can it be shorter?

[Dorren]

doing access control on rails controller,

------ I have this input --------------
hash = {"index" => "list",
["edit", "update"] => "manage_one",
["new", "create", "destroy"] => "manage_all"}

------- I want this output ----------
{"index"=>"list",
"edit"=>"manage_one",
"update"=>"manage_one",
"new"=>"manage_all",
"create"=>"manage_all",
"destroy"=>"manage_all"
}

------ I have this code ------------
hash = Hash[*hash.to_a.collect{|x|
Array === x[0] ? x[0].zip([x[1]]*x[0].size) : x
}.flatten]

----- I want shorter code ------

thanks.

 

[Dorren]

a little clearer than previous one.

hash = Hash[*hash.inject([]){|arr, (k, v)|
arr += Array === k ? k.zip([v]* k.size)
: [k, v]
}.flatten]

 

[Vincent Fourmond]

doing access control on rails controller,

------ I have this input --------------
hash = {"index" => "list",
["edit", "update"] => "manage_one",
["new", "create", "destroy"] => "manage_all"}

------- I want this output ----------
{"index"=>"list",
"edit"=>"manage_one",
"update"=>"manage_one",
"new"=>"manage_all",
"create"=>"manage_all",
"destroy"=>"manage_all"
}

------ I have this code ------------
hash = Hash[*hash.to_a.collect{|x|
Array === x[0] ? x[0].zip([x[1]]*x[0].size) : x
}.flatten]

----- I want shorter code ------

What about this:

h = {}
for k,v in hash
[k].flatten.each {|l| h[l] = v}
end
hash = h

=> {"new"=>"manage_all", "edit"=>"manage_one",

"destroy"=>"manage_all", "create"=>"manage_all", "index"=>"list",
"update"=>"manage_one"}

Cheers,

Vincent

 

[Vincent Fourmond]

h = {}
for k,v in hash
[k].flatten.each {|l| h[l] = v}
end
hash = h

Actually, I've got shorter ;-):

h = {}
for k,v in hash
[*k].each {|l| h[l] = v}
end
hash = h

Vince

 

[Dorren]

better, thank tou.

doing access control on rails controller,

------ I have this input --------------
hash = {"index" => "list",
["edit", "update"] => "manage_one",
["new", "create", "destroy"] => "manage_all"}

------- I want this output ----------
{"index"=>"list",
"edit"=>"manage_one",
"update"=>"manage_one",
"new"=>"manage_all",
"create"=>"manage_all",
"destroy"=>"manage_all"
}

------ I have this code ------------
hash = Hash[*hash.to_a.collect{|x|
Array === x[0] ? x[0].zip([x[1]]*x[0].size) : x
}.flatten]

----- I want shorter code ------ What about this:

h = {}
for k,v in hash
[k].flatten.each {|l| h[l] = v}
end
hash = h> => {"new"=>"manage_all", "edit"=>"manage_one","destroy"=>"manage_all", "create"=>"manage_all", "index"=>"list",
"update"=>"manage_one"}

Cheers,

Vincent

 

[William James]

a little clearer than previous one.

hash = Hash[*hash.inject([]){|arr, (k, v)|
arr += Array === k ? k.zip([v]* k.size)
: [k, v]
}.flatten]

Hash[ *hash.map{|k,v| k=[*k]; k.zip([v]*k.size) }.flatten ]

 

[Daniel Martin]

Actually, I've got shorter ;-):

h = {}
for k,v in hash
[*k].each {|l| h[l] = v}
end
hash = h

Well, if we're golfing:

h={};hash.map{|k,v|[*k].map{|t|h[t]=v}};h

 

[Robert Klemme]

Actually, I've got shorter ;-):

h = {}
for k,v in hash
[*k].each {|l| h[l] = v}
end
hash = h

Well, if we're golfing:

h={};hash.map{|k,v|[*k].map{|t|h[t]=v}};h

Not really shorter but I though there should be at least one solution
with #inject:

hash.inject({}){|h,(k,v)| k.to_a.each {|x| h[x]=v};h}

=> {"new"=>"manage_all", "edit"=>"manage_one", "destroy"=>"manage_all",
"create"=>"manage_all", "index"=>"list", "update"=>"manage_one"}

hash.inject({}){|h,(k,v)| k.each {|x| h[x]=v} rescue h[k]=v;h}

=> {"new"=>"manage_all", "edit"=>"manage_one", "destroy"=>"manage_all",
"create"=>"manage_all", "index"=>"list", "update"=>"manage_one"}



robert

 

