I am trying to print a ruby hash:
opts = {
'one' => '1',
'two' => '1',
'three' => '0'
}
I want the output to be
one=1
two=1
three=0
This works fine with this code on one machine which runs ruby 1.8.7
print opts.map{|k,v| k + '=' + v + "\n"}.to_s
But on a different machine which runs ruby 1.9, it prints
["one=1\n", "two=1\n", "three=0\n"]
What is going wrong?
Try
print opts.map{|k,v| k + '=' + v + "\n"}.join
The explanation is easy: With ruby 1.9 Array.to_s changed its behaviour.
An alternative:
puts opts.map{|k,v| k + '=' + v }.join("\n")
or
puts opts.map{|k,v| "#{k}=#{v}" }.join("\n")
I would prefer:
opts.each{|k,v| puts "#{k}=#{v}" }
And another version, but with another look:
opts.each{|k,v| puts "%-10s= %s" % [k,v]}
The result is:
one = 1
two = 1
three = 0
(But the keys should be not longer then the length in %-10s.)
It's working as expected. Give this a try:
a={:one=>1, :two=>2, :three=>3}
a.each {|k,v| puts "#{k}=>#{v}" }
Try:
res = ""
opts.map{|k,v| res += k + '=' + v + "\n"}
puts res
Related
Based on the link
I tried to delete "" in the array on ruby
However still not get what I want, if anyone knows, please advice me
a = gets
lines = []
aaa = []
b = []
bb =[]
while line = gets do
lines << line.chomp.split(' ')
end
for k in 0..(lines.size - 1) do
b << lines[k][1].to_i + 1
end
for i in 0..(lines.size - 1)do
bb << lines[i][0] + ' ' + b[i].to_s
end
for l in 0..(lines.size - 1)do
p bb[l]
end
Input
3
Tanaka 18
Sato 50
Suzuki 120
Output
[["Tanaka", "18"], ["Sato", "50"], ["Suzuki", "120"]]
"Tanaka 19"
"Tanaka 19"
"Sato 51"
"Suzuki 121"
As pointed out in the comments, you can get rid of the quotation marks by replacing p (Ruby's inspect/print) with puts.
While we're at it, you can make this much more "Ruby-ish" by using .readlines to scoop up all the input into an array, and by replacing the multiple counting loops with .map or .each iterators. The following is more concise, and allows you to lose the first input line which you're just throwing away anyway.
lines = STDIN.readlines(chomp: true).map do |line|
l = line.split(' ')
[l[0], l[1].to_i + 1].join(' ')
# or
# "#{l[0]} #{l[1].to_i + 1}"
end
lines.each { |line| puts line }
With Ruby 3, you can use rightward-assignment for the first part if you find it more readable:
STDIN.readlines(chomp: true).map do |line|
l = line.split(' ')
"#{l[0]} #{l[1].to_i + 1}"
end => lines
# Character Counter
class String
def count_lcases
count(('a'..'z').to_a.join(''))
end
def count_upcases
count(('A'..'Z').to_a.join(''))
end
def count_num
count((0..9).to_a.join(''))
end
def count_spl_chars
length - count_lcases - count_upcases - count_num
end
end
input = ARGV[0]
if ARGV.empty?
puts 'Please provide an input'
exit
end
puts 'Lowercase characters = %d' % [input.count_lcases]
puts 'Uppercase characters = %d' % [input.count_upcases]
puts 'Numeric characters = %d' % [input.count_num]
puts 'Special characters = %d' % [input.count_spl_chars]
I used ranges to count characters but count function is called 3 times.
I can always use loops and count it one by one.I was wondering is there any way to optimize this?...
If you are using Ruby 2.7 you could use tally; the string's chars are just iterated one time.
def classify_char(c)
case c
when /[a-z]/ then :lcase
when /[A-Z]/ then :ucase
when /\d/ then :digit
else :other
end
end
p "asg3456 ERTYaeth".chars.map{|c| classify_char(c) }.tally
# => {:lcase=>7, :digit=>4, :other=>2, :ucase=>4}
If Ruby 2.3...2.7, this will work:
CHAR_CLASSES = {
lcase: ?a..?z,
ucase: ?A..?Z,
digit: ?0..?9,
}
p "asg3456 ERTYaeth".each_char.with_object(Hash.new(0)) { |c, o|
o[CHAR_CLASSES.find { |label, group| group === c }&.first || :other] += 1
}
For < 2.3,
p "asg3456 ERTYaeth".each_char.with_object(Hash.new(0)) { |c, o|
p = CHAR_CLASSES.find { |label, group| group === c }
o[p ? p.first : :other] += 1
}
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
Given a (multiline) string, where each line is separated by "\n" and may not be necessarily of the same length, what is the best way to transpose it into another string as follows? Lines shorter than the longest one should be padded with space (right padding in terms of the original, or bottom padding in terms of the output). Applying the operation on a string twice should be idempotent modulo padding.
