to the regex without any change to the block.
irb(main):040:0> s = 'a2b4c156'
irb(main):041:0> s.gsub(/[a-zA-Z0-9]\d/) { |x| x[0] * x[1..-1].to_i }
=> "aabbbbc555555"
Using your approach, it's important that we increment the indices up by two, so let's generate an array of indexes.
irb(main):032:0> str = "h5d3"
irb(main):033:0> (0 ... str.length).step(2).to_a
=> [0, 2]
Then we can loop over those:
def uncompress(str)
new_str = ""
(0 ... str.length).step(2) do |i|
new_str += str[i] * str[i + 1].to_i
end
new_str
end
And now:
irb(main):042:0> uncompress("a4d2b1")
=> "aaaaddb"
Scanning, Splitting, and Integer Coercion
Rather than trying to debug your code, I'll just offer what I consider a simpler and more idiomatic solution:
str = "a2b4c1"
str.scan(/\p{Alpha}\d/).map(&:chars).map { _1 * _2.to_i }
#=> ["aa", "bbbb", "c"]
By scanning for your letter + number pattern and then splitting the matched patterns into an array of characters, you create a collection of array elements that are easy to iterate over. You can easily see the interim steps here by putting a comment character after any part of the method chain to inspect what's being returned.
Lastly, you just need to ensure that the second element of each character array is coerced from a string to an integer, as otherwise you are trying to multiply strings with other strings. For example "a" * "2" is not going to accomplish what you want, while "a" * 2 will.
In Ruby You could easily do it like I have shown below. We don't have to write code like C or other language.
Input
a="a2b4c1"
Program
p a.chars
.slice_after(/\d+/)
.map{|x|x.last.to_i.times.map{x.first}.join}
.join
Output
"aabbbbc"
new to Ruby, new to coding in general...
I'm trying to add new elements into my hash, incrementing the value when necessary. So I used Hash.new(0) and I'm trying to add new values using the "+=" symbol, but when I do this I get an error message -
"/tmp/file.rb:6:in `+': String can't be coerced into Integer (TypeError)
from /tmp/file.rb:6:in `block in stockList'
from /tmp/file.rb:3:in `each'
from /tmp/file.rb:3:in `each_with_index'
from /tmp/file.rb:3:in `stockList'
from /tmp/file.rb:24:in `<main>'
"
Here's my code:
def stockList(stock, cat)
hash = Hash.new(0)
stock.each_with_index do |word, i|
if cat.include?(word[i])
char = word[i]
hash[char] += num(word)
end
end
new_arr = []
hash.each do |k, v|
new_arr.push(k,v)
end
return new_arr
end
def num(word)
nums = "1234567890"
word.each_char.with_index do |char, i|
if nums.include?(char)
return word[i..-1]
end
end
end
puts stockList(["ABAR 200", "CDXE 500", "BKWR 250", "BTSQ 890", "DRTY 600"], ["A", "B"])
Does anyone know why this is happening?
It's a codewars challenge -- I'm basically given two arrays and am meant to return a string that adds the numbers associated with the word that starts with the letter(s) listed in the second array.
For this input I'm meant to return " (A : 200) - (B : 1140) "
Your immediate problem is that num(word) returns a string, and a string can't be added to a number in the line hash[char] += num(word). You can convert the string representation of a numeric value using .to_i or .to_f, as appropriate for the problem.
For the overall problem I think you've added too much complexity. The structure of the problem is:
Create a storage object to tally up the results.
For each string containing a stock and its associated numeric value (price? quantity?), split the string into its two tokens.
If the first character of the stock name is one of the target values,
update the corresponding tally. This will require conversion from string to integer.
Return the final tallies.
One minor improvement is to use a Set for the target values. That reduces the work for checking inclusion from O(number of targets) to O(1). With only two targets, the improvement is negligible, but would be useful if the list of stocks and targets increase beyond small test-case problems.
I've done some renaming to hopefully make things clearer by being more descriptive. Without further ado, here it is in Ruby:
require 'set'
def get_tallies(stocks, prefixes)
targets = Set.new(prefixes) # to speed up .include? check below
tally = Hash.new(0)
stocks.each do |line|
name, amount = line.split(/ +/) # one or more spaces is token delimiter
tally[name[0]] += amount.to_i if targets.include?(name[0]) # note conversion to int
end
tally
end
stock_list = ["ABAR 200", "CDXE 500", "BKWR 250", "BTSQ 890", "DRTY 600"]
prefixes = ["A", "B"]
p get_tallies(stock_list, prefixes)
which prints
{"A"=>200, "B"=>1140}
but that can be formatted however you like.
