I have a Clojure function autocomplete that takes in a prefix and returns the possible autocompletion of that string. For example, if the input is abs, I get absolute as the possible complete word.
The problem is that the function does a case-insentitive match for the prefix so if I have Abs as the prefix input, I still get absolute instead of Absolute.
What would be the correct way to get the final string completion that matches the case of the original prefix entered?
So, a function case-match that could work like
(case-match "Abs" "absolute") => "Absolute"
You can use the prefix string as the prefix to the case-insensitive search result. Just use subs to drop the length of the prefix from the search result:
(defn case-match [prefix s]
(str prefix (subs s (count prefix))))
(case-match "Abs" "absolute")
=> "Absolute"
This assumes your autocomplete function stays separate, and case-match would be applied to its result.
also, you can go with startsWith
(defn case-match
[prefix s]
(when (.startsWith (clojure.string/lower-case s)
(clojure.string/lower-case prefix))
s))
(case-match "Abs" "absolute")
#=> "absolute"
(case-match "Absence" "absolute")
#=> nil
Related
Basically I have a problem, here is the information needed to solve the problem.
PigLatin. Pig Latin is a way of rearranging letters in English words for fun. For example, the sentence “pig latin is stupid” becomes “igpay atinlay isway upidstay”.
Vowels(‘a’,‘e’,‘i’,‘o’,and‘u’)are treated separately from the consonants(any letter that isn’t a vowel).
For simplicity, we will consider ‘y’ to always be a consonant. Although various forms of Pig Latin exist, we will use the following rules:
(1) Words of two letters or less simply have “way” added on the end. So “a” becomes “away”.
(2) In any word that starts with consonants, the consonants are moved to the end, and “ay” is added. If a word begins with more than two consonants, move only the first two letters. So “hello” becomes “ellohay”, and “string” becomes “ringstay”.
(3) Any word which begins with a vowel simply has “way” added on the end. So “explain” becomes “explainway”.
Write a function (pig-latin L) that consumes a non-empty (listof Str) and returns a Str containing the words in L converted to Pig Latin.
Each value in L should contain only lower case letters and have a length of at least 1.
I understand that i need to set three main conditions here, i'm struggling with Racket and learning the proper syntax to write out my solutions. first I need to make a conditions that looks at a string and see if it's length is 2 or less to meet the (1) condition. For (2) I need to look at the first two characters in a string, i'm assuming I have to convert the string into a list of char(string->list). For (3) I understand I just have to look at the first character in the string, i basically have to repeat what I did with (2) but just look at the first character.
I don't know how to manipulate a list of char though. I also don't know how to make sure string-length meets a criteria. Any assistance would be appreciated. I basically have barely any code for my problem since I am baffled on what to do here.
An example of the problem is
(pig-latin (list "this" "is" "a" "crazy" "exercise")) =>
"isthay isway away azycray exerciseway"
The best strategy to solve this problem is:
Check in the documentation all the available string procedures. We don't need to transform the input string to a list of chars to operate upon it, and you'll find that there are existing procedures that meet all of our needs.
Write helper procedures. In fact, we only need a procedure that tells us if a string contains a vowel at a given position; the problem states that only a-z characters are used so we can negate this procedure to also find consonants.
It's also important to identify the best order to write the conditions, for example: conditions 1 and 3 can be combined in a single case. This is my proposal:
(define (vowel-at-index? text index)
(member (string-ref text index)
'(#\a #\e #\i #\o #\u)))
(define (pigify text)
; cases 1 and 3
(cond ((or (<= (string-length text) 2)
(vowel-at-index? text 0))
(string-append text "way"))
; case 2.1
((and (not (vowel-at-index? text 0))
(vowel-at-index? text 1))
(string-append (substring text 1)
(substring text 0 1)
"ay"))
; case 2.2
(else
(string-append (substring text 2)
(substring text 0 2)
"ay"))))
(define (pig-latin lst)
(string-join (map pigify lst)))
For the final step, we only need to apply the pigify procedure to each element in the input, and that's what map does. It works as expected:
(pig-latin '("this" "is" "a" "crazy" "exercise"))
=> "isthay isway away azycray exerciseway"
I want to convert a string into a list of one strings in Racket:
(string-split-wishful "abcd" "") => (list "a" "b" "c" "d")
This is the function that I wish for. The closest thing is string-split which doesn't do what I want:
(string-split "abcd" "") => (list "" "a" "b" "c" "d" "")
How do I get rid of the superfluous empty strings at the beginning and the end? I know that I can do something like (reverse (cdr (reverse (cdr (string-split "abcd" ""))))) but I want to know if there's a more idiomatic way of doing this.
Try this:
(string-split "abcd" #rx"(?<=.)(?=.)")
; ==> ("a" "b" "c" "d")
It uses a regular expression instead of string and the regular expression consist of a zero-width positive look-behind assertion such that it only matches after a character and one zero-width positive look-ahead assertion such that the match need one char in its right side to match.
