Ruby find number of occurance within range - ruby

I have the following data where each row tells me a start and finish time of a process.
I would like to know from 12:20:00 to 14:00:00 with a step of 5 mins, I'd like to know how many processes running at each time instance. For example, there are 2 and 1 processes running at 12:30 and 12:35 respectively.
I'd like to implement this in Ruby 1.8 and what's the efficient Rubyiest way of doing this?
12:28:08, 12:33:29
12:28:20, 12:33:41
12:32:32, 12:32:44
12:36:56, 12:42:31
13:08:55, 13:09:08
14:09:00, 14:09:12
14:59:19, 15:04:37
15:41:40, 15:41:52
(Comments)
Ps: I have already got an array for the start time, sTime and end time, eTime. I want to do something like this:
(sTime..eTime).step($time_interval) do |cTime| # Current Time
cnt = 0
(0..(sessionstarttime.length-1)).each {|i| if cTime.between? (sessionstarttime[i], sessionendtime[i]); cnt += 1}
printf "%s, %d\n", cTime.strftime("%d/%m/%Y %H:%M:%S"), cnt
end

You can try this code (developed on 1.9 but should work on 1.8 as well):
a = %Q{
12:28:08, 12:33:29
12:28:20, 12:33:41
12:32:32, 12:32:44
12:36:56, 12:42:31
13:08:55, 13:09:08
14:09:00, 14:09:12
14:59:19, 15:04:37
15:41:40, 15:41:52
}
start = '12:20:00'
stop = '14:00:00'
require 'stringio'
def time_to_sec(time)
a = time.split(':').map(&:to_i)
a[0] * 3600 + a[1] * 60 + a[2]
end
def sec_to_time(sec)
h, n = sec.divmod 3600
m, s = n.divmod 60
"%02d:%02d:%02d" % [h, m, s]
end
rows = StringIO.new(a).read.delete(",").split("\n").reject{ |i| i.empty? }.map do |range|
range.split.map{ |time| time_to_sec(time) }
end
ranges = rows.map{ |i| i[0]..i[1] }
(time_to_sec(start)..time_to_sec(stop)).step(5*60) do |time|
cnt = ranges.count{|i| i.include? time}
puts "#{sec_to_time(time)}: #{cnt}"
end
Of course you don't need 'a' variable or StringIO if working with real files.

If you convert the values to a Time object (note I've assumed a date of 2000-01-01 for this example), you can do the following:
a= [
{ :s=> Time.utc(2000, 1, 1, 12, 28, 8), :e=> Time.utc(2000, 1, 1, 12, 33, 29) },
{ :s=> Time.utc(2000, 1, 1, 12, 28, 20), :e=> Time.utc(2000, 1, 1, 12, 33, 41) },
{ :s=> Time.utc(2000, 1, 1, 12, 32, 32), :e=> Time.utc(2000, 1, 1, 12, 32, 44) },
{ :s=> Time.utc(2000, 1, 1, 12, 36, 56), :e=> Time.utc(2000, 1, 1, 12, 42, 31) },
{ :s=> Time.utc(2000, 1, 1, 13, 8, 55), :e=> Time.utc(2000, 1, 1, 13, 9, 8) },
{ :s=> Time.utc(2000, 1, 1, 14, 9, 0), :e=> Time.utc(2000, 1, 1, 14, 9, 12) },
{ :s=> Time.utc(2000, 1, 1, 14, 59, 19), :e=> Time.utc(2000, 1, 1, 15, 4, 37) },
{ :s=> Time.utc(2000, 1, 1, 15, 41, 40), :e=> Time.utc(2000, 1, 1, 15, 41, 52) }
]
checkTime = Time.utc(2000, 1, 1, 12, 32, 40)
a.delete_if{|b| #b[:s] is start time, b[:e] is end time
(b[:s] > checkTime) || (b[:e] < checkTime)
}

Here are a couple of simple objects that model something that should calculate what you need. This gives you a start to an interface you can use to do more complex logic if you need it.
require 'time'
# Object Definitions
class ProcessTimelineEntry
def initialize(start_time, end_time)
#start_time = start_time
#end_time = end_time
end
def running_at?(time)
time >= #start_time && time < #end_time
end
end
class ProcessTimeline
def initialize()
#entries = []
end
def add_entry(start_time, end_time)
#entries << ProcessTimelineEntry.new(start_time, end_time)
end
def process_count_at(time)
#entries.count { |e| e.running_at?(time) }
end
end
# Example Usage
timeline = ProcessTimeline.new
DATA.readlines.each do |line|
start_time, end_time = line.split(', ')
timeline.add_entry(Time.parse(start_time), Time.parse(end_time))
end
puts timeline.process_count_at(Time.parse("12:30"))
puts timeline.process_count_at(Time.parse("12:35"))
__END__
12:28:08, 12:33:29
12:28:20, 12:33:41
12:32:32, 12:32:44
12:36:56, 12:42:31
13:08:55, 13:09:08
14:09:00, 14:09:12
14:59:19, 15:04:37
15:41:40, 15:41:52

