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Hello I have been doing some research for sometime on this particular project I have been working on and I am at a loss. What I am looking to do is use information from a file and convert that to a hash using some of those components for my key. Within the file I have:1,Foo,20,Smith,40,John,55
An example of what I am looking for I am looking for an output like so {1 =>[Foo,20], 2 =>[Smith,40] 3 => [John,55]}
Here is what I got.
h = {}
people_file = File.open("people.txt") # I am only looking to read here.
until people_file.eof?
i = products_file.gets.chomp.split(",")
end
people_file.close
FName = 'test'
str = "1,Foo,20,Smith, 40,John,55"
File.write(FName, str)
#=> 26
base, *arr = File.read(FName).
split(/\s*,\s*/)
enum = (base.to_i).step
arr.each_slice(2).
with_object({}) {|pair,h| h[enum.next]=pair}
#=> {1=>["Foo", "20"], 2=>["Smith", "40"],
# 3=>["John", "55"]}
The steps are as follows.
s = File.read(FName)
#=> "1,Foo,20,Smith, 40,John,55"
base, *arr = s.split(/\s*,\s*/)
#=> ["1", "Foo", "20", "Smith", "40", "John", "55"]
base
#=> "1"
arr
#=> ["Foo", "20", "Smith", "40", "John", "55"]
a = base.to_i
#=> 1
I assume the keys are to be sequential integers beginning with a #=> 1.
enum = a.step
#=> (1.step)
enum.next
#=> 1
enum.next
#=> 2
enum.next
#=> 3
Continuing,
enum = a.step
b = arr.each_slice(2)
#=> #<Enumerator: ["Foo", "20", "Smith", "40", "John", "55"]:each_slice(2)>
Note I needed to redefine enum (or execute enum.rewind) to reinitialize it. We can see the elements that will be generated by this enumerator by converting it to an array.
b.to_a
#=> [["Foo", "20"], ["Smith", "40"], ["John", "55"]]
Continuing,
c = b.with_object({})
#=> #<Enumerator: #<Enumerator: ["Foo", "20", "Smith", "40", "John", "55"]
# :each_slice(2)>:with_object({})>
c.to_a
#=> [[["Foo", "20"], {}], [["Smith", "40"], {}], [["John", "55"], {}]]
The now-empty hashes will be constructed as calculations progress.
c.each {|pair,h| h[enum.next]=pair}
#=> {1=>["Foo", "20"], 2=>["Smith", "40"], 3=>["John", "55"]}
To see how the last step is performed, each initially directs the enumerator c to generate the first value, which it passes to the block. The block variables are assigned to that value, and the block calculation is performed.
enum = a.step
b = arr.each_slice(2)
c = b.with_object({})
pair, h = c.next
#=> [["Foo", "20"], {}]
pair
#=> ["Foo", "20"]
h #=> {}
h[enum.next]=pair
#=> ["Foo", "20"]
Now,
h#=> {1=>["Foo", "20"]}
The calculations are similar for the remaining two elements generated by the enumerator c.
See IO::write, IO::read, Numeric#step, Enumerable#each_slice, Enumerator#with_object, Enumerator#next and Enumerator#rewind. write and read respond to File because File is a subclass of IO (File.superclass #=> IO). split's argument, the regular expression, /\s*,\s*/, causes the string to be split on commas together with any spaces that surround the commas. Converting [["Foo", "20"], {}] to pair and h is a product of Array Decompostion.
I want to create an instance method like Array#my_map and that method should behavior of the original Array#map method.
I want to same output from new method as below:
arr = [1, 2, 3, 4]
arr.new_map_method do |x|
x + 1
end # => [2, 3, 4, 5]
arr.new_map_method(&:to_s) # => ["1", "2", "3", "4"]
The easiest way of creating a method with the same behavior of a different method is a simple alias:
class Array
alias_method :new_map_method, :map
end
If, for whatever strange reason, you don't want to use map, you can use inject instead:
class Array
def new_map_method
return enum_for(__callee__) unless block_given?
inject([]) {|acc, el| acc << yield(el) }
end
end
Thanks guys! I also made one solution for my qeestion
class Array
def my_map(&block)
result = []
each do |element|
result << block.call(element)
end
result
end
end
function calling
[1,2,3].my_map(&:to_s)
output => ["1", "2", "3"]
[1, 2, 3, 4, 5].my_map do |x|
x
end
output => [1, 2, 3, 4, 5]
I have a set of array like this:
[["1","2"],["1","3"],["2","3"],["2","5"]]
I want to find the union of first values like
["1","2"],["1","3"] matches so i need to create new array like ["1","2,3"]
so the resulting array will be like
[["1","2,3"],["2","3,5"]]
Like most problems in Ruby, the Enumerable module does the job:
input = [["1","2"],["1","3"],["2","3"],["2","5"]]
result = input.group_by do |item|
# Group by first element
item[0]
end.collect do |key, items|
# Compose into new format
[
key,
items.collect do |item|
item[1]
end.join(',')
]
end
puts result.inspect
# => [["1", "2,3"], ["2", "3,5"]]
The group_by method comes in very handy when aggregating things like this, and collect is great for rewriting how the elements appear.