The particular issue triggering this error is that your def num(word) is essentially a no-op, returning the word without any change.
But you actually don't need this function: this...
word.delete('^0-9').to_i
... gives you back the word with all non-digit characters stripped, cast to integer.
Note that without to_i you'll still receive the "String can't be coerced into Integer" error: Ruby is not as forgiving as JavaScript, and tries to protect you from results that might surprise you.
It's a codewars challenge -- I'm basically given two arrays and am
meant to return a string that adds the numbers associated with the
word that starts with the letter(s) listed in the second array.
For this input I'm meant to return " (A : 200) - (B : 1140) "
This is one way to get there:
def stockList(stock, cat)
hash = Hash.new(0)
stock.each do |word|
letter = word[0]
if cat.include?(letter)
hash[letter] += word.delete('^0-9').to_i
end
end
hash.map { |k, v| "#{k}: #{v}" }
end
Besides type casting, there's another difference here: always choosing the initial letter of the word. With your code...
stock.each_with_index do |word, i|
if cat.include?(word[i])
char = word[i]
... you actually took the 1st letter of the 1st ticker, the 2nd letter of the 2nd ticker and so on. Don't use indexes unless your results depend on them.
stock = ["ABAR 200", "CDXE 500", "BKWR 250", "BTSQ 890", "DRTY 600"]
cat = ["A", "B"]
I concur with your decision to create a hash h with the form of Hash::new that takes an argument (the "default value") which h[k] returns when h does not have a key k. As a first step we can write:
h = stock.each_with_object(Hash.new(0)) { |s,h| h[s[0]] += s[/\d+/].to_i }
#=> {"A"=>200, "C"=>500, "B"=>1140, "D"=>600}
Then Hash#slice can be used to extract the desired key-value pairs:
h = h.slice(*cat)
#=> {"A"=>200, "B"=>1140}
At this point you have all the information you need to display the result any way you like. For example,
" " << h.map { |k,v| "(#{k} : #{v})" }.join(" - ") << " "
#=> " (A : 200) - (B : 1140) "
If h before h.slice(*cat) is large relative to h.slice(*cat) you can reduce memory requirements and probably speed things somewhat by writing the following.
require 'set'
cat_set = cat.to_set
#=> #<Set: {"A", "B"}>
h = stock.each_with_object(Hash.new(0)) do |s,h|
h[s[0]] += s[/\d+/].to_i if cat_set.include?(s[0])
end
#=> {"A"=>200, "B"=>1140}
Let's say I have a string, like string= "aasmflathesorcerersnstonedksaottersapldrrysaahf". If you haven't noticed, you can find the phrase "harry potter and the sorcerers stone" in there (minus the space).
I need to check whether string contains all the elements of the string.
string.include? ("sorcerer") #=> true
string.include? ("harrypotterandtheasorcerersstone") #=> false, even though it contains all the letters to spell harrypotterandthesorcerersstone
Include does not work on shuffled string.
How can I check if a string contains all the elements of another string?
Sets and array intersection don't account for repeated chars, but a histogram / frequency counter does:
require 'facets'
s1 = "aasmflathesorcerersnstonedksaottersapldrrysaahf"
s2 = "harrypotterandtheasorcerersstone"
freq1 = s1.chars.frequency
freq2 = s2.chars.frequency
freq2.all? { |char2, count2| freq1[char2] >= count2 }
#=> true
Write your own Array#frequency if you don't want to the facets dependency.
class Array
def frequency
Hash.new(0).tap { |counts| each { |v| counts[v] += 1 } }
end
end
I presume that if the string to be checked is "sorcerer", string must include, for example, three "r"'s. If so you could use the method Array#difference, which I've proposed be added to the Ruby core.
class Array
def difference(other)
h = other.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
reject { |e| h[e] > 0 && h[e] -= 1 }
end
end
str = "aasmflathesorcerersnstonedksaottersapldrrysaahf"
target = "sorcerer"
target.chars.difference(str.chars).empty?