Here is a solution which will scale better to large numbers of start-stop pairs or time steps than the other posted answers (given that you want to know the number of processes running during each time step, not just 1 or 2 selected time steps):
START = Time.utc(2000,1,1, 12,20,0).to_i
FINISH = Time.utc(2000,1,1, 14,0,0).to_i
STEP = 60*5 # 5 minutes
result = Array.new(((FINISH-START).to_f/STEP).ceil, 0)
processes = %Q{
12:28:08, 12:33:29
12:28:20, 12:33:41
12:32:32, 12:32:44
12:36:56, 12:42:31
13:08:55, 13:09:08
14:09:00, 14:09:12
14:59:19, 15:04:37
15:41:40, 15:41:52 }
processes.each_line do |times|
times =~ /(\d\d):(\d\d):(\d\d), (\d\d):(\d\d):(\d\d)/
st = Time.utc(2000,1,1, $1.to_i,$2.to_i,$3.to_i).to_i
fin = Time.utc(2000,1,1, $4.to_i,$5.to_i,$6.to_i).to_i
st = START if st < START
fin = END if fin > END
(st..fin).step(STEP) do |t|
result[(t-START)/STEP] += 1
end
end
The count of how many processes were running during each time step will be left in result. You can put an object wrapper around it if desired to provide a nice interface.

Related

Need a Ruby method to convert a binary array to ASCII alpha-numeric string

I have an array of [1, 0, 11, 0, 4, 0, 106, 211, 169, 1, 0, 12, 0, 8, 0, 1, 26, 25, 32, 189, 77, 216, 1, 0, 1, 0, 4, 0, 0, 0, 0, 12, 15].
I would love to create a string version mostly for logging purposes. My end result would be "01000B0004006AD3..."
I could not find a simple way to take each array byte value and pack a string with an ASCII presentation of the byte value.
My solution is cumbersome. I appreciate advice on making the solution slick.
array.each {|x|
value = (x>>4)&0x0f
if( value>9 ) then
result_string.concat (value-0x0a + 'A'.ord).chr
else
result_string.concat (value + '0'.ord).chr
end
value = (x)&0x0f
if( value>9 ) then
result_string.concat (value-0x0a + 'A'.ord).chr
else
result_string.concat (value + '0'.ord).chr
end
}
Your question isn't very clear, but I guess something like this is what you are looking for:
array.map {|n| n.to_s(16).rjust(2, '0').upcase }.join
#=> "01000B0004006AD3A901000C000800011A1920BD4DD80100010004000000000C0F"
or
array.map(&'%02X'.method(:%)).join
#=> "01000B0004006AD3A901000C000800011A1920BD4DD80100010004000000000C0F"
Which one of the two is more readable depends on how familiar your readers are with sprintf-style format strings, I guess.
It's actually pretty simple:
def hexpack(data)
data.pack('C*').unpack('H*')[0]
end
That packs your bytes using integer values (C) and unpacks the resulting string to hex (H). In practice:
hexpack([1, 0, 11, 0, 4, 0, 106, 211, 169, 1, 0, 12, 0, 8, 0, 1, 26, 25, 32, 189, 77, 216, 1, 0, 1, 0, 4, 0, 0, 0, 0, 12, 15])
# => "01000b0004006ad3a901000c000800011a1920bd4dd80100010004000000000c0f"
I might suggest you stick to hex or base64 instead of making your own formatting
dat = [1, 0, 11, 0, 4, 0, 106, 211, 169, 1, 0, 12, 0, 8, 0, 1, 26, 25, 32, 189, 77, 216, 1, 0, 1, 0, 4, 0, 0, 0, 0, 12, 15]
Hexadecimal
hex = dat.map { |x| sprintf('%02x', x) }.join
# => 01000b0004006ad3a901000c000800011a1920bd4dd80100010004000000000c0f
Base64
require 'base64'
base64 = Base64.encode64(dat.pack('c*'))
# => AQALAAQAatOpAQAMAAgAARoZIL1N2AEAAQAEAAAAAAwP\n
Proquints
What? Proquints are pronounceable unique identifiers which makes them great for reading/communicating binary data. In your case, maybe not the best because you're dealing with 30+ bytes here, but they're very suitable for smaller byte strings
# proquint.rb
# adapted to ruby from https://github.com/deoxxa/proquint
module Proquint
C = %w(b d f g h j k l m n p r s t v z)
V = %w(a i o u)
def self.encode (bytes)
bytes << 0 if bytes.size & 1 == 1
bytes.pack('c*').unpack('S*').reduce([]) do |acc, n|
c1 = n & 0x0f
v1 = (n >> 4) & 0x03
c2 = (n >> 6) & 0x0f
v2 = (n >> 10) & 0x03
c3 = (n >> 12) & 0x0f
acc << C[c1] + V[v1] + C[c2] + V[v2] + C[c3]
end.join('-')
end
def decode str
# learner's exercise
# or see some proquint library (eg) https://github.com/deoxxa/proquint
end
end
Proquint.encode dat
# => dabab-rabab-habab-potat-nokab-babub-babob-bahab-pihod-bohur-tadot-dabab-dabab-habab-babab-babub-zabab
Of course the entire process is reversible too. You might not need it, so I'll leave it as an exercise for the learner
It's particularly nice for things like IP address, or any other short binary blobs. You gain familiarity too as you see common byte strings in their proquint form
Proquint.encode [192, 168, 11, 51] # bagop-rasag
Proquint.encode [192, 168, 11, 52] # bagop-rabig
Proquint.encode [192, 168, 11, 66] # bagop-ramah
Proquint.encode [192, 168, 22, 19] # bagop-kisad
Proquint.encode [192, 168, 22, 20] # bagop-kibid

