I am trying to get a list of colors to print after the color blue without using the index. I did this
colors = ["Red", "Blue", "Green", "Purple", "White", "Black"]
colors.each { |item| print item if item > "Blue" }
but the output is this
Red Green Purple White
Does anyone know why?
The question "Why" is already answered in the comment of davidhu2000:
Because you are printing all items that is alphabetically after blue
If you don't want to use the index, then you can use a solution with a help variable:
colors = ["Red", "Blue", "Green", "Purple", "White", "Black"]
blue_found = false
colors.each { |item|
if blue_found
print item
else
blue_found ||= item == "Blue"
end
}
If you prefer a one-liner you could use
blue_found ? (print item) : (blue_found ||= item == "Blue" )
Another possibility:
colors = ["Red", "Blue", "Green", "Purple", "White", "Black"]
blue_found = false
colors.each { |item|
print item if blue_found
blue_found ||= item == "Blue"
}
I was curious, why you don't want to use the index, so I made some more research outside your question.
The solution with index would be:
colors.each_with_index { |item,i|
print item if i > colors.index("Blue")
}
The index-method may need some more runtime. To test it, I made a little benchmark:
colors = ["Red", "Blue", "Green", "Purple", "White", "Black"]
require 'benchmark'
TEST_LOOPS = 100_000
def action(item)
#Just a no-action.
end
Benchmark.bm(10) {|b|
b.report('test/variable') {
TEST_LOOPS.times {
blue_found = false
colors.each { |item|
action(item) if blue_found
blue_found ||= item == "Blue"
}
} #Testloops
} #b.report
b.report('test/index') {
TEST_LOOPS.times {
colors.each_with_index { |item,i|
action(item) if i > colors.index("Blue")
}
} #Testloops
} #b.report
b.report('test/index2') {
TEST_LOOPS.times {
index_blue = colors.index("Blue")
colors.each_with_index { |item,i|
action(item) if i > index_blue
}
} #Testloops
} #b.report
b.report('test/dogbert') {
TEST_LOOPS.times {
# Drop all items until you find "Blue", then drop the "Blue".
colors.drop_while { |item| item != "Blue" }.drop(1).each do |item|
action(item)
end
} #Testloops
} #b.report
b.report('test/pjs') {
TEST_LOOPS.times {
after_blue = colors.dup
loop do
break if after_blue.shift == 'Blue'|| after_blue.length < 1
end
after_blue.each{|item| action(item) }
} #Testloops
} #b.report
} #Benchmark
The result:
user system total real
test/variable 0.187000 0.000000 0.187000 ( 0.179010)
test/index 0.250000 0.000000 0.250000 ( 0.254015)
test/index2 0.140000 0.000000 0.140000 ( 0.136008)
test/dogbert 0.110000 0.000000 0.110000 ( 0.111006)
test/pjs 0.327000 0.000000 0.327000 ( 0.332019)
So the version with a help variable is (as expected) faster. But if you define the index once outside the loop, then the solution with index (test/index2) is nearly as fast as test/variable.
Disclaimer: I made not many thoughts for faster solutions. So maybe there is a more efficient way. Thanks to dogbert for his hint.
Array#index call outside the .each_with_index block. Right now that implementation is O(n^2), which would be O(n) if you move that out. (But it's still slower than the first implementation since it has to traverse the array twice.)
Another variant that runs neck and neck with @knut's test/variable benchmark is:
def values_after(target, arr)
result = arr.dup
loop do
break if result.shift == target || result.length < 1
end
result
end
colors = ['Red', 'Blue', 'Green', 'Purple', 'White', 'Black']
puts *values_after('Blue', colors)
# Or, if you prefer all on one comma-separated line:
# puts values_after('Blue', colors).join(', ')
The || result.length < 1 is a guard clause to prevent an infinite loop if target is not an element in arr.
after_blue.each{|item| print item } to give the result.
after_blue array contains only those elements after the "Blue" entry, and you can do whatever you want with them, however you want.|| after_blue.length < 1Although the two answers already present give the correct answer, I'd use a more functional style, which in my opinion is much more readable.
colors = ["Red", "Blue", "Green", "Purple", "White", "Black"]
# Drop all items until you find "Blue", then drop the "Blue".
colors.drop_while { |item| item != "Blue" }.drop(1).each do |item|
puts item
end
Since @knut mentioned benchmarks, this code is only a tiny bit slower than @knut's test/variable benchmark:
user system total real
test/variable 0.790000 0.000000 0.790000 ( 0.790960)
test/drop 0.900000 0.000000 0.900000 ( 0.898137)