#=> true
target = "harrypotterandtheasorcerersstone"
target.chars.difference(str.chars).empty?
#=> true
If the characters of target must not only be in str, but must be in the same order, we could write:
target = "sorcerer"
r = Regexp.new "#{ target.chars.join "\.*" }"
#=> /s.*o.*r.*c.*e.*r.*e.*r/
str =~ r
#=> 2 (truthy)
(or !!(str =~ r) #=> true)
target = "harrypotterandtheasorcerersstone"
r = Regexp.new "#{ target.chars.join "\.*" }"
#=> /h.*a.*r.*r.*y* ... o.*n.*e/
str =~ r
#=> nil
A different albeit not necessarily better solution using sorted character arrays and sub-strings:
Given your two strings...
subject = "aasmflathesorcerersnstonedksaottersapldrrysaahf"
search = "harrypotterandthesorcerersstone"
You can sort your subject string using .chars.sort.join...
subject = subject.chars.sort.join # => "aaaaaaacddeeeeeffhhkllmnnoooprrrrrrssssssstttty"
And then produce a list of substrings to search for:
search = search.chars.group_by(&:itself).values.map(&:join)
# => ["hh", "aa", "rrrrrr", "y", "p", "ooo", "tttt", "eeeee", "nn", "d", "sss", "c"]
You could alternatively produce the same set of substrings using this method
search = search.chars.sort.join.scan(/((.)\2*)/).map(&:first)
And then simply check whether every search sub-string appears within the sorted subject string:
search.all? { |c| subject[c] }
Create a 2 dimensional array out of your string letter bank, to associate the count of letters to each letter.
Create a 2 dimensional array out of the harry potter string in the same way.
Loop through both and do comparisons.
I have no experience in Ruby but this is how I would start to tackle it in the language I know most, which is Java.
I am trying to change numbers up to 100 from integers into words, but have run into some trouble, can anyone point out what is missing with my code:
def in_words(integer)
numWords = {
0=>"zero",
1=>"one",
2=>"two",
3=>"three",
4=>"four",
5=>"five",
6=>"six",
7=>"seven",
8=>"eight",
9=>"nine",
10=>"ten",
11=>"eleven",
12=>"twelve",
13=>"thirteen",
14=>"fourteen",
15=>"fifteen",
16=>"sixteen",
17=>"seventeen",
18=>"eighteen",
19=>"nineteen",
20=>"twenty",
30=>"thirty",
40=>"fourty",
50=>"fifty",
60=>"sixty",
70=>"seventy",
80=>"eighty",
90=>"ninety",
100=>"one hundred"
}
array = integer.to_s.split('')
new_array = []
numWords.each do |k,v|
array.each do |x|
if x = k
new_array.push(v)
end
end
end
new_array.join('')
end
Right now when I do:
inwords(0)
I get the following:
=>"zeroonetwothreefourfivesixseveneightnineteneleventwelvethirteenfourteenfiftee nsixteenseventeeneighteennineteentwentythirtyfourtyfiftysixtyseventyeightyninetyone hundred"
Edit
I noticed your code iterates through the array a lot of times and uses the = instead of the == in your if statements.
Your code could be more efficient using the Hash's #[] method in combination with the #map method.., here's a one-line alternative:
integer.to_s.split('').map {|i| numWords[i.to_i]} .join ' '
Also, notice that the integer.to_s.split('') will split the array into one-digit strings, so having numbers up to a hundred isn't relevant for the code I proposed.
To use all the numbers in the Hash, you might want to use a Regexp to identify the numbers you have. One way is to do the following (I write it in one line, but it's easy to break it down using variable names for each step):
integer.to_s.gsub(/(\d0)|([1]?\d)/) {|v| v + " "} .split.map {|i| numWords[i.to_i]} .join ' '
# or:
integer.to_s.gsub(/(#{numWords.keys.reverse.join('|')})/) {|v| v + " "} .split.map {|i| numWords[i.to_i]} .join ' '
# out = integer.to_s
# out = out.gsub(/(#{numWords.keys.reverse.join('|')})/) {|v| v + " "}
# out = out.split
# out = out.map {|i| numWords[i.to_i]}
# out = out.join ' '
Edit 2
Since you now mention that you want the method to accept numbers up to a hundred and return the actual number (23 => twenty three), maybe a different approach should be taken... I would recommend that you update your question as well.