Ruby 100 doors returning 100 nil

I'm solving the '100 doors' problem from Rosetta Code in Ruby. Briefly,
there are 100 doors, all closed, designated 1 to 100
100 passes are made, designated 1 to 100
on the ith pass, every ith door is "toggled": opened if it's closed, closed if it's open
determine the state of each door after 100 passes have been completed.
Therefore, on the first pass all doors are opened. On the second pass even numbered doors are closed. On the third pass doors i for which i%3 == 0 are toggled, and so on.
Here is my attempt at solving the problem.
visit_number = 0
door_array = []
door_array = 100.times.map {"closed"}
until visit_number == 100 do
door_array = door_array.map.with_index { |door_status, door_index|
if (door_index + 1) % (visit_number + 1) == 0
if door_status == "closed"
door_status = "open"
elsif door_status == "open"
door_status = "closed"
end
end
}
visit_number += 1
end
print door_array
But it keeps printing me an array of 100 nil when I run it: Look at all this nil !
What am I doing wrong?
That's what your if clauses return. Just add a return value explicitly.
until visit_number == 100 do
door_array = door_array.map.with_index { |door_status, door_index|
if (door_index + 1) % (visit_number + 1) == 0
if door_status == "closed"
door_status = "open"
elsif door_status == "open"
door_status = "closed"
end
end
door_status
}
visit_number += 1
end
OR:
1.upto(10) {|i| door_array[i*i-1] = 'open'}
The problem is the outer if block doesn't explicitly return anything (thus returns nil implicitly) when the condition does not meet.
A quick fix:
visit_number = 0
door_array = []
door_array = 100.times.map {"closed"}
until visit_number == 100 do
door_array = door_array.map.with_index { |door_status, door_index|
if (door_index + 1) % (visit_number + 1) == 0
if door_status == "closed"
door_status = "open"
elsif door_status == "open"
door_status = "closed"
end
else #<------------- Here
door_status #<------------- And here
end
}
visit_number += 1
end
print door_array
Consider these approaches:
door_array.map { |door|
case door
when "open"
"closed"
when "closed"
"open"
end
}
or
rule = { "open" => "closed", "closed" => "open" }
door_array.map { |door| rule[door] }
or
door_array.map { |door| door == 'open' ? 'closed' : 'open' }
Code
require 'prime'
def even_nbr_divisors?(n)
return false if n==1
arr = Prime.prime_division(n).map { |v,exp| (0..exp).map { |i| v**i } }
arr.shift.product(*arr).map { |a| a.reduce(:*) }.size.even?
end
closed, open = (1..100).partition { |n| even_nbr_divisors?(n) }
closed #=> [ 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22,
# 23, 24, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40,
# 41, 42, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 57,
# 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74,
# 75, 76, 77, 78, 79, 80, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91,
# 92, 93, 94, 95, 96, 97, 98, 99],
open #= [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
