Is there an easier transposition of the following hash? I have a solution that works but the transpose method is hard to read. The use case is a hash of jobs (j1, j2, j3, etc..) and the dates they occur on (d1, d2, d3, etc...). The requirement is to take a hash of events (e1, e2, e3 etc..) that are grouped by date then job and transpose them to events grouped by job then date.
I experimented with the #zip method but that would require injecting nil at j1:d1 and then stripping them back out of the results. I also struggled to find an example usage of #zip with a block argument. I understand that #zip with a block always returns nil but as a result I couldn't understand how you would ever practically use it.
require 'rspec'
require 'pry'
# | d1 | d2 | d3 |
# ----------------------
# j1 | | e2 | e3 |
# --------------------------
# j2 | e4 | e5 | e6 |
# --------------------------
# j3 | e7 | | e9 |
# --------------------------
def transpose(h)
Hash[
dates(h).map do |d|
[
d,
Hash[ h.keys.map do |j|
h[j][d] ? [j, h[j][d]] : nil
end.compact ]
]
end
]
end
def dates(h)
h.values.map(&:keys).reduce(:|).sort
end
describe "transpose" do
let(:starting) {
{
j1: { d2: :e2, d3: :e3 },
j2: { d1: :e4, d2: :e5, d3: :e6 },
j3: { d1: :e7, d3: :e9 }
}
}
let(:transposed) {
{
d1: { j2: :e4, j3: :e7 },
d2: { j1: :e2, j2: :e5 },
d3: { j1: :e3, j2: :e6, j3: :e9 }
}
}
it { expect(dates(starting)).to eq([:d1, :d2, :d3]) }
it { expect(transpose(starting)).to eq(transposed) }
end
I've rewritten your transpose
method a bit, it should be faster and cleaner:
def transpose(h)
h.each_with_object({}) do |(outer, data), ret|
data.each do |inner, event|
ret[inner] = {} unless ret[inner]
ret[inner][outer] = event
end
end
end
It doesn't use unnecessary mapping of dates, and works both ways (changes inner and outer keys). Let me know what you think :)
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