def in_words(integer)
numWords = {
0=>"zero",
1=>"one",
2=>"two",
3=>"three",
4=>"four",
5=>"five",
6=>"six",
7=>"seven",
8=>"eight",
9=>"nine",
10=>"ten",
11=>"eleven",
12=>"twelve",
13=>"thirteen",
14=>"fourteen",
15=>"fifteen",
16=>"sixteen",
17=>"seventeen",
18=>"eighteen",
19=>"nineteen",
20=>"twenty",
30=>"thirty",
40=>"fourty",
50=>"fifty",
60=>"sixty",
70=>"seventy",
80=>"eighty",
90=>"ninety",
100=>"one hundred"
}
raise "cannot accept such large numbers" if integer > 100
raise "cannot accept such small numbers" if integer < 0
return "one hundred" if integer == 100
if integer < 20 || integer %10 == 0
numWords[integer]
else
[numWords[integer / 10 * 10], numWords[integer % 10]].join ' '
end
end
the integer / 10 * 10 makes the number a round number (ten, twenty, etc') because integers don't have fractions (so, 23/10 == 2 and 2 * 10 == 20). The same could be achieved using integer.round(-1), which is probably better.
It seems like all you're trying to do is find a mapping from an implicit hash
module NumWords
INT2STR = {
0=>"zero",
1=>"one",
2=>"two",
3=>"three",
4=>"four",
5=>"five",
6=>"six",
7=>"seven",
8=>"eight",
9=>"nine",
10=>"ten",
11=>"eleven",
12=>"twelve",
13=>"thirteen",
14=>"fourteen",
15=>"fifteen",
16=>"sixteen",
17=>"seventeen",
18=>"eighteen",
19=>"nineteen",
20=>"twenty",
30=>"thirty",
40=>"fourty",
50=>"fifty",
60=>"sixty",
70=>"seventy",
80=>"eighty",
90=>"ninety",
100=>"one hundred"
}
module_function
def in_words(integer)
INT2STR[integer]
end
end
The above code separates the hash definition from the method call so that the hash doesn't get recreated every time you call in_words.
You can also use Hash#fetch instead of Hash#[] as Andrey pointed out.
Your test whether x = k is your first problem (in two ways).
Firstly, if x = k means assign the value of k to x and then execute the if block if that value is true (basically anything other than false or nil).
What you should actually be testing is x == k which will return true if x is equal to k.
The second problem is that you converted your number into an array of string representation so you are comparing, for example, if "0" == 0. This won't return true because they are different types.
If you convert it to if x.to_i == k then your if block will be executed and you'll get:
> in_words(0)
=> "zero"
Then you get to move onto the next problem which is that you're looking at your number digit by digit and some of the values you are testing against need two digits to be recognised:
> in_words(10)
=> "zeroone"
You might be in looking at a different question then - or maybe that is the question you wanted answered all along!
Here's another way you might do it:
ONES_TO_TEXT = { 0=>"zero", 1=>"one", 2=>"two", 3=>"three", 4=>"four",
5=>"five", 6=>"six", 7=>"seven", 8=>"eight", 9=>"nine" }
TEENS_TO_TEXT = { 10=>"ten", 11=>"eleven", 12=>"twelve",
13=>"thirteen", 15=>"fifteen" }
TENS_TO_TEXT = { 2=>"twenty", 3=>"thirty", 5=>"fifty", 8=>"eighty" }
def in_words(n)
raise ArgumentError, "#{n} is out-of_range" unless (0..100).cover?(n)
case n.to_s.size
when 1 then ONES_TO_TEXT[n]
when 3 then "one hundred"
else
case n
when (10..19)
TEENS_TO_TEXT.key?(n) ? TEENS_TO_TEXT[n] : ONES_TO_TEXT[n]+"teen"
else
t,o = n.divmod(10)
(TENS_TO_TEXT.key?(t) ? TENS_TO_TEXT[t] : ONES_TO_TEXT[t]+"ty") +
(o.zero? ? '' : "-#{ONES_TO_TEXT[o]}")
end
end
end
Let's try it:
in_words(5) #=> "five"
in_words(10) #=> "ten"
in_words(15) #=> "fifteen"
in_words(20) #=> "twenty"
in_words(22) #=> "twenty-two"
in_words(30) #=> "thirty"
in_words(40) #=> "fourty"
in_words(45) #=> "fourty-five"
in_words(50) #=> "fifty"
in_words(80) #=> "eighty"
in_words(99) #=> "ninety-nine"
in_words(100) #=> "one hundred"
Here the increased complexity may not be justified, but this approach may in fact simplify the calculations when the maximum permitted value of n is much greater than 100.