Explanation
All doors are initially closed. Consider the 24th door. It is toggled during the following passes:
pass 1: opened
pass 2: closed
pass 3: opened
pass 4: closed
pass 6: opened
pass 8: closed
pass 12: opened
pass 24: closed
Notice that the door is toggled once for each of 24's divisors. Therefore, if we had a method divisors(n) that returned an array of n's divisors, we could determine the number of toggles as follows:
nbr_toggles = divisors(24).size
#=> [1,2,3,4,6,8,12,24].size
#=> 8
Since the door is toggled an even number of times, we conclude that it will be in its original state (closed) after all the dust has settled. Similarly, for n = 9,
divisors(9).size
#=> [1,3,9].size
#=> 3
We therefore conclude door #9 will be open at the end, since 3 is odd.
divisors can be defined as follows.
def divisors(n)
arr = Prime.prime_division(n).map { |v,exp| (0..exp).map { |i| v**i } }
arr.first.product(*arr[1..-1]).map { |a| a.reduce(:*) }
end
For example,
divisors 24
#=> [1, 3, 2, 6, 4, 12, 8, 24]
divisors 9
#=> [1, 3, 9]
divisors 1800
#=> [1, 5, 25, 3, 15, 75, 9, 45, 225, 2, 10, 50, 6, 30, 150, 18, 90, 450,
# 4, 20, 100, 12, 60, 300, 36, 180, 900, 8, 40, 200, 24, 120, 600, 72,
# 360, 1800]
Since we only care if there are an odd or even number of divisors, we can instead write
def even_nbr_divisors?(n)
return false if n==1
arr = Prime.prime_division(n).map { |v,exp| (0..exp).map { |i| v**i } }
arr.shift.product(*arr).map { |a| a.reduce(:*) }.size.even?
end
For n = 24, the steps are as follows:
n = 24
a = Prime.prime_division(n)
#=> [[2, 3], [3, 1]]
arr = a.map { |v,exp| (0..exp).map { |i| v**i } }
#=> [[1, 2, 4, 8], [1, 3]]
b = arr.shift
#=> [1, 2, 4, 8]
arr
#=> [[1, 3]]
c = b.product(*arr)
#=> [[1, 1], [1, 3], [2, 1], [2, 3], [4, 1], [4, 3], [8, 1], [8, 3]]
d = c.map { |a| a.reduce(:*) }
#=> [1, 3, 2, 6, 4, 12, 8, 24]
e = d.size
#=> 8
e.even?
#=> true
Lastly,
(1..100).partition { |n| even_nbr_divisors?(n) }
returns the result shown above.

Ruby from-to-step Sequence

def generator(from, to, step)
ary = [from]
nex = from += step
min = from += step
while from != to
if from < to
from += step
ary.push(nex)
nex += step
elsif from > to
from -= step
ary.push(min)
min -= step
else
return nil
end
end
return ary
end
can Someone help explain to me why this only returns up to the 'to'element minus 2
for example when
generator(10,20,1) it will return [10,11,12..18] instead of going all the way to 20
Change
nex = from += step
min = from += step
To
nex = from + step
min = from + step
Your from is already being incremented twice with the step because of that (so it loops less than intended).
If you want to write it in a one-liner you could do something like this using Numeric#step
2.3.0 > 10.step(20).to_a
#=> [10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]
2.3.0 > 20.step(10, -1).to_a
#=> [20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10]

