**I wanted to find the prevalence of smoking in urban and in rural area.
I used this code**
I found this result:
We see that it results a little different, calculating all responses a whole. How can I use svyciprop especificially for one type of response. For example, how can I get result the prevalence of smoking in only rural area or in only urban area?
looks like smoking is zero/one and urban_rural is either 'urban' or 'rural' so
svyciprop( ~ smoking , subset( svs1 , urban_rural == 'rural' ) )
svyciprop( ~ smoking , subset( svs1 , urban_rural == 'urban' ) )
Related
is there anybody can tell me how to define the formula in the rweka?
A<- InfoGainAttributeEval(formula ~ . , data = TrainDataLSVT,na.action=NULL )
there are 310 features in the TrainDataLSVT.
Since you do not provide your data, I will illustrate with the built-in iris data (see ?iris). For this data, the goal is to predict the Species as a function of the other variables. You can express that as a formula for InfoGainAttributeEval like this:
InfoGainAttributeEval(Species ~ ., data=iris)
Sepal.Length Sepal.Width Petal.Length Petal.Width
0.6982615 0.3855963 1.4180030 1.3784027
The values returned are the scores for each variable. The key part is the formula Species ~ . You should read this as "Species as a function of all other variables". Details on how to write a formula are available on the help page ?formula.
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.
I know nothing about Lua, but I was able to modify a script I wanted to. I'm having troubles sorting a table though.
I've found table utils (convert table to string), here's my table:
{{line="(Golden Aura) Challenging An owl would be either very brave or very stupid.",range="(+16 to +21)",message="(Golden Aura) Challenging An owl would be either very brave or very stupid.",colour="crimson",srt=9,keyword="owl",name="An owl"},
{line="(Golden Aura) A busy squirrel chuckles at the thought of you fighting him.",range="(+3 to +8)",message="(Golden Aura) A busy squirrel chuckles at the thought of you fighting him.",colour="gold",srt=7,keyword="squirrel",name="(Golden Aura) A busy squirrel"},
{line="(Red Aura) A parakeet should be a fair fight!",range="(-2 to +2)",message="(Red Aura) A parakeet should be a fair fight!",colour="springgreen",srt=5,keyword="parakeet",name="(Red Aura) A parakeet"},
{line="(Golden Aura) Challenging A cat would be either very brave or very stupid.",range="(+16 to +21)",message="(Golden Aura) Challenging A cat would be either very brave or very stupid.",colour="crimson",srt=9,keyword="cat",name="A cat"}}
I was able to add srt key and I want to sort a table by that. Can someone kind tell me how to do that, please?
table.sort( table:t [, function( left, right ):sorting function ] )
So, since you want to sort by v.srt, you would do something like:
table.sort( t, function( a, b ) return a.srt < b.srt end )
for k, v in pairs( t ) do
print( v.srt, v.name )
end
Which should sort them in ascending order and then display them.
I'm thinking about an algorithm that will create X most unique concatenations of Y parts, where each part can be one of several items. For example 3 parts:
part #1: 0,1,2
part #2: a,b,c
part #3: x,y,z
And the (random, one case of some possibilities) result of 5 concatenations:
0ax
1by
2cz
0bz (note that '0by' would be "less unique " than '0bz' because 'by' already was)
2ay (note that 'a' didn't after '2' jet, and 'y' didn't after 'a' jet)
Simple BAD results for next concatenation:
1cy ('c' wasn't after 1, 'y' wasn't after 'c', BUT '1'-'y' already was as first-last
Simple GOOD next result would be:
0cy ('c' wasn't after '0', 'y' wasn't after 'c', and '0'-'y' wasn't as first-last part)
1az
1cx
I know that this solution limit possible results, but when all full unique possibilities will gone, algorithm should continue and try to keep most avaible uniqueness (repeating as few as possible).
Consider real example:
Boy/Girl/Martin
bought/stole/get
bottle/milk/water
And I want results like:
Boy get milk
Martin stole bottle
Girl bought water
Boy bought bottle (not water, because of 'bought+water' and not milk, because of 'Boy+milk')
Maybe start with a tree of all combinations, but how to select most unique trees first?
Edit: According to this sample data, we can see, that creation of fully unique results for 4 words * 3 possibilities, provide us only 3 results:
Martin stole a bootle
Boy bought an milk
He get hard water
But, there can be more results requested. So, 4. result should be most-available-uniqueness like Martin bought hard milk, not Martin stole a water
Edit: Some start for a solution ?
Imagine each part as a barrel, wich can be rotated, and last item goes as first when rotates down, first goes as last when rotating up. Now, set barells like this:
Martin|stole |a |bootle
Boy |bought|an |milk
He |get |hard|water
Now, write sentences as We see, and rotate first barell UP once, second twice, third three and so on. We get sentences (note that third barell did one full rotation):
Boy |get |a |milk
He |stole |an |water
Martin|bought|hard|bootle
And we get next solutions. We can do process one more time to get more solutions:
He |bought|a |water
Martin|get |an |bootle
Boy |stole |hard|milk
The problem is that first barrel will be connected with last, because rotating parallel.
I'm wondering if that will be more uniqe if i rotate last barrel one more time in last solution (but the i provide other connections like an-water - but this will be repeated only 2 times, not 3 times like now). Don't know that "barrels" are good way ofthinking here.
