Please help in this. I have the 2 arrays below. Array a contains hours and array b contain the same hours and then values that happened at those hours.
a = ["1015","1240","1732"]
b = ["1015","X|2","D|5","1240","B|11","F|8","X|7","1732","D|9","X|1","B|3"]
So in array b:
Elements "X|2","D|5" happened at hour 10:15
Elements "B|11","F|8","X|7" happened at hour 12:40
Elements "D|9","X|1","B|3" happened at hour 17:32
First part of each element in B could be repeated, for example, X happened in 3 hours with different values, so in the output, I´d like to print hours and unique values, this is X, D, B and F
The output I´m looking for is:
HOUR X D B F
1015 2 5
1240 7 11 8
1732 1 9 3
The code I have so far is below but I still not able to organize the output in desired order.
val=[]
headers=[]
b.each{|v|
if v.include? "|"
headers << v.split("|")[0]
val << v.split("|")[1]
else
val << ["HOUR",v]
end
}
puts ["HOURS",headers.uniq].join(" ")
puts val
Current output of my code:
HOURS X D B F
HOUR
1015
2
5
HOUR
1240
11
8
7
HOUR
1732
9
1
3
I have assumed that a merely contains the times in b, sorted. As that can be computed there is no need to provide that information as an input.
Code
def print_table(data, time_label, column_spacing)
h = data.slice_before { |s| !s.include?('|') }.
each_with_object({}) { |(t,*a),h|
h[t] = a.map { |s| s.split('|') }.to_h.tap { |g| g.default = '' } }
row_labels = h.keys.sort
column_labels = h.values_at(*row_labels).reduce([]) { |a,g| a | g.keys }
image = [[time_label, *column_labels],
*row_labels.map { |time| [time, *h[time].values_at(*column_labels)] }]
row_label_width, *column_widths = image.transpose.map { |r| r.map(&:size).max }
print_image(image, row_label_width, column_widths, column_spacing)
end
def print_image(image, row_label_width, column_widths, column_spacing)
image.each do |time, *values|
print time.ljust(row_label_width)
values.zip(column_widths).each { |s,width| print s.rjust(width + column_spacing) }
puts
end
end
Example
b = ["1240", "B|11", "F|8", "X|7",
"1015", "X|2", "D|5",
"1732", "D|9", "X|1", "B|3"]
time_label = "HOUR"
column_spacing = 2
print_table(b, time_label, column_spacing)
prints
HOUR X D B F
1015 2 5
1240 7 11 8
1732 1 9 3
Note that the times-of-day in b are not in sorted order.
Explanation
For the values in the Example section, the first step is to group the elements of the array b into groupings (arrays) by time-of-day.
groups = b.slice_before { |s| !s.include?('|') }
#=> #<Enumerator: #<Enumerator::Generator:0x000000022b2490>:each>
See Enumerable#slice_before. We can see the objects that will be generated by this enumerator by converting it to an array.
groups.to_a
#=> [["1240", "B|11", "F|8", "X|7"],
# ["1015", "X|2", "D|5"],
# ["1732", "D|9", "X|1", "B|3"]]
Next, let's convert groups to a hash.
h = groups.each_with_object({}) { |(t,*a),h|
h[t] = a.map { |s| s.split('|') }.
to_h.
tap { |g| g.default = '' } }
#=> {"1240"=>{"B"=>"11", "F"=>"8", "X"=>"7"},
# "1015"=>{"X"=>"2", "D"=>"5"},
# "1732"=>{"D"=>"9", "X"=>"1", "B"=>"3"}}
See Enumerable#each_with_object, Array#to_h, Object#tap and Hash#default=. g.default = '' assigns the hash a default value of an empty space. This means that g[k] returns an empty space if g does not have a key k. For example, h["1015"]["B"] #=> "". g.default = '' returns '', which is why it is is enclosed in a tap block, which returns g with the default defined.
This article provides an explanation of the use of the splat operator. (Here, in a nutshell: [1, *[2, 3]] #=> [1, 2, 3]).
For the column labels we have a few options. Regardless, we first need the unique keys in the values (hashes) of h corresponding to the keys in row_labels.
row_labels = h.keys.sort
#=> ["1015", "1240", "1732"]
column_labels = h.values_at(*row_labels)
#=> [{"X"=>"2", "D"=>"5"},
# {"B"=>"11", "F"=>"8", "X"=>"7"},
# {"D"=>"9", "X"=>"1", "B"=>"3"}]
column_labels = column_labels.reduce([]) { |a,g| a | g.keys }
#=> ["X", "D", "B", "F"]
See Enumerable#values_at, Enumerable#reduce (aka inject) and Array#|. I have assumed this gives the desired column order, but the elements of column_labels could be reordered if desired. I present two possible options in the last section of my answer.
We next construct an array containing all the values in the table to be printed.
image = [[time_label, *column_labels],
*row_labels.map { |time| [time, *h[time].values_at(*column_labels)] }]
#=> [["HOUR", "X", "D", "B", "F"],
# ["1015", "2", "5", "", ""],
# ["1240", "7", "", "11", "8"],
# ["1732", "1", "9", "3", ""]]
Enumerable#values_at pulls out the values (hashes) in h[time] that correspond to each rows of the table, in their desired order.
We may then print the table as follows.
row_label_width, *column_widths = image.transpose.map { |r| r.map(&:size).max }
# => [4, 1, 1, 2, 1]
so that
row_label_width
#=> 4
column_widths
#=> [1, 1, 2, 1]
image.each do |time, *values|
print time.ljust(row_label_width)
values.zip(column_widths).each { |s,width| print s.rjust(width + column_spacing) }
puts
end
prints the table shown in the Example section.
Column order
As I said earlier the elements of column_labels could be reordered if desired. One possibility is to sort the labels alphabetically.
column_labels = h.values_at(*row_labels).reduce([]) { |a,g| a | g.keys }.sort!
#=> ["B", "D", "F", "X"]
Another is that we are given a desired order of all possible column labels.
desired = ["Y", "F", "D", "Z", "B", "X"]
Then compute the following.
column_labels = desired & h.values_at(*row_labels).reduce([]) { |a,g| a | g.keys }
#=> ["F", "D", "B", "X"]