What you are asking for is not a true union for a true union of each 2 it would be:
data = [["1","2"],["1","3"],["2","3"],["2","5"]]
data.each_slice(2).map{|a,b| a | b.to_a }
#=> [["1","2","3"],["2","3","5"]]
Here is a very simple solution that modifies this concept to fit your needs:
data = [["1","2"],["1","3"],["2","3"],["2","5"]]
data.each_slice(2).map do |a,b|
unified = (a | b.to_a)
[unified.shift,unified.join(',')]
end
#=> [["1", "2,3"], ["2", "3,5"]]
Added to_a to piped variable b in the event that there are an uneven number of arrays. eg.
data = [["1","2"],["1","3"],["2","3"],["2","5"],["4","7"]]
data.each_slice(2).map do |a,b|
unified = (a | b.to_a)
[unified.shift,unified.join(',')]
end
#=> [["1", "2,3"], ["2", "3,5"], ["4","7"]]
If you meant that you want this to happen regardless of order then this will work but will destroy the data object
data.group_by(&:shift).map{|k,v| [k,v.flatten.join(',')]}
#=> [["1", "2,3"], ["2", "3,5"], ["4","7"]]
Non destructively you could call
data.map(&:dup).group_by(&:shift).map{|k,v| [k,v.flatten.join(',')]}
#=> [["1", "2,3"], ["2", "3,5"], ["4","7"]]
Here's another way.
Code
def doit(arr)
arr.each_with_object({}) { |(i,*rest),h| (h[i] ||= []).concat(rest) }
.map { |i,rest| [i, rest.join(',')] }
end
Examples
arr1 = [["1","2"],["1","3"],["2","3"],["2","5"]]
doit(arr1)
#=> [["1", "2,3"], ["2", "3,5"]]
arr2 = [["1","2","6"],["2","7"],["1","3"],["2","3","9","4","cat"]]
doit(arr2)
# => [["1", "2,6,3"], ["2", "7,3,9,4,cat"]]
Explanation
For arr1 above, we obtain:
enum = arr1.each_with_object({})
#=> #<Enumerator: [["1", "2"], ["1", "3"], ["2", "3"],
# ["2", "5"]]:each_with_object({})>
We can convert enum to an array see its elements:
enum.to_a
#=> [[["1", "2"], {}], [["1", "3"], {}],
# [["2", "3"], {}], [["2", "5"], {}]]
These elements will be passed into the block, and assigned to the block variables, by Enumerator#each, which will invoke Array#each. The first of these elements ([["1", "2"], {}]) can be obtained by invoking Enumerator#next on enum:
(i,*rest),h = enum.next
#=> [["1", "2"], {}]
i #=> "1"
rest #=> ["2"]
h #=> {}
We then execute:
(h[i] ||= []).concat(rest)
#=> (h["1"] ||= []).concat(["2"])
#=> (nil ||= []).concat(["2"])
#=> [].concat(["2"])
#=> ["2"]
each then passes the next element of enum to the block:
(i,*rest),h = enum.next
#=> [["1", "3"], {"1"=>["2"]}]
i #=> "1"
rest #=> ["3"]
h #=> {"1"=>["2"]}
(h[i] ||= []).concat(rest)
#=> (h["1"] ||= []).concat(["3"])
#=> (["2"] ||= []).concat(["3"])
#=> ["2"].concat(["3"])
#=> ["2", "3"]
After passing the last two elements of enum into the block, we obtain:
h=> {"1"=>["2", "3"], "2"=>["3", "5"]}
map creates an enumerator:
enum_h = h.each
#=> > #<Enumerator: {"1"=>["2", "3"]}:each>
and calls Enumerator#each (which calls Hash#each) to pass each element of enum_h into the block:
i, rest = enum_h.next
#=> ["1", ["2", "3"]]
then computes:
[i, rest.join(',')]
#=> ["1", ["2", "3"].join(',')]
#=> ["1", "2,3"]
The other element of enum_h is processed similarly.
