Separate characters and numbers following specific rules - ruby

I am trying to distinguish flight numbers.
Example:
flightno = "FR556"
split_data = flightno.upcase.match(/([A-Za-z]+)(\d+)/)
first = split_data[1] # FR
second = split_data[1] # 556
I then go on to query the database to find an airline based on the FR in this example and apply some logic with the result which is Ryanair.
My problem is when the flight number might be:
flightno = "U21920"
split_data = flightno.upcase.match(/([A-Za-z]+)(\d+)/)
first = split_data[1] # U
second = split_data[1] # 21920
i basically want first to be U2 not just U. This is used to search the database of airlines by their IATA code in this case is U2
****EDIT**
In the interest of clarity i made some mistakes in terminology when asking my question. Due to the complexities of booking reference numbers, the input is taken from whatever the passenger provides. For an easyJet flight for example, the passenger may input EZY1920 or U21920 only the airline provides either so the passenger is ignorant really.
"EZY" = ICAO
"U2" = IATA
I take the input from the user and try to separate the ICAO or IATA from the flight number "1920" but there is no way of determining that without searching the database or separating the input which i feel is cumbersome from a user experience point of view.
Using a regex to separate characters from numbers works until the user inputs an IATA as part of their flight number (the passenger won't know the difference) and as you can see in the example above this confuses the regex.**
The trouble is i cant think of any other pattern with flight numbers. They always have at least two characters made up of just letters or a mixture of a letter and a number and can be 3 characters in length. The numbers part can be as short as 1 but can also be as long as 4 - always numbers.
****edit**
As has been mentioned in the comments, there is no fixed size however one thing that is always true (at least so far) is the first character will always be a letter regardless if it is ICAO or IATA.
After considering every bodies input so far i'm wondering if searching the database and returning airlines with an IATA or ICAO that matches the first two letters provided by the user (U2), (FR), (EZ) might be one way to go, however this is subject to obvious problems should an ICAO or IATA be released that matches another airline, for example "EZY" & "EZT". This is not future proof and i'm looking for better ruby or regex solutions.**
Appreciate your input.
EDIT
I have answered my own question below. While other answers provide a solution for handling some conditions they would fall down if the flight number began with a number so i worked out a crass but to date stable way to analyse the string for digits and then work out if it is an ICAO or IATA from that.

A solution I think of is that you match your given flight number against a complete list of ICAO/IATA codes: https://raw.githubusercontent.com/datasets/airport-codes/master/data/airport-codes.csv
Spending some time with google might give you a more appropriate list.
Then use the first three characters (if that is the maximum) of your flight number to find a match within the icao codes. If you find one, you will know where to seperate your string.
Here a minimal ugly example that should set you on a track. Feel free to update!
ICAOCODES = %w(FR DEU U21) # grab your data here
def retrieve_flight_information(flightnumber)
ICAOCODES.each do |icao|
co = flightnumber.match(icao).to_s
if co.length > 0
# airline
puts co
# flight number
puts flightnumber.gsub(co,'')
end
end
end
retrieve_flight_information("FR556")
#=> FR
#=> 556
retrieve_flight_information("U21214123")
#=> U21
#=> 214123
The biggest flaw lies in using .gsub() as it might mess up your flightnumber in case it looks like this: "FR21413FR2"
However you will find plenty of solutions to this problem on so.
As mentioned in the comments, a list of icao codes is not what you are looking for. But what is relevant here, is that you somehow need a list of strings that you can securely compare against.

