'delete_at' and 'slice' remove the item at the index and return that item. But I don't really care about the removed item. I just want a new array with that item removed. Ruby's Array class doesn't seem to provide such a method.
Example would be:
a = ['a','b','c','d']
b = a.remove(2) #b => ['a','b','d']
Here 'remove' is a fictitious method that does what I want. I need the original array, so I want a new array return. I wonder if Ruby already has some built-in like this?
class Array
def remove(idx)
self[0...idx] + self[idx+1..-1]
end
end
a = ['a','b','c','d']
a.reject {|i| i == a[2] }
#=> ["a", "b", "d"]
irb(main):001:0> a = %w[ a b c d ]
#=> ["a", "b", "c", "d"]
irb(main):002:0> a.reject.with_index{ |o,i| i==2 }
#=> ["a", "b", "d"]
irb(main):003:0> a
#=> ["a", "b", "c", "d"]
Requires Ruby 1.9
With some monkey-patching:
irb(main):013:0> class Array
irb(main):014:1> def remove_at(i)
irb(main):015:2> self.dup.tap{ |clone| clone.delete_at(i) }
irb(main):016:2> end
irb(main):017:1> end
#=> nil
irb(main):018:0> a.remove_at(2)
#=> ["a", "b", "d"]
irb(main):019:0> a
#=> ["a", "b", "c", "d"]
it's quite hacky:
a = ['a','b','c','d']
b, = [a, a.delete_at(0)] # => ['b','c','d']
but it's faster(on my eeepc)
require 'benchmark'
n = 5000
Benchmark.bm do |x|
a = (1..5000).to_a
b = nil
x.report { n.times do; b, = [a, a.delete_at(0)]; end }
a = (1..5000).to_a
b = nil
x.report { n.times do; b = a.reject.with_index{ |o,i| i == 0 }; end }
end
user system total real
0.032000 0.000000 0.032000 ( 0.034002)
21.808000 0.156000 21.964000 ( 22.696298) OMG!
Related
I am trying to transform a given string into a hash with each its character = key and index = value.
For example, if I have str = "hello", I would like it to transform into {"h"=>0, "e"=>1, "l"=>2, "l"=>3, "o"=>4}.
I created a method as such:
def map_indices(arr)
arr.map.with_index {|el, index| [el, index]}.to_h
end
#=> map_indices('hello'.split(''))
#=> {"h"=>0, "e"=>1, "l"=>3, "o"=>4}
The problem is it skips the first l. If I reverse the order of el and index: arr.map.with_index {|el, index| [index, el]}.to_h, I get all the letters spelled out: {0=>"h", 1=>"e", 2=>"l", 3=>"l", 4=>"o"}
But when I invert it, I get the same hash that skips one of the l's.
map_indices('hello'.split('')).invert
#=> {"h"=>0, "e"=>1, "l"=>3, "o"=>4}
Why is this behaving like such? How can I get it to print {"h"=>0, "e"=>1, "l"=>2, "l"=>3, "o"=>4}?
It can be done, but will confuse other Ruby programmers.A normal hash treats a key "a" as identical to another "a". Unless a little known feature .compare_by_identity is used:
h = {}.compare_by_identity
"hello".chars.each_with_index{|c,i| h[c] = i}
p h # => {"h"=>0, "e"=>1, "l"=>2, "l"=>3, "o"=>4}
Any of the following could be used. For
str = "hello"
all return
{"h"=>[0], "e"=>[1], "l"=>[2, 3], "o"=>[4]}
str.each_char
.with_index
.with_object({}) { |(c,i),h| (h[c] ||= []) << i }
See String#each_char, Enumerator#with_index and Enumerator#with_object. The block variables have been written to exploit array decomposition.
str.each_char
.with_index
.with_object(Hash.new { |h,k| h[k] = [] }) { |(c,i),h| h[c] << i }
See the form of Hash::new that takes a block and no argument. If a hash has been defined
h = Hash.new { |h,k| h[k] = [] }
and later
h[c] << i
is executed, h[c] is first set equal to an empty array if h does not have a key c.
str.size
.times
.with_object(Hash.new { |h,k| h[k] = [] }) { |i,h| h[str[i]] << i }
str.each_char
.with_index
.group_by(&:first)
.transform_values { |a| a.flat_map(&:last) }
See Enumerable#group_by, Hash#transform_values (introduced in Ruby v2.5) and Enumerable#flat_map.