Array deducting 1 from previous element when repeated more than once

I have the following array:
Input:
array = [211, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200]
Ouput:
array = [211, 200, 199, 198, 197, 196 ... ]
I've tried each_with_index but couldn't get the desired result.
I don't understand what are to be done with nils, so I haven't addressed that. Let arr be array or array.sort.reverse, depending on requirements. I think this is want you want to do? (See my comment on the question.)
def change_em(arr)
dup_indices = arr.each_index
.group_by { |i| arr[i] }
.values
.flat_map { |a| a.drop(1) }
puts "dup_indices = #{dup_indices}"
last = 0 # anything '-' responds to
arr.each_index.map { |i| last = dup_indices.include?(i) ? last-1 : arr[i] }
end
I've included the puts just to clarify what I'm doing here.
change_em [10, 8, 5, 5, 7]
#=> dup_indices = [3]
#=> [10, 8, 5, 4, 7]
change_em [10, 8, 7, 5, 5]
#=> dup_indices = [4]
#=> [10, 8, 7, 5, 4]
change_em [10, 9, 9, 8, 8, 8]
#=> dup_indices = [2, 4, 5]
#=> [10, 9, 8, 8, 7, 6]
change_em [211, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 196]
#=> dup_indices = [2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
#=> [211, 200, 199, 198, 197, 196, 195, 194, 193, 192, 191, 190, 196]
Notice that the statement
last = dup_indices.include?(i) ? last-1 : arr[i]
is doing double-duty: it updates the value of last and returns the mapped value for the index i. Note also that dup_indices cannot contain 0.
Not sure I fully understand your requirements, but here's my attempt:
# Transforms an array of numbers into a sorted array of the same length, where
# each successive element is always smaller than the preceding element.
def force_descending(array)
array.sort.reverse.each_with_object([]) do |element, collection|
collection << if collection.empty? || element < collection.last
element
else
collection.last-1
end
end
end
Sample inputs/outputs:
force_descending [211, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200]
#=> [211, 200, 199, 198, 197, 196, 195, 194, 193, 192, 191, 190]
force_descending [10, 8, 5, 5, 7]
#=> [10, 8, 7, 5, 4]
force_descending [211, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 196]
#=> [211, 200, 199, 198, 197, 196, 195, 194, 193, 192, 191, 190, 189]
Wrote this is a more functional style.
def f(arr, dup_element = nil, dup_count = 0)
return generate_dup_array(dup_element, dup_count) if arr.empty?
if arr.head != arr.tail.head # Not duplicates
if dup_count == 0 # No duplicates to insert
[arr.head] + f(arr.tail)
else # There are duplicates to insert
generate_dup_array(dup_element, dup_count) + f(arr.tail)
end
else # Duplicate found, continue with tail of array and increase dup_count
f(arr.tail, arr.head, dup_count + 1)
end
end
def generate_dup_array(dup_element, dup_count)
return [] if dup_count == 0
(dup_element - dup_count..dup_element).to_a.reverse
end
class Array
def head; self.first; end
def tail; self[1..-1]; end
end
p f [211, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200]
# => [211, 200, 199, 198, 197, 196, 195, 194, 193, 192, 191, 190]
p f [10, 8, 5, 5, 7].sort.reverse
# => [10, 8, 7, 5, 4]
p f [9, 6, 6, 5, 5, 4, 4, 3, 3, 3, 2, 2, 1]
# => [9, 6, 5, 5, 4, 4, 3, 3, 2, 1, 2, 1, 1]
its already in decsending order
For the one-liner crowd:
results = numbers.chunk {|num| num}.flat_map {|num, group| (group.length == 1) ? num : ((num - (group.length-1))..num).to_a.reverse}
For sane programmers:
numbers = [211, 200, 200, 200]
start_of_dups = "_START_" #Something not in the array
dup_count = 0
results = numbers.map do |num|
if start_of_dups == num
dup_count += 1
num - dup_count
else
dup_count = 0
start_of_dups = num
end
end
p results
--output:--
[211, 200, 199, 198]
But if:
array = [10, 10, 10, 9]
--output:--
[10, 9, 8, 9]

How can I convert this array so that each element represents the cumulative value of the previous elements? [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Cumulative array sum in Ruby
I have an array of integers like this
[20, 25, 40, 60]
How can I turn it into an array with each element representing the cumulative value of the elements before it, including itself?
[20, 45, 85, 145]
I'm using Rails 3.2.0 & ruby 1.9.3
s = 0
[20, 25, 40, 60].map{|e| s += e}
[20, 25, 40, 60].reduce([]) do |arr, v|
arr << (arr.last || 0) + v
end
Or an ugly one liner.
[20, 25, 40, 60].reduce([0]){ |a, v| a << a[-1] + v }[1..-1]
array = [20, 25, 40, 60]
(array.size - 1).times { |i| array[i + 1] += array[i] }
puts array
# => [20, 45, 85, 145]
arr = [20, 25, 40, 60]
first = []
sum = 0
arr.each do |e|
sum += e
first << sum
end
puts first
arr.each_with_index.map{|x, i| x + (i==0 ? 0 : arr[0..i-1].inject(:+))}
=> [20, 45, 85, 145]
Matlab:
B = cumsum(A)
Ruby:
class Array
def ruby_cumsum!
(1..size-1).each {|i| self[i] += self[i-1] }
self
end
end

Resources