I have a fairly crass solution that seems to be working in all scenarios i can throw at it to date. I wanted to make this available to anybody else that might find it useful?
The general rule of thumb for flight codes/numbers seems to be:
IATA: two characters made up of any combination letters and digits
ICAO: three characters made up of letters only (to date)
With that in mind we should be able to work out if we need to search the database by IATA or ICAO depending on the condition of the first three characters.
First we take the flight number and convert to uppercase
string = "U21920".upcase
Next we analyse the first three characters to check for any numbers.
first_three = string[0,3] # => U21
Is there a digit in first_three?
if first_three =~ /\d/ # => true
iata = first_three[0,2] # => If true lets get rid of the last character
# Now we go to the database searching IATA (U2)
search = Airline.where('iata LIKE ?', "#{iata}%") # => Starts with search, just in case
Otherwise if there isnt a digit found in the string
else
icao = string.match(/([A-Za-z]+)(\d+)/)
search = Airline.where('icao LIKE ?', "#{icao[1]}%")
This seems to work for the random flight numbers ive tested it with today from a few of the major airport live departure/arrival boards. Its an interesting problem because some airlines issue tickets with either an ICAO or IATA code as part of the flight number which means passengers won't know any different, not to mention, some airports provide flight information in their own format so assumign there isnt a change to the ICAO and IATA build then the above should work.
Here is an example script you can run
test.rb
puts "What is your flight number?"
string = gets.upcase
first_three = string[0,3]
puts "Taking first three from #{string} is #{first_three}"
if first_three =~ /\d/ # Calling String's =~ method.
puts "The String #{first_three} DOES have a number in it."
iata = first_three[0,2]
search = Airline.where('iata LIKE ?', "#{iata}%")
puts "Searching Airlines starting with IATA #{iata} = #{search.count}"
puts "Found #{search.first.name} from IATA #{iata}"
else
puts "The String #{first_three} does not have a number in it."
icao = string.match(/([A-Za-z]+)(\d+)/)
search = Airline.where('icao LIKE ?', "#{icao[1]}%")
puts "Searching Airlines starting with ICAO #{icao[1]} = #{search.count}"
puts "Found #{search.first.name} from IATA #{icao[1]}"
end
Airline
Airline(id: integer, name: string, iata: string, icao: string, created_at: datetime, updated_at: datetime )
stick this in your lib folder and run
rails runner lib/test.rb
Obviously you can remove all of the puts statements to get straight to the result. I'm using rails runner to include access to my Airline model when running the script.