Note that
str.each_char
.with_index
.group_by(&:first)
#=> {"h"=>[["h", 0]], "e"=>[["e", 1]], "l"=>[["l", 2], ["l", 3]],
# "o"=>[["o", 4]]}
Another option you can use is zipping two enumerations together.
s = "hello"
s.chars.zip(0..s.size)
This yields: [["h", 0], ["e", 1], ["l", 2], ["l", 3], ["o", 4]]
I am new to Ruby and I am sure this can be refactored, but another alternative might be:
arr1 = "Hello".split(%r{\s*})
arr2 = []
for i in 0..arr1.size - 1
arr2 << i
end
o = arr1.zip(arr2)
a_h = []
o.each do |i|
a_h << Hash[*i]
end
p a_h.each_with_object({}) { |k, v| k.each { |kk,vv| (v[kk] ||= []) << vv } }
=> {"H"=>[0], "e"=>[1], "l"=>[2, 3], "o"=>[4]}
let's have this hash:
hash = {"a" => 1, "b" => {"c" => 3}}
hash.get_all_keys
=> ["a", "b", "c"]
how can i get all keys since hash.keys returns just ["a", "b"]
This will give you an array of all the keys for any level of nesting.
def get_em(h)
h.each_with_object([]) do |(k,v),keys|
keys << k
keys.concat(get_em(v)) if v.is_a? Hash
end
end
hash = {"a" => 1, "b" => {"c" => {"d" => 3}}}
get_em(hash) # => ["a", "b", "c", "d"]
I find grep useful here:
def get_keys(hash)
( hash.keys + hash.values.grep(Hash){|sub_hash| get_keys(sub_hash) } ).flatten
end
p get_keys my_nested_hash #=> ["a", "b", "c"]
I like the solution as it is short, yet it reads very nicely.
Version that keeps the hierarchy of the keys
Works with arrays
Works with nested hashes
keys_only.rb
# one-liner
def keys_only(h); h.map { |k, v| v = v.first if v.is_a?(Array); v.is_a?(Hash) ? [k, keys_only(v)] : k }; end
# nicer
def keys_only(h)
h.map do |k, v|
v = v.first if v.is_a?(Array);
if v.is_a?(Hash)
[k, keys_only(v)]
else
k
end
end
end
hash = { a: 1, b: { c: { d: 3 } }, e: [{ f: 3 }, { f: 5 }] }
keys_only(hash)
# => [:a, [:b, [[:c, [:d]]]], [:e, [:f]]]
P.S.: Yes, it looks like a lexer :D
Bonus: Print the keys in a nice nested list
# one-liner
def print_keys(a, n = 0); a.each { |el| el.is_a?(Array) ? el[1] && el[1].class == Array ? print_keys(el, n) : print_keys(el, n + 1) : (puts " " * n + "- #{el}") }; nil; end
# nicer
def print_keys(a, n = 0)
a.each do |el|
if el.is_a?(Array)
if el[1] && el[1].class == Array
print_keys(el, n)
else
print_keys(el, n + 1)
end
else
puts " " * n + "- #{el}"
end
end
nil
end
> print_keys(keys_only(hash))
- a
- b
- c
- d
- e
- f
def get_all_keys(hash)
hash.map do |k, v|
Hash === v ? [k, get_all_keys(v)] : [k]
end.flatten
end
Please take a look of following code:
hash = {"a" => 1, "b" => {"c" => 3}}
keys = hash.keys + hash.select{|_,value|value.is_a?(Hash)}
.map{|_,value| value.keys}.flatten
p keys
result:
["a", "b", "c"]
New solution, considering #Bala's comments.
class Hash
def recursive_keys
if any?{|_,value| value.is_a?(Hash)}
keys + select{|_,value|value.is_a?(Hash)}
.map{|_,value| value.recursive_keys}.flatten
else
keys
end
end
end
hash = {"a" => 1, "b" => {"c" => {"d" => 3}}, "e" => {"f" => 3}}
p hash.recursive_keys
result:
["a", "b", "e", "c", "d", "f"]
Also deal with nested arrays that include hashes
def all_keys(items)
case items
when Hash then items.keys + items.values.flat_map { |v| all_keys(v) }
when Array then items.flat_map { |i| all_keys(i) }
else []
end
end
class Hash
def get_all_keys
[].tap do |result|
result << keys
values.select { |v| v.respond_to?(:get_all_keys) }.each do |value|
result << value.get_all_keys
end
end.flatten
end
end
hash = {"a" => 1, "b" => {"c" => 3}}
puts hash.get_all_keys.inspect # => ["a", "b", "c"]
Here is another approach :
def get_all_keys(h)
h.each_with_object([]){|(k,v),a| v.is_a?(Hash) ? a.push(k,*get_all_keys(v)) : a << k }
end
hash = {"a" => 1, "b" => {"c" => {"d" => 3}}}
p get_all_keys(hash)
# >> ["a", "b", "c", "d"]
I'm sure there is a more elegant solution, but this option works:
blah = {"a" => 1, "b" => {"c" => 3}}
results = []
blah.each do |k,v|
if v.is_a? Hash
results << k
v.each_key {|key| results << key}
else
results << k
end
end
puts results
hash.keys is the simplest one I have seen to return an array of the key values in a hash.