[Eric Hodel]

--Apple-Mail-2-287622042
Content-Transfer-Encoding: 7bit
Content-Type: text/plain;
charset=US-ASCII;
format=flowed

h = {}
for k,v in hash
[k].flatten.each {|l| h[l] = v}
end
hash = h

Actually, I've got shorter ;-):

h = {}
for k,v in hash
[*k].each {|l| h[l] = v}
end
hash = h

but fastest (I assume all entries worked correctly):

$ ruby bm.rb
Rehearsal --------------------------------------------------------
original 10.440000 0.010000 10.450000 ( 10.492738)
original clean 11.480000 0.040000 11.520000 ( 11.632237)
vince 5.210000 0.010000 5.220000 ( 5.264394)
william 12.760000 0.020000 12.780000 ( 12.814230)
daniel 6.900000 0.010000 6.910000 ( 6.924297)
robert inject 6.560000 0.000000 6.560000 ( 6.588160)
robert inject rescue 6.080000 0.010000 6.090000 ( 6.094140)
---------------------------------------------- total: 59.530000sec

user system total real
original 10.450000 0.010000 10.460000 ( 10.467971)
original clean 11.490000 0.030000 11.520000 ( 11.600672)
vince 5.210000 0.000000 5.210000 ( 5.219863)
william 12.820000 0.010000 12.830000 ( 12.839722)
daniel 6.930000 0.010000 6.940000 ( 6.945329)
robert inject 6.580000 0.010000 6.590000 ( 6.591095)
robert inject rescue 6.080000 0.000000 6.080000 ( 6.117226)


--Apple-Mail-2-287622042
Content-Transfer-Encoding: 7bit
Content-Type: text/x-ruby-script;
x-unix-mode=0644;
name=bm.rb
Content-Disposition: attachment;
filename=bm.rb

require 'benchmark'

N = 500_000

hash = {
"index" => "list",
["edit", "update"] => "manage_one",
["new", "create", "destroy"] => "manage_all"
}

Benchmark.bmbm do |bm|

bm.report 'original' do
N.times do
Hash[*hash.to_a.collect{|x|
Array === x[0] ? x[0].zip([x[1]]*x[0].size) : x
}.flatten]
end
end

bm.report 'original clean' do
N.times do
Hash[*hash.inject([]){|arr, (k, v)|
arr += Array === k ? k.zip([v]* k.size) : [k, v]
}.flatten]
end
end

bm.report 'vince' do
N.times do
h = {}
for k,v in hash
[*k].each {|l| h[l] = v}
end
end
end

bm.report 'william' do
N.times do
Hash[ *hash.map{|k,v| k=[*k]; k.zip([v]*k.size) }.flatten ]
end
end

bm.report 'daniel' do
N.times do
h={};hash.map{|k,v|[*k].map{|t|h[t]=v}}
end
end

bm.report 'robert inject' do
N.times do
hash.inject({}){|h,(k,v)| k.to_a.each {|x| h[x]=v};h}
end
end

bm.report 'robert inject rescue' do
N.times do
hash.inject({}){|h,(k,v)| k.each {|x| h[x]=v} rescue h[k]=v;h}
end
end

end


--Apple-Mail-2-287622042
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Content-Type: text/plain;
charset=US-ASCII;
format=flowed


--
Eric Hodel - (e-mail address removed) - http://blog.segment7.net

I LIT YOUR GEM ON FIRE!


--Apple-Mail-2-287622042--

 

[Erik Veenstra]

There's a glitch, which is not yet mentioned...

Consider this old hash: {["A", "B"]=>"C", ["B", "A"]=>"D"}

What's the value of new_hash["A"]? Is it "C", or is it "D"?

That depends on the order on which you build the new hash.
Which depends on the order in which you walk through the old
hash. Which isn't deterministic, AFAIK. Maybe it is
determinstic if you know in which order the old hash was built.
But, given _a_ hash, you simply don't know the order in which
you walk through it, so you don't know the order in which the
new hash will be built, so you don't know what the result is
going to be. Cool...

Adding a sort to the algorithm fixes this.

gegroet,
Erik V. - http://www.erikveen.dds.nl/

----------------------------------------------------------------

hash.collect do |k,v|
[[k].flatten, v]
end.sort.inject({}) do |h,(ks,v)|
ks.each do |k|
h[k] = v
end
h
end

----------------------------------------------------------------