Related

Python Birthday paradox math not working

it run corectly but it should have around 500 matches but it only has around 50 and I dont know why!
This is a probelm for my comsci class that I am having isues with
we had to make a function that checks a list for duplication I got that part but then we had to apply it to the birthday paradox( more info here http://en.wikipedia.org/wiki/Birthday_problem) thats where I am runing into problem because my teacher said that the total number of times should be around 500 or 50% but for me its only going around 50-70 times or 5%
duplicateNumber=0
import random
def has_duplicates(listToCheck):
for i in listToCheck:
x=listToCheck.index(i)
del listToCheck[x]
if i in listToCheck:
return True
else:
return False
listA=[1,2,3,4]
listB=[1,2,3,1]
#print has_duplicates(listA)
#print has_duplicates(listB)
for i in range(0,1000):
birthdayList=[]
for i in range(0,23):
birthday=random.randint(1,365)
birthdayList.append(birthday)
x= has_duplicates(birthdayList)
if x==True:
duplicateNumber+=1
else:
pass
print "after 1000 simulations with 23 students there were", duplicateNumber,"simulations with atleast one match. The approximate probibilatiy is", round(((duplicateNumber/1000)*100),3),"%"
This code gave me a result in line with what you were expecting:
import random
duplicateNumber=0
def has_duplicates(listToCheck):
number_set = set(listToCheck)
if len(number_set) is not len(listToCheck):
return True
else:
return False
for i in range(0,1000):
birthdayList=[]
for j in range(0,23):
birthday=random.randint(1,365)
birthdayList.append(birthday)
x = has_duplicates(birthdayList)
if x==True:
duplicateNumber+=1
print "after 1000 simulations with 23 students there were", duplicateNumber,"simulations with atleast one match. The approximate probibilatiy is", round(((duplicateNumber/1000.0)*100),3),"%"
The first change I made was tidying up the indices you were using in those nested for loops. You'll see I changed the second one to j, as they were previously bot i.
The big one, though, was to the has_duplicates function. The basic principle here is that creating a set out of the incoming list gets the unique values in the list. By comparing the number of items in the number_set to the number in listToCheck we can judge whether there are any duplicates or not.
Here is what you are looking for. As this is not standard practice (to just throw code at a new user), I apologize if this offends any other users. However, I believe showing the OP a correct way to write a program should be could all do us a favor if said user keeps the lack of documentation further on in his career.
Thus, please take a careful look at the code, and fill in the blanks. Look up the python doumentation (as dry as it is), and try to understand the things that you don't get right away. Even if you understand something just by the name, it would still be wise to see what is actually happening when some built-in method is being used.
Last, but not least, take a look at this code, and take a look at your code. Note the differences, and keep trying to write your code from scratch (without looking at mine), and if it messes up, see where you went wrong, and start over. This sort of practice is key if you wish to succeed later on in programming!
def same_birthdays():
import random
'''
This is a program that does ________. It is really important
that we tell readers of this code what it does, so that the
reader doesn't have to piece all of the puzzles together,
while the key is right there, in the mind of the programmer.
'''
count = 0
#Count is going to store the number of times that we have the same birthdays
timesToRun = 1000 #timesToRun should probably be in a parameter
#timesToRun is clearly defined in its name as well. Further elaboration
#on its purpose is not necessary.
for i in range(0,timesToRun):
birthdayList = []
for j in range(0,23):
random_birthday = random.randint(1,365)
birthdayList.append(random_birthday)
birthdayList = sorted(birthdayList) #sorting for easier matching
#If we really want to, we could provide a check in the above nester
#for loop to check right away if there is a duplicate.
#But again, we are here
for j in range(0, len(birthdayList)-1):
if (birthdayList[j] == birthdayList[j+1]):
count+=1
break #leaving this nested for-loop
return count
If you wish to find the percent, then get rid of the above return statement and add:
return (count/timesToRun)
Here's a solution that doesn't use set(). It also takes a different approach with the array so that each index represents a day of the year. I also removed the hasDuplicate() function.
import random
sim_total=0
birthdayList=[]
#initialize an array of 0's representing each calendar day
for i in range(365):
birthdayList.append(0)
for i in range(0,1000):
first_dup=True
for n in range(365):
birthdayList[n]=0
for b in range(0, 23):
r = random.randint(0,364)
birthdayList[r]+=1
if (birthdayList[r] > 1) and (first_dup==True):
sim_total+=1
first_dup=False
avg = float(sim_total) / 1000 * 100
print "after 1000 simulations with 23 students there were", sim_total,"simulations with atleast one duplicate. The approximate problibility is", round(avg,3),"%"