Simple ruby question. Lets say I have an array of 10 strings and I want to move elements at array[3] and array[5] into a totally new array. The new array would then only have the two elements I moved from the first array, AND the first array would then only have 8 elements since two of them have been moved out.
Use Array#slice! to remove the elements from the first array, and append them to the second array with Array#<<:
arr1 = ['Foo', 'Bar', 'Baz', 'Qux']
arr2 = []
arr2 << arr1.slice!(1)
arr2 << arr1.slice!(2)
puts arr1.inspect
puts arr2.inspect
Output:
["Foo", "Baz"]
["Bar", "Qux"]
Depending on your exact situation, you may find other methods on array to be even more useful, such as Enumerable#partition:
arr = ['Foo', 'Bar', 'Baz', 'Qux']
starts_with_b, does_not_start_with_b = arr.partition{|word| word[0] == 'B'}
puts starts_with_b.inspect
puts does_not_start_with_b.inspect
Output:
["Bar", "Baz"]
["Foo", "Qux"]
a = (0..9).map { |i| "el##{i}" }
x = [3, 5].sort_by { |i| -i }.map { |i| a.delete_at(i) }
puts x.inspect
# => ["el#5", "el#3"]
puts a.inspect
# => ["el#0", "el#1", "el#2", "el#4", "el#6", "el#7", "el#8", "el#9"]
As noted in comments, there is some magic to make indices stay in place. This can be avoided by first getting all the desired elements using a.values_at(*indices), then deleting them as above.
Code:
arr = ["null","one","two","three","four","five","six","seven","eight","nine"]
p "Array: #{arr}"
third_el = arr.delete_at(3)
fifth_el = arr.delete_at(4)
first_arr = arr
p "First array: #{first_arr}"
concat_el = third_el + "," + fifth_el
second_arr = concat_el.split(",")
p "Second array: #{second_arr}"
Output:
c:\temp>C:\case.rb
"Array: [\"null\", \"one\", \"two\", \"three\", \"four\", \"five\", \"six\", \"s
even\", \"eight\", \"nine\"]"
"First array: [\"null\", \"one\", \"two\", \"four\", \"six\", \"seven\", \"eight
\", \"nine\"]"
"Second array: [\"three\", \"five\"]"
Why not start deleting from the highest index.
arr = ['Foo', 'Bar', 'Baz', 'Qux']
index_array = [2, 1]
new_ary = index_array.map { |index| arr.delete_at(index) }
new_ary # => ["Baz", "Bar"]
arr # => ["Foo", "Qux"]
Here's one way:
vals = arr.values_at *pulls
arr = arr.values_at *([*(0...arr.size)] - pulls)
Try it.
arr = %w[Now is the time for all Rubyists to code]
pulls = [3,5]
vals = arr.values_at *pulls
#=> ["time", "all"]
arr = arr.values_at *([*(0...arr.size)] - pulls)
#=> ["Now", "is", "the", "for", "Rubyists", "to", "code"]
arr = %w[Now is the time for all Rubyists to code]
pulls = [5,3]
vals = arr.values_at *pulls
#=> ["all", "time"]
arr = arr.values_at *([*(0...arr.size)] - pulls)
#=> ["Now", "is", "the", "for", "Rubyists", "to", "code"]
The following code:
str = "1, hello,2"
puts str
arr = str.split(",")
puts arr.inspect
arr.collect { |x| x.strip! }
puts arr.inspect
produces the following result:
1, hello,2
["1", " hello", "2"]
["1", "hello", "2"]
This is as expected. The following code:
str = "1, hello,2"
puts str
arr = (str.split(",")).collect { |x| x.strip! }
puts arr.inspect
Does however produce the following output:
1, hello,2
[nil, "hello", nil]
Why do I get these "nil"? Why can't I do the .collect immediately on the splitted-array?
Thanks for the help!
The #collect method will return an array of the values returned by each block's call. In your first example, you're modifying the actual array contents with #strip! and use those, while you neglect the return value of #collect.
In the second case, you use the #collect result. Your problem is that #strip! will either return a string or nil, depending on its result – especially, it'll return nil if the string wasn't modified.
Therefore, use #strip (without the exclamation mark):
1.9.3-p194 :005 > (str.split(",")).collect { |x| x.strip }
=> ["1", "hello", "2"]
Because #strip! returns nil if the string was not altered.
In your early examples you were not using the result of #collect, just modifying the strings with #strip!. Using #each in that case would have made the non-functional imperative loop a bit more clear. One normally uses #map / #collect only when using the resulting new array.
You last approach looks good, you wrote a functional map but you left the #strip! in ... just take out the !.