Suppose I have the following code:
#!/usr/bin/env ruby -wKU
h = {}
h[[1, "a"]] = "first"
h[[2, "b"]] = "second"
puts h.to_yaml
# case 1 - works fine
h.each do |k, v|
num, char = k
puts "key = #{[num, char]}; value = #{v}"
end
# case 2 - works fine
h.each_key do | num, char |
puts "key = #{[num, char]}; value = #{h[[num, char]]}"
end
# case 3 - Doesn't work
# how can I get all three values in one go?
h.each do | [num, char], v |
puts "key = #{[num, char]}; value = #{v}"
end
How would I create an iterator where I could get all 3 values (key[0], key[1], value) assigned in the block parameters? Is this even possible?
h.each do | (num, char), v |
puts "key = #{[num, char]}; value = #{v}"
end
Related
If I have a string like this
str =<<END
7312357006,1.121
3214058234,3456
7312357006,1234
1324958723,232.1
3214058234,43.2
3214173443,234.1
6134513494,23.2
7312357006,11.1
END
If a number in the first value shows up again, I want to add their second values together. So the final string would look like this
7312357006,1246.221
3214058234,3499.2
1324958723,232.1
3214173443,234.1
6134513494,23.2
If the final output is an array that's fine too.
There are lots of ways to do this in Ruby. One particularly terse way is to use String#scan:
str = <<END
7312357006,1.121
3214058234,3456
7312357006,1234
1324958723,232.1
3214058234,43.2
3214173443,234.1
6134513494,23.2
7312357006,11.1
END
data = Hash.new(0)
str.scan(/(\d+),([\d.]+)/) {|k,v| data[k] += v.to_f }
p data
# => { "7312357006" => 1246.221,
# "3214058234" => 3499.2,
# "1324958723" => 232.1,
# "3214173443" => 234.1,
# "6134513494" => 23.2 }
This uses the regular expression /(\d+),([\d.]+)/ to extract the two values from each line. The block is called with each pair as arguments, which are then merged into the hash.
This could also be written as a single expression using each_with_object:
data = str.scan(/(\d+),([\d.]+)/)
.each_with_object(Hash.new(0)) {|(k,v), hsh| hsh[k] += v.to_f }
# => (same as above)
There are likewise many ways to print the result, but here are a couple I like:
puts data.map {|kv| kv.join(",") }.join("\n")
# => 7312357006,1246.221
# 3214058234,3499.2
# 1324958723,232.1
# 3214173443,234.1
# 6134513494,23.2
# or:
puts data.map {|k,v| "#{k},#{v}\n" }.join
# => (same as above)
You can see all of these in action on repl.it.
Edit: Although I don't recommend either of these for the sake of readability, here's more just for kicks (requires Ruby 2.4+):
data = str.lines.group_by {|s| s.slice!(/(\d+),/); $1 }
.transform_values {|a| a.sum(&:to_f) }
...or, to going straight to a string:
puts str.lines.group_by {|s| s.slice!(/(\d+),/); $1 }
.map {|k,vs| "#{k},#{vs.sum(&:to_f)}\n" }.join
Since repl.it is stuck on Ruby 2.3: Try it online!
You could achieve this using each_with_object, as below:
str = "7312357006,1.121
3214058234,3456
7312357006,1234
1324958723,232.1
3214058234,43.2
3214173443,234.1
6134513494,23.2
7312357006,11.1"
# convert the string into nested pairs of floats
# to briefly summarise the steps: split entries by newline, strip whitespace, split by comma, convert to floats
arr = str.split("\n").map(&:strip).map { |el| el.split(",").map(&:to_f) }
result = arr.each_with_object(Hash.new(0)) do |el, hash|
hash[el.first] += el.last
end
# => {7312357006.0=>1246.221, 3214058234.0=>3499.2, 1324958723.0=>232.1, 3214173443.0=>234.1, 6134513494.0=>23.2}
# You can then call `to_a` on result if you want:
result.to_a
# => [[7312357006.0, 1246.221], [3214058234.0, 3499.2], [1324958723.0, 232.1], [3214173443.0, 234.1], [6134513494.0, 23.2]]
each_with_object iterates through each pair of data, providing them with access to an accumulator (in this the hash). By following this approach, we can add each entry to the hash, and add together the totals if they appear more than once.
Hope that helps - let me know if you've any questions.
def combine(str)
str.each_line.with_object(Hash.new(0)) do |s,h|
k,v = s.split(',')
h.update(k=>v.to_f) { |k,o,n| o+n }
end.reduce('') { |s,kv_pair| s << "%s,%g\n" % kv_pair }
end
puts combine str
7312357006,1246.22
3214058234,3499.2
1324958723,232.1
3214173443,234.1
6134513494,23.2
Notes:
using String#each_line is preferable to str.split("\n") as the former returns an enumerator whereas the latter returns a temporary array. Each element generated by the enumerator is line of str that (unlike the elements of str.split("\n")) ends with a newline character, but that is of no concern.
see Hash::new, specifically when a default value (here 0) is used. If a hash has been defined h = Hash.new(0) and h does not have a key k, h[k] returns the default value, zero (h is not changed). When Ruby encounters the expression h[k] += 1, the first thing she does is expand it to h[k] = h[k] + 1. If h has been defined with a default value of zero, and h does not have a key k, h[k] on the right of the equality (syntactic sugar1 for h.[](k)) returns zero.
see Hash#update (aka merge!). h.update(k=>v.to_f) is syntactic sugar for h.update({ k=>v.to_f })
see Kernel#sprint for explanations of the formatting directives %s and %g.
the receiver for the expression reduce('') { |s,kv_pair| s << "%s,%g\n" % kv_pair } (in the penultimate line), is the following hash.