How to match between two arrays and update one based on criteria

I'm trying to match two supplier csv's and update one based on the results of the other; things like if price is different, update one file with the matching item of the other. If the product is in the first csv but not in the other, update it. Once the data set is adjusted, I'll write it back to the csv which I'm ok with. Each supplier file is about 9000 lines long. Sample data from the two Puts lines in the code are:
#<struct RecordBUY item_type=nil, buy_product_id="1000", product_name="Plastic Jeweled Crown", product_type=nil, product_code_SKU="105238", option_set=nil, duplicate={"1000"=>["105238"]}, brand_name="Rubies Costumes", prod_desc="This plastic crown has six large jewel stones accross the top. Adjustable headband. (Colors of the jewel stones may vary, our choice please.)", cost_price="$3.76", prod_weight="00.14", prod_width="5.75", prod_height="0.5", prod_depth="23.5", prod_category="Hats, Wigs & Masks", prod_upn="082686025935", prod_size="One Size", prod_color="Gold">
#<struct BCRecord item_type="Product", bc_product_id="620", product_name="Dollar Ring", product_type=nil, product_code_SKU="109624", option_set=nil, duplicate=nil, brand_name="Rubies Costumes", prod_desc="Ring has three large glittery Dollar Signs '$' that extend over your fingers.", cost_price="3.20", prod_weight="0.7200", prod_width="4.0000", prod_height="1.0000", prod_depth="7.0000", prod_category="Accessories & Makeup", prod_upn="82686006996", prod_size=nil, prod_color=nil, option_set=nil, price="5.60", allow_purchases=[21]>
I read the csv data into arrays against respective objects, but don't know how to do searching and updating efficiently. I did not come across concepts to avoid the bad ones (or whether doing a bad one on 9k lines is actually bad or just frowned upon). What I have is:
puts records[0]
puts recordsBC[1]
#start script
records.each do | buyline |
recordsBC.each do | bcline |
if bcline.product_code_SKU == buyline.product_code_SKU
##update pricing (brute force);
#bcline.price = buyline.cost_price * 1.75 #this fails with undefined method `price=' for #<Record:0x007fbb9088b960>
bcline.cost_price = buyline.cost_price
end
##if product is in BC currently, but not in buy - needs to be marked as inactive in BC
if bcline.product_code_SKU.include? buyline.product_code_SKU
#bcline.allow_purchases = "N" # this fails with undefined method `allow_purchases=' for #<Record:0x007fb2878822c8>
end
#if product is in Buy but not in BC then add it into BC
if buyline.product_code_SKU.include? bcline.product_code_SKU
recordsBC.push buyline
end
end
end
I can't figure out a better way, nor understand why I'm getting the undefined method errors on some but not all lines. I'm not after complete answers, just enough to figure out the rest of the solution.
I'd start by reducing the number of iterations. At the moment you are iterating through all of recordsBC for each buyline. So I'd start with:
records.each do | buyline |
record_subset = recordsBC.select{|r|!(r.product_code_SKU.split & buyling.product_code_SKU.split).empty?}
record_subset.each do |bcline|
.....
end
end
That should mean you only iterate through bcline items that have a matching product_code_SKU. You may have to modify the split as your example doesn't show how multiple SKUs are separated (e.g. '123 456', '123,456', or '123/456')

How to Connect Logic with Objects

I have a system that contains x number of strings. These string are shown in a UI based on some logic. For example string number 1 should only show if the current time is past midday and string 3 only shows if a randomly generated number between 0-1 is less than 0.5.
How would be the best way to model this?
Should the logic just be in code and be linked to a string by some sort or ID?
Should the logic be some how stored with the strings?
NOTE The above is a theoretical example before people start questioning my logic.
It's usually better to keep resources (such as strings) separate from logic. So referring strings by IDs is a good idea.
It seems that you have a bunch of rules which you have to link to the display of strings. I'd keep all three as separate entities: rules, strings, and the linking between them.
An illustration in Python, necessarily simplified:
STRINGS = {
'morning': 'Good morning',
'afternoon': 'Good afternoon',
'luck': 'you must be lucky today',
}
# predicates
import datetime, random
def showMorning():
return datetime.datetime.now().hour < 12
def showAfternoon():
return datetime.datetime.now().hour >= 12
def showLuck():
return random.random() > 0.5
# interconnection
RULES = {
'morning': showMorning,
'afternoon': showAfternoon,
'luck': showLuck,
}
# usage
for string_id, predicate in RULES.items():
if predicate():
print STRINGS[string_id]

English words (ONE, TWO, THREE) to number(1,2,3) in ruby

I am trying to convert words/strings to numbers in Ruby for example:-
ONE => 1
TWO => 2
THREE => 3
FOUR => 4
etc...
I have seen many examples other way around (numbers to words); however, I am not been able to find an example of how to convert words to numbers. It would be great if I can get any insight or help on this.
This might help: http://www.rubyquiz.com/quiz25.html
EDIT: read the question incorrectly. As per my comment below, here's a way you could approach this.
If num_to_word(number) is your method of converting from a number to a word:
def number_hash_creator(min, max)
number_hash = {}
for num in (min..max)
number_hash[num_to_word(num)] == num
end
number_hash
end
Then do something like:
number_hash = number_hash_creator(min, max) # min and max are whatever you need them to be
number_hash['three']
=> 3
You could also append a similar method to the string class so that you could do things like "three".to_number
I created a hash up-to twenty for all words and only for tens(thirty, forty, fifty etc.). Using Reg-ex took off two words and added them for example twenty two is addition of 20+2=22; right now my script only works till hundred, but it can be extended for numbers over 100 and so on.
I think you will find this tutorial very interesting particularly the code near the bottom:
http://pine.fm/LearnToProgram/?Chapter=08