Input string
abc
def ghi
jk lm no
Output string
adj
bek
cf
l
gm
h
in
o
Here are five approaches. (Yes, I got a bit carried away, but I find that trying to think of different ways to accomplish the same task is good exercise for the grey cells.)
#1
An uninteresting, brute-force method:
a = str.split("\n")
l = a.max_by(&:size).size
puts a.map { |b| b.ljust(l).chars }
.transpose
.map { |c| c.join.rstrip }.join("\n")
adj
bek
cf
l
gm
h
in
o
#2
This method and all that follow avoid the use of ljust and transpose, and make use of the fact that if e is an empty array, e.shift returns nil and leaves e an empty array. (Aside: I am often reaching for the non-existent method String#shift. Here it would have avoided the need to convert each line to an array of characters.)
a = str.split("\n").map(&:chars)
a.max_by(&:size).size.times.map { a.map { |e| e.shift || ' ' }.join.rstrip }
#3
This and the remaining methods avoid the need to compute the length of the longest string:
a = str.split("\n").map(&:chars)
a_empty = Array(a.size, [])
[].tap { |b| b << a.map { |e| e.shift || ' ' }.join.rstrip while a != a_empty }
#4
This method makes use of Enumerator#lazy, which has been available since v2.0.
a = str.split("\n").map(&:chars)
(0..Float::INFINITY).lazy.map do |i|
a.each { |e| e.shift } if i > 0
a.map { |e| e.first || ' ' }.join.rstrip
end.take_while { c = a.any? { |e| !e.empty? } }.to_a
(I initially had a problem getting this to work, as I was not getting the element of the output (" o"). The fix was adding the third line and changing the line that follows from a.map { |e| e.shift || ' ' }.join.rstrip to what I have now. I mention this because it seems like it may be common problem when using lazy.)
#5
Lastly, use recursion:
def recurse(a, b=[])
return b[0..-2] if a.last.empty?
b << a.map { |e| e.shift || ' ' }.join.rstrip
recurse(a, b)
end
a = str.split("\n").map(&:chars)
recurse(a)
I would write it like this:
def transpose s
lines = s.split(?\n)
longest = lines.map { |l| l.length }.max
(0..longest).map do |index|
lines.map { |l| l[index] || ' ' }.join
end * ?\n
end
This one works
s = "abc\ndef ghi\njk lm no\n"
s = s.split("\n")
s2 = ''
i = 0
while true
line = ''
s.each do |row|
line += (row[i] or ' ')
end
if line.strip == ''
break
end
s2 += line + "\n"
i += 1
end
puts s2
This one also works
s = "abc\ndef ghi\njk lm no\n"
s = s.split("\n")
maxlen = s.inject(0) {|m,r| m=[m, r.length].max}
s.map! {|r| r.ljust(maxlen).split(//)}
s = s.transpose.map {|r| r.join('')}.join("\n")
puts s
A play on what Chron did for an earlier version of ruby (e.g., 1.8.x). Example based on your original input that showed newline characters
str="abc\\n
def ghi\\n
jk lm no\\n"
def transpose s
lines = s.gsub("\\n","").split("\n")
longest = lines.map { |line| line.length }.max
(0..longest).map do |char_index|
lines.map { |line| line.split('')[char_index] || ' ' }.join
end * "\\n\n"
end
puts transpose(str)
I would write it like this:
def transpose_text(text)
# split the text into lines
text = text.split("\n")
# find the length of the longest line
max_line_length = text.map(&:size).max
# pad each line with white space and convert them to character arrays
text.map! { |line| line.ljust(max_line_length).chars }
#transpose the character arrays and then join them all into one string
text.transpose.map(&:join).join("\n")
end