{"7312357006"=>1246.221, "3214058234"=>3499.2, "1324958723"=>232.1,
"3214173443"=>234.1, "6134513494"=>23.2}
1 Syntactic sugar is a shortcut allowed by Ruby.
Implemented this solution as hash was giving me issues:
d = []
s.split("\n").each do |line|
x = 0
q = 0
dup = false
line.split(",").each do |data|
if x == 0 and d.include? data then dup = true ; q = d.index(data) elsif x == 0 then d << data end
if x == 1 and dup == false then d << data end
if x == 1 and dup == true then d[q+1] = "#{'%.2f' % (d[q+1].to_f + data.to_f).to_s}" end
if x == 2 and dup == false then d << data end
x += 1
end
end
x = 0
s = ""
d.each do |val|
if x == 0 then s << "#{val}," end
if x == 1 then s << "#{val}\n ; x = 0" end
x += 1
end
puts(s)
I'm writing a program which takes input, stores it as a hash and sorts the values.
I'm having trouble comparing a current hash value with a variable.
Sample Input:
3
A 1
B 3
C 5
A 2
B 7
C 2
Sample Output:
A 1 2
B 3 7
C 2 5
Everything works apart from this part, and I'm unsure why.
if values.key?(:keys)
if values[keys] >= val
values.store(keys,val.prepend(val + " "))
else
values.store(keys,val.concat(" " + val))
end
else
values.store(keys,val)
end
i = i + 1
end
Rest of code:
#get amount of records
size = gets.chomp
puts size
size = size.to_i
values = Hash.new(0)
i = 0
while i < (size * 2)
text = gets.chomp
#split string and remove space
keys = text.split[0]
val = text.split[1]
#check if key already exists,
# if current value is greater than new value append new value to end
# else put at beginning of current value
if values.key?(:keys)
if values[keys] >= val
values.store(keys,val.prepend(val + " "))
else
values.store(keys,val.concat(" " + val))
end
else
values.store(keys,val)
end
i = i + 1
end
#sort hash by key
values = values.sort_by { |key, value| key}
#output hash values
values.each{|key, value|
puts "#{key}:#{value}"
}
Could anyone help me out? It would be most appreciated.
The short answer is that there are two mistakes in your code. Here is the fixed version:
if values.key?(keys)
if values[keys] >= val
values.store(keys,values[keys].prepend(val + " "))
else
values.store(keys,values[keys].concat(" " + val))
end
else
values.store(keys,val)
end
The if statement was always evaluating as false, because you were looking for hash key named :keys (which is a Symbol), not the variable you've declared named keys.
Even with that fixed, there was a second hidden bug: You were storing a incorrect new hash value. val.concat(" " + val) would give you results like A 2 2, not A 1 2, since it's using the new value twice, not the original value.
With that said, you code is still very confusing to read... Your variables are size, i, text, val, values, key and keys. It would have been a lot easier to understand with clearer variable names, if nothing else :)
Here is a slightly improved version, without changing the overall structure of your code:
puts "How may variables to loop through?"
result_length = gets.chomp.to_i
result = {}
puts "Enter #{result_length * 2} key-value pairs:"
(result_length * 2).times do
input = gets.chomp
input_key = input.split[0]
input_value = input.split[1]
#check if key already exists,
# if current value is greater than new value append new value to end
# else put at beginning of current value
if result.key?(input_key)
if result[input_key] >= input_value
result[input_key] = "#{input_value} #{result[input_key]}"
else
result[input_key] = "#{result[input_key]} #{input_value}"
end
else
result[input_key] = input_value
end
end
#sort hash by key
result.sort.to_h
#output hash result
result.each{|key, value|
puts "#{key}:#{value}"
}
h = Hash.new { |h,k| h[k] = [] }
input = ['A 1', 'B 3', 'C 5', 'A 2', 'B 7', 'C 2'].join("\n")
input.each_line { |x| h[$1] << $2 if x =~ /^(.*?)\s+(.*?)$/ }
h.keys.sort.each do |k|
puts ([k] + h[k].sort).join(' ')
end
# A 1 2
# B 3 7
# C 2 5
This would be a more Ruby-ish way to write your code :
input = "A 1
B 3
C 5
A 2
B 7
C 2"
input.scan(/[A-Z]+ \d+/)
.map{ |str| str.split(' ') }
.group_by{ |letter, _| letter }
.each do |letter, pairs|
print letter
print ' '
puts pairs.map{ |_, number| number }.sort.join(' ')
end
#=>
# A 1 2
# B 3 7
# C 2 5
I need to compare 2 hashes in ruby where one of the hash contains the numeric values in quoted string, which makes that a string. Consider the following 2 hashes:
hash1 = {"A"=>"0", "B"=>"1", "SVHTID"=>"VH", "D"=>"0", "E"=>"19930730", "F"=>"TEST - DEPOSIT", "G"=>"2.25000000"}
hash2 = {"a"=>"0", "b"=>1, "c"=>"VH", "d"=>0,"e"=>19930730, "f"=>"TEST - DEPOSIT", "g"=>2.25}
Now the code i have written so far is as follows:
hash2 = Hash[hash2.map {|key, value| [key.upcase, value] }]
hash1.each{|k,v| hash1[k] = hash1[k].to_i if hash1[k].match(/-?\d+(?:\.\d+)?/)}
hash1.keys.select { |key| hash1[key] != hash2[key] }.each { |key|
puts "expected #{key} => #{hash1[key]}, but found #{key} => #{hash2[key]}"
}
what is does is that it also converts the float value to integer and the original value is lost.