I need to know if all letters in a string are unique. For a string to be unique, a letter can only appear once. If all letters in a string are distinct, the string is unique. If one letter appears multiple times, the string is not unique.
"Cwm fjord veg balks nth pyx quiz."
# => All 26 letters are used only once. This is unique
"This is a string"
# => Not unique, i and s are used more than once
"two"
# => unique, each letter is shown only once
I tried writing a function that determines whether or not a string is unique.
def unique_characters(string)
for i in ('a'..'z')
if string.count(i) > 1
puts "This string is unique"
else
puts "This string is not unique"
end
end
unique_characters("String")
I receive the output
"This string is unique" 26 times.
Edit:
I would like to humbly apologize for including an incorrect example in my OP. I did some research, trying to find pangrams, and assumed that they would only contain 26 letters. I would also like to thank you guys for pointing out my error. After that, I went on wikipedia to find a perfect pangram (I wrongly thought the others were perfect).
Here is the link for reference purposes
http://en.wikipedia.org/wiki/List_of_pangrams#Perfect_pangrams_in_English_.2826_letters.29
Once again, my apologies.
s = "The quick brown fox jumps over the lazy dog."
.downcase
("a".."z").all?{|c| s.count(c) <= 1}
# => false
Another way to do it is:
s = "The quick brown fox jumps over the lazy dog."
(s.downcase !~ /([a-z]).*\1/)
# => false
I would solve this in two steps: 1) extract the letters 2) check if there are duplicates:
letters = string.scan(/[a-z]/i) # append .downcase to ignore case
letters.length == letters.uniq.length
Here is a method that does not convert the string to an array:
def dupless?(str)
str.downcase.each_char.with_object('') { |c,s|
c =~ /[a-z]/ && s.include?(c) ? (return false) : s << c }
true
end
dupless?("Cwm fjord veg balks nth pyx quiz.") #=> true
dupless?("This is a string.") #=> false
dupless?("two") #=> true
dupless?("Two tubs") #=> false
If you want to actually keep track of the duplicate characters:
def is_unique?(string)
# Remove whitespaces
string = string.gsub(/\s+/, "")
# Build a hash counting all occurences of each characters
h = Hash.new { |hash, key| hash[key] = 0 }
string.chars.each { |c| h[c] += 1 }
# An array containing all the repetitions
res = h.keep_if {|k, c| c > 1}.keys
if res.size == 0
puts "All #{string.size} characters are used only once. This is unique"
else
puts "Not unique #{res.join(', ')} are used more than once"
end
end
is_unique?("This is a string") # Not unique i, s are used more than once
is_unique?("two") # All 3 characters are used only once. This is unique
To check if a string is unique or not, you can try out this:
string_input.downcase.gsub(/[^a-z]/, '').split("").sort.join('') == ('a' .. 'z').to_a.join('')
This will return true, if all the characters in your string are unique and if they include all the 26 characters.
def has_uniq_letters?(str)
letters = str.gsub(/[^A-Za-z]/, '').chars
letters == letters.uniq
end
If this doesn't have to be case sensitive,
def has_uniq_letters?(str)
letters = str.downcase.gsub(/[^a-z]/, '').chars
letters == letters.uniq
end
In your example, you mentioned you wanted additional information about your string (list of unique characters, etc), so this example may also be useful to you.
# s = "Cwm fjord veg balks nth pyx quiz."
s = "This is a test string."
totals = Hash.new(0)
s.downcase.each_char { |c| totals[c] += 1 if ('a'..'z').cover?(c) }
duplicates, uniques = totals.partition { |k, v| v > 1 }
duplicates, uniques = Hash[duplicates], Hash[uniques]
# duplicates = {"t"=>4, "i"=>3, "s"=>4}
# uniques = {"h"=>1, "a"=>1, "e"=>1, "r"=>1, "n"=>1, "g"=>1}