I have
a = ["a", "d", "c", "b", "b", "c", "c"]
and need to print something like (sorted descending by number of occurrences):
c:3
b:2
I understand first part (finding NON-unique) is:
b = a.select{ |e| a.count(e) > 1 }
=> ["c", "b", "b", "c", "c"]
or
puts b.select{|e, c| [e, a.count(e)] }.uniq
c
b
How to output each non-unique with number of occurrences sorted backwards?
puts a.uniq.
map { | e | [a.count(e), e] }.
select { | c, _ | c > 1 }.
sort.reverse.
map { | c, e | "#{e}:#{c}" }
The group_by method is used for this often:
a.group_by{ |i| i }
{
"a" => [
[0] "a"
],
"d" => [
[0] "d"
],
"c" => [
[0] "c",
[1] "c",
[2] "c"
],
"b" => [
[0] "b",
[1] "b"
]
}
I like:
a.group_by{ |i| i }.each_with_object({}) { |(k,v), h| h[k] = v.size }
{
"a" => 1,
"d" => 1,
"c" => 3,
"b" => 2
}
Or:
Hash[a.group_by{ |i| i }.map{ |k,v| [k, v.size] }]
{
"a" => 1,
"d" => 1,
"c" => 3,
"b" => 2
}
One of those might scratch your itch. From there you can reduce the result using a little test:
Hash[a.group_by{ |i| i }.map{ |k,v| v.size > 1 && [k, v.size] }]
{
"c" => 3,
"b" => 2
}
If you just want to print the information use:
puts a.group_by{ |i| i }.map{ |k,v| "#{k}: #{v.size}" }
a: 1
d: 1
c: 3
b: 2
From Ruby 2.7, you can utilise Enumerable#tally and numbered block arguments:
a = ["a", "d", "c", "b", "b", "c", "c"]
puts a.tally.filter { _2 > 1 }.sort_by { -_2 }.map &:first
Here, Enumerable#tally returns a hash like { 'a' => 1, 'b' => 2, ... }, which you then have to filter and sort. After sorting, the hash would've collapsed to a nested array, e.g. [['b', 2], ...]. The last step is to take the first argument of each array element, using &:first.
How about:
a.sort.chunk{|x| a.count(x)}.sort.reverse.each do |n, v|
puts "#{v[0]}:#{n}" if n > 1
end
I personally like this solution:
a.inject({}) {|hash, val| hash[val] ||= 0; hash[val] += 1; hash}.
reject{|key, value| value == 1}.sort.reverse.
each_pair{|k,v| puts("#{k}:#{v}")}
a.reduce(Hash.new(0)) { |memo,x| memo[x] += 1; memo } # Frequency count.
.select { |_,count| count > 1 } # Choose non-unique items.
.sort_by { |x| -x[1] } # Sort by number of occurrences descending.
# => [["c", 3], ["b", 2]]
Also:
a.group_by{|x|x}.map{|k,v|[k,v.size]}.select{|x|x[1]>1}.sort_by{|x|-x[1]}
# => [["c", 3], ["b", 2]]
This will give you a hash with element => occurrences:
b.reduce(Hash.new(0)) do |hash, element|
hash[element] += 1
hash
end
puts a.uniq.
map { |e| a.count(e) > 1 ? [e, a.count(e)] : nil }.compact.
sort { |a, b| b.last <=> a.last }
I have some array
>> a = ["a..c", "0..2"]
=> ["a..c", "0..2"]
I need convert this array to another array
>> b = ("a".."c").to_a + (0..2).to_a
=> ["a", "b", "c", 0, 1, 2]
How I can do it?
a.flat_map do |string_range|
from, to = string_range.split("..", 2)
(from =~ /^\d+$/ ? (from.to_i..to.to_i) : (from..to)).to_a
end
#=> => ["a", "b", "c", 0, 1, 2]
what about this?
a = ["a..c", "0..2"]
b = a.map { |e| Range.new( *(e).split('..') ).to_a }.flatten
no flat_map used so it works the same on all versions
as #steenslag correctly mentioned, this version does not convert to integers.
here is a version that does:
b = a.map do |e|
Range.new( *(e).split('..').map{ |c| c =~ /\A\d+\Z/ ? c.to_i : c } ).to_a
end.flatten
see it in action here
a = ["a..c", "0..2"]
b = a.flat_map{|str| Range.new(*str.split('..')).to_a} # => ["a", "b", "c", "0", "1", "2"]
p b.map!{|v| Integer(v) rescue v} # => ["a", "b", "c", 0, 1, 2]
Write a method, which given an array, returns a hash whose keys are words in the array and whose values are the number of times each word appears.
arr=["A", "man", "a", "plan", "a", "canal","Panama"]
# => {'a' => 3, 'man' => 1, 'canal' => 1, 'panama' => 1, 'plan' => 1}
How do I achieve that? Here's my code:
hash={}
arr.each do |i|
hash.each do |c,v|
hash[c]=v+1
end
end
hash = arr.inject({}) do |hash, element|
element.downcase!
hash[element] ||= 0
hash[element] += 1
hash
end
arr.inject(Hash.new(0)){|h,k| k.downcase!; h[k] += 1; h}
arr = ["A", "man", "a", "plan", "a", "canal","Panama"]
r = {}
arr.each { |e| e.downcase!; r[e] = arr.count(e) if r[e].nil? }
Output
p r
#==> {"a"=>3, "man"=>1, "plan"=>1, "canal"=>1, "panama"=>1}