What i want is that when the above 2 hashes are compared the output should contain only G as mismatched type and following should be printed:
Expected: G=>2.25000000 but found G=>2.25
# normalization
h1, h2 = [hash1, hash2].map do |h|
h.map do |k, v|
[k.to_s.downcase, v.to_s.to_i.to_s == v.to_s ? v.to_i : v]
end.to_h
end
Now we are ready to compare:
h1.reject { |k, v| h2[k].nil? || h2[k] == v }
#⇒ { "g" => "2.25000000" }
This might be printed, formatted etc as you want.
I have an Array-1 say
arr1 =['s','a','sd','few','asdw','a','sdfeg']
And a second Array
arr2 = ['s','a','d','f','w']
I want to take arr1 and sort the frequency of letters by inputting arr2 with result
[s=> 4, a=> 2, d => 3] So on and so forth.
As far as I can muddle around.. Nothing below works, Just my thoughts on it?
hashy = Hash.new
print "give me a sentance "
sentance = gets.chomp.downcase.delete!(' ')
bing = sentance.split(//)
#how = sentance.gsub!(/[^a-z)]/, "") #Remove nil result
#chop = how.to_s.split(//).uniq
#hashy << bing.each{|e| how[e] }
#puts how.any? {|e| bing.count(e)}
#puts how, chop
bing.each {|v| hashy.store(v, hashy[v]+1 )}
puts bing
Thank you for your time.
I assumed that you want to count all letters in the sentence you put in, and not array 1. Assuming that, here's my take on it:
hashy = Hash.new()
['s','a','d','f','w'].each {|item| hashy[item.to_sym] = 0}
puts "give me a sentence"
sentence = gets.chomp.downcase.delete!(' ')
sentence_array = []
sentence.each_char do |l|
sentence_array.push(l)
end
hashy.each do |key, value|
puts "this is key: #{key} and value #{hashy[key]}"
sentence_array.each do |letter|
puts "letter: #{letter}"
if letter.to_sym == key
puts "letter #{letter} equals key #{key}"
value = value + 1
hashy[key] = value
puts "value is now #{value}"
end
end
end
puts hashy
How can I have a hash of hash of hash?
My test returns
undefined method `[]' for nil:NilClass (NoMethodError)
Any tips?
found = Hash.new()
x = 1;
while x < 4 do
found[x] = Hash.new()
y = 1
while y < 4 do
found[x][y] = Hash.new()
found[x][y]['name1'] = 'abc1'
found[x][y]['name2'] = 'abc2'
found[x][y]['name3'] = 'abc3'
y += 1
end
x += 1
end
found.each do |k, v, y|
puts "k : #{k}"
puts " : #{v[y['name1']]}"
puts " : #{v[y['name2']]}"
puts " : #{v[y['name3']]}"
puts
end
I think you want something like this:
First of all create the data structure. You want nested hashes so you need to define default values for each hash key.
found = Hash.new do |hash,key|
hash[key] = Hash.new do |hash,key|
hash[key] = Hash.new
end
end
Run the search
(1..3).each do |x|
(1..3).each do |y|
found[x][y]['name1'] = 'abc1'
found[x][y]['name2'] = 'abc1'
found[x][y]['name3'] = 'abc1'
end
end
Then display the results
found.each do |x, y_hash|
y_hash.each do |y, name_hash|
name_hash.each do |name, value|
puts "#{x} => #{y} => #{name} => #{value}"
end
end
end
The way you build the hash seems to be functional. What probably causes the error is this loop:
found.each do |k, v, y|
Hash#each yields key/value pairs, so y will be assigned nil, thus causing the error two lines below. What you probably meant is a nested loop like
found.each do |x, h1|
h1.each do |y, h2|
puts h2['name1']
end
end
You should also be aware that you can write these kinds of counting loops more concisely in Ruby:
found = Hash.new { |h,k| h[k] = {} }
1.upto(3) do |x|
1.upto(3) do |y|
found[x][y] = {
'name1' => 'abc1',
'name2' => 'abc2',
'name3' => 'abc3',
}
end
end