Alexis' suggestion is nice too, might even perform better too:
(map string (string->list "abcd"))
How can i retrieve a list of items (string) which contain a specific word from another list. Here is an Example :
(setq l '("word1_jj" "word2_mm" "word3_jj" "word4_kk"))
I wanna extract all string in which figure "_jj.
You should make ends-with-p that takes the word and an ending. To do that you find out how many characters there is in the two strings and use subseq to make a string consisting of the last letters of the word. You can use equal to check it with the supplied argument it should match.
When you have that you can do this:
(remove-if-not (lambda (x) (ends-with-p x "_jj"))
'("word1_jj" "word2_mm" "word3_jj" "word4_kk"))
; ==> ("word1_jj" "word3_jj")
Alternatively you could make a make-end-predicate that returns a lambda that takes a word:
(remove-if-not (make-end-predicate "_jj")
'("word1_jj" "word2_mm" "word3_jj" "word4_kk"))
; ==> ("word1_jj" "word3_jj")
This code does not work as I expected. Could you please explain why?
(defn make-str [s c]
(let [my-str (ref s)]
(dosync (alter my-str str c))))
(defn make-str-from-chars
"make a string from a sequence of characters"
([chars] make-str-from-chars chars "")
([chars result]
(if (== (count chars) 0) result
(recur (drop 1 chars) (make-str result (take 1 chars))))))
Thank you!
This is very slow & incorrect way to create string from seq of characters. The main problem, that changes aren't propagated - ref creates new reference to existing string, but after it exits from function, reference is destroyed.
The correct way to do this is:
(apply str seq)
for example,
user=> (apply str [\1 \2 \3 \4])
"1234"
If you want to make it more effective, then you can use Java's StringBuilder to collect all data in string. (Strings in Java are also immutable)
You pass a sequence with one character in it to your make-str function, not the character itself. Using first instead of take should give you the desired effect.
Also there is no need to use references. In effect your use of them is a gross misuse of them. You already use an accumulator in your function, so you can use str directly.
(defn make-str-from-chars
"make a string from a sequence of characters"
([chars] (make-str-from-chars chars ""))
([chars result]
(if (zero? (count chars))
result
(recur (drop 1 chars) (str result (first chars))))))
Of course count is not very nice in this case, because it always has to walk the whole sequence to figure out its length. So you traverse the input sequence several times unnecessarily. One normally uses seq to identify when a sequence is exhausted. We can also use next instead of drop to save some overhead of creating unnecessary sequence objects. Be sure to capture the return value of seq to avoid overhead of object creations later on. We do this in the if-let.
(defn make-str-from-chars
"make a string from a sequence of characters"
([chars] (make-str-from-chars chars ""))
([chars result]
(if-let [chars (seq chars)]
(recur (next chars) (str result (first chars)))
result)))
Functions like this, which just return the accumulator upon fully consuming its input, cry for reduce.
(defn make-str-from-chars
"make a string from a sequence of characters"
[chars]
(reduce str "" chars))
This is already nice and short, but in this particular case we can do even a little better by using apply. Then str can use the underlying StringBuilder to its full power.
(defn make-str-from-chars
"make a string from a sequence of characters"
[chars]
(apply str chars))
Hope this helps.
You can also use clojure.string/join, as follows:
(require '[clojure.string :as str] )
(assert (= (vec "abcd") [\a \b \c \d] ))
(assert (= (str/join (vec "abcd")) "abcd" ))
There is an alternate form of clojure.string/join which accepts a separator. See:
http://clojuredocs.org/clojure_core/clojure.string/join
I'm just starting off with Lisp and need some help. This is technically homework, but I gave it a try and am getting somewhat what I wanted:
(defun speed (kmp)
(cond ((> kmp 100) "Fast")
((< kmp 40) "Slow")
(t "Average")))
However, if I run the program it displays "Average" instead of just Average (without the quotes).
How can I get it to display the string without quotes?
You can use symbols instead of strings. But keep in mind that symbols will be converted to uppercase:
> 'Average
AVERAGE
If you care about case or want to embed spaces, use format:
> (format t "Average")
Average
The read-eval-print loop displays the return value of your function, which is one of the strings in a cond branch. Strings are printed readably by surrounding them with double-quotes.
You could use (write-string (speed 42)). Don't worry that it also shows the string in double-quotes - that's the return value of write-string, displayed after the quoteless output.
You can also use symbols instead of strings:
(defun speed (kmp)
(cond ((> kmp 100) 'fast)
((< kmp 40) 'slow)
(t 'average)))
Symbols are uppercased by default, so internally fast is then FAST.
You can write any symbol in any case and with any characters using escaping with vertical bars:
|The speeed is very fast!|
Above is a valid symbol in Common Lisp and is stored internally just as you write it with case preserved.