Ruby, Count syllables

I am using ruby to calculate the Gunning Fog Index of some content that I have, I can successfully implement the algorithm described here:
Gunning Fog Index
I am using the below method to count the number of syllables in each word:
Tokenizer = /([aeiouy]{1,3})/
def count_syllables(word)
len = 0
if word[-3..-1] == 'ing' then
len += 1
word = word[0...-3]
end
got = word.scan(Tokenizer)
len += got.size()
if got.size() > 1 and got[-1] == ['e'] and
word[-1].chr() == 'e' and
word[-2].chr() != 'l' then
len -= 1
end
return len
end
It sometimes picks up words with only 2 syllables as having 3 syllables. Can anyone give any advice or is aware of a better method?
text = "The word logorrhoea is often used pejoratively to describe prose that is highly abstract and contains little concrete language. Since abstract writing is hard to visualize, it often seems as though it makes no sense and all the words are excessive. Writers in academic fields that concern themselves mostly with the abstract, such as philosophy and especially postmodernism, often fail to include extensive concrete examples of their ideas, and so a superficial examination of their work might lead one to believe that it is all nonsense."
# used to get rid of any puncuation
text = text.gsub!(/\W+/, ' ')
word_array = text.split(' ')
word_array.each do |word|
puts word if count_syllables(word) > 2
end
"themselves" is being counted as 3 but it's only 2
The function I give you before is based upon these simple rules outlined here:
Each vowel (a, e, i, o, u, y) in a
word counts as one syllable subject to
the following sub-rules:
Ignore final -ES, -ED, -E (except
for -LE)
Words of three letters or
less count as one syllable
Consecutive vowels count as one
syllable.
Here's the code:
def new_count(word)
word.downcase!
return 1 if word.length <= 3
word.sub!(/(?:[^laeiouy]es|ed|[^laeiouy]e)$/, '')
word.sub!(/^y/, '')
word.scan(/[aeiouy]{1,2}/).size
end
Obviously, this isn't perfect either, but all you'll ever get with something like this is a heuristic.
EDIT:
I changed the code slightly to handle a leading 'y' and fixed the regex to handle 'les' endings better (such as in "candles").
Here's a comparison using the text in the question:
# used to get rid of any puncuation
text = text.gsub!(/\W+/, ' ')
words = text.split(' ')
words.each do |word|
old = count_syllables(word.dup)
new = new_count(word.dup)
puts "#{word}: \t#{old}\t#{new}" if old != new
end
The output is:
logorrhoea: 3 4
used: 2 1
makes: 2 1
themselves: 3 2
So it appears to be an improvement.
One thing you ought to do is teach your algorithm about diphthongs. If I'm reading your code correctly, it would incorrectly flag "aid" as having two syllables.
You can also add "es" and the like to your special-case endings (you already have "ing") and just not count it as a syllable, but that might still result in some miscounts.
Finally, for best accuracy, you should convert your input to a spelling scheme or alphabet that has a definite relationship to the word's pronunciation. With your "themselves" example, the algorithm has no reliable way to know that the "e" "ves" is dropped. However, if you respelled it as "themselvz", or taught the algorithm the IPA and fed it [ðəmsɛlvz], it becomes very clear that the word is only pronounced with two syllables. That, of course, assumes you have control over the input, and is probably more work than just counting the syllables yourself.
To begin with it seems you should decrement len for the suffixes that should be excluded.
len-=1 if /.*[ing,es,ed]$/.match(word)
You could also check out Lingua::EN::Readability.
It can also calculate several readability measures, such as a Fog Index and a Flesch-Kincaid level.
PS. I think I know where you got the function from. DS.
There is also a rubygem called Odyssey that calculates Gunning Fog, along with some of the other popular ones (Flesch-Kincaid, SMOG, etc.)

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