I think that we should first found a definition for uniqueness
For example, what is changing uniqueness to drop ? If we use word that was already used ? Do repeating 2 words close to each other is less uniqe that repeating a word in some gap of other words ? So, this problem can be subjective.
But I think that in lot of sequences, each word should be used similar times (like selecting word randomly and removing from a set, and after getting all words refresh all options that they can be obtained next time) - this is easy to do.
But, even if we get each words similar number od times, we should do something to do-not-repeat-connections between words. I think, that more uniqe is repeating words far from each other, not next to each other.
Anytime you need a new concatenation, just generate a completely random one, calculate it's fitness, and then either accept that concatenation or reject it (probabilistically, that is).
const C = 1.0
function CreateGoodConcatenation()
{
for (rejectionCount = 0; ; rejectionCount++)
{
candidate = CreateRandomConcatination()
fitness = CalculateFitness(candidate) // returns 0 < fitness <= 1
r = GetRand(zero to one)
adjusted_r = Math.pow(r, C * rejectionCount + 1) // bias toward acceptability as rejectionCount increases
if (adjusted_r < fitness)
{
return candidate
}
}
}
CalculateFitness should never return zero. If it does, you might find yourself in an infinite loop.
As you increase C, less ideal concatenations are accepted more readily.
As you decrease C, you face increased iterations for each call to CreateGoodConcatenation (plus less entropy in the result)
I am doing a CSV Import tool for the project I'm working on.
The client needs to be able to enter the data in excel, export them as CSV and upload them to the database.
For example I have this CSV record:
1, John Doe, ACME Comapny (the typo is on purpose)
Of course, the companies are kept in a separate table and linked with a foreign key, so I need to discover the correct company ID before inserting.
I plan to do this by comparing the company names in the database with the company names in the CSV.
the comparison should return 0 if the strings are exactly the same, and return some value that gets bigger as the strings get more different, but strcmp doesn't cut it here because:
"Acme Company" and "Acme Comapny" should have a very small difference index, but
"Acme Company" and "Cmea Mpnyaco" should have a very big difference index
Or "Acme Company" and "Acme Comp." should also have a small difference index, even though the character count is different.
Also, "Acme Company" and "Company Acme" should return 0.
So if the client makes a type while entering data, i could prompt him to choose the name he most probably wanted to insert.
Is there a known algorithm to do this, or maybe we can invent one :)
?
You might want to check out the Levenshtein Distance algorithm as a starting point. It will rate the "distance" between two words.
This SO thread on implementing a Google-style "Do you mean...?" system may provide some ideas as well.
I don't know what language you're coding in, but if it's PHP, you should consider the following algorithms:
levenshtein(): Returns the minimal number of characters you have to replace, insert or delete to transform one string into another.
soundex(): Returns the four-character soundex key of a word, which should be the same as the key for any similar-sounding word.
metaphone(): Similar to soundex, and possibly more effective for you. It's more accurate than soundex() as it knows the basic rules of English pronunciation. The metaphone generated keys are of variable length.
similar_text(): Similar to levenshtein(), but it can return a percent value instead.
I've had some success with the Levenshtein Distance algorithm, there is also Soundex.
What language are you implementing this in? we may be able to point to specific examples
I have actually implemented a similar system. I used the Levenshtein distance (as other posters already suggested), with some modifications. The problem with unmodified edit distance (applied to whole strings) is that it is sensitive to word reordering, so "Acme Digital Incorporated World Company" will match poorly against "Digital Incorporated World Company Acme" and such reorderings were quite common in my data.
I modified it so that if the edit distance of whole strings was too big, the algorithm fell back to matching words against each other to find a good word-to-word match (quadratic cost, but there was a cutoff if there were too many words, so it worked OK).
I've taken SoundEx, Levenshtein, PHP similarity, and double metaphone and packaged them up in C# in one set of extension methods on String.
Entire blog post here.
There's multiple algorithms to do just that, and most databases even include one by default. It is actually a quite common concern.
If its just about English words, SQL Server for example includes SOUNDEX which can be used to compare on the resulting sound of the word.
http://msdn.microsoft.com/en-us/library/aa259235%28SQL.80%29.aspx
I'm implementing it in PHP, and I am now writing a piece of code that will break up 2 strings in words and compare each of the words from the first string with the words of the second string using levenshtein and accept the lowes possible values. Ill post it when I'm done.
Thanks a lot.
Update: Here's what I've come up with:
function myLevenshtein( $str1, $str2 )
{
// prepare the words
$words1 = explode( " ", preg_replace( "/\s+/", " ", trim($str1) ) );
$words2 = explode( " ", preg_replace( "/\s+/", " ", trim($str2) ) );
$found = array(); // array that keeps the best matched words so we don't check them again
$score = 0; // total score
// In my case, strings that have different amount of words can be good matches too
// For example, Acme Company and International Acme Company Ltd. are the same thing
// I will just add the wordcount differencre to the total score, and weigh it more later if needed
$wordDiff = count( $words1 ) - count( $words2 );
foreach( $words1 as $word1 )
{
$minlevWord = "";
$minlev = 1000;
$return = 0;
foreach( $words2 as $word2 )
{
$return = 1;
if( in_array( $word2, $found ) )
continue;
$lev = levenshtein( $word1, $word2 );
if( $lev < $minlev )
{
$minlev = $lev;
$minlevWord = $word2;
}
}
if( !$return )
break;
$score += $minlev;
array_push( $found, $minlevWord );
}
return $score + $wordDiff;
}