I'm fairly new to Scheme and I'm using DrRacket and I hope to get some much needed assistance.
I was first tasked to extract the first character from a non-empty string, which I was successful in doing so.
;;Signature: string-first: string -> string
;;Purpose: To extract the first character from a non-empty string
;;Tests
(check-expect (string-first "cookies") "c")
(check-expect (string-first "party") "p")
;;Definition: string-first
(define (string-first str)
(string-ith str 0))
Now, I am tasked with extracting the last character from a non-empty string. I cannot figure this out. My guess is that it involves "string-length" in combination with "string-ith."
Hopefully someone can help me out. Thank you in advance.
You're on the right track, assuming that the string is non-empty the implementation is as simple as this:
(define (string-last str)
(string-ref str (sub1 (string-length str))))
Explanation:
I'm using the built-in string-ref procedure, which I'm guessing does the same as your implementation of string-ith (always try to use existing procedures)
A string's valid indexes start at zero and end at the string's length minus one (use sub1 for subtracting a single unit)
So finding the last character is a simple matter of combining string-ref and string-length, remembering that the last character will be at the index length-1
Related
in my program I (after many procedures) get tokenized words. Unfortunately due to reversing them they hold punctuation characters at the beginning of a word eg. "BARGE "UR 106
How to move that " from the beginning to the end -> BARGE "UR 106"
another example: (.REIS (HASAN M should be -> REIS (HASAN M.)
Up to now I've tried:
{DOCHS14.SHEM__ONIA} startswith ["\"","\(","\."\,"\(\."]
then
Local StringVar str:={DOCHS14.SHEM__ONIA}[0]
TrimLeft ({DOCHS14.SHEM__ONIA})
{DOCHS14.SHEM__ONIA}&str;
But that gives me errors:
A number, currency amount, boolean, date, time, date-time, or string is expected here.
How to fix that? or is there another way to solve this problem?
There are multiple issues in your formula.
startswith expects a string not an array of strings.
Only the double qoute must be escaped, but you did it wrong. (See here)
While a solution with startswith is also possible, I have used the Left-function instead. In your third example one have to check two characters, so this must be checked first and output another result.
if Left({DOCHS14.SHEM__ONIA}, 2) = "(." Then
Mid({DOCHS14.SHEM__ONIA}, 3) + ".)"
else if Left({DOCHS14.SHEM__ONIA}, 1) in ["""", "(", ".", ","] Then
Mid({DOCHS14.SHEM__ONIA}, 2) + Left({DOCHS14.SHEM__ONIA}, 1)
else
{DOCHS14.SHEM__ONIA}
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"
Suppose we are given two strings s1 and s2(both lowercase). We have two find the minimal lexographic string that can be formed by merging two strings.
At the beginning , it looks prettty simple as merge of the mergesort algorithm. But let us see what can go wrong.
s1: zyy
s2: zy
Now if we perform merge on these two we must decide which z to pick as they are equal, clearly if we pick z of s2 first then the string formed will be:
zyzyy
If we pick z of s1 first, the string formed will be:
zyyzy which is correct.
As we can see the merge of mergesort can lead to wrong answer.
Here's another example:
s1:zyy
s2:zyb
Now the correct answer will be zybzyy which will be got only if pick z of s2 first.
There are plenty of other cases in which the simple merge will fail. My question is Is there any standard algorithm out there used to perform merge for such output.
You could use dynamic programming. In f[x][y] store the minimal lexicographical string such that you've taken x charecters from the first string s1 and y characters from the second s2. You can calculate f in bottom-top manner using the update:
f[x][y] = min(f[x-1][y] + s1[x], f[x][y-1] + s2[y]) \\ the '+' here represents
\\ the concatenation of a
\\ string and a character
You start with f[0][0] = "" (empty string).
For efficiency you can store the strings in f as references. That is, you can store in f the objects
class StringRef {
StringRef prev;
char c;
}
To extract what string you have at certain f[x][y] you just follow the references. To udapate you point back to either f[x-1][y] or f[x][y-1] depending on what your update step says.
It seems that the solution can be almost the same as you described (the "mergesort"-like approach), except that with special handling of equality. So long as the first characters of both strings are equal, you look ahead at the second character, 3rd, etc. If the end is reached for some string, consider the first character of the other string as the next character in the string for which the end is reached, etc. for the 2nd character, etc. If the ends for both strings are reached, then it doesn't matter from which string to take the first character. Note that this algorithm is O(N) because after a look-ahead on equal prefixes you know the whole look-ahead sequence (i.e. string prefix) to include, not just one first character.
EDIT: you look ahead so long as the current i-th characters from both strings are equal and alphabetically not larger than the first character in the current prefix.
I am trying to practice recursion, but at the moment I don't quite understand it well...
I want to write a recursive Boolean function which takes 2 strings as arguments, and returns true if the second string can be made equal to the first by replacing some letters with a certain special character.
I'll demonstrate what I mean:
Let s1 = "hello", s2 = "h%lo", where '%' is the special character.
The function will return true since '%' can replace "el", causing the two strings to be equal.
Another example:
Let s1 = "hello", s2 = "h%l".
The function will return false since an 'o' is lacking in the second string, and there is no special character that can replace the 'o' (h%l% would return true).
Now the problem isn't so much with writing the code, but with understanding how to solve the problem in general, I don't even know where to begin.
If someone could guide me in the right direction I would be very grateful, even by just using English words, I'll try to translate it to code (Java)...
Thank you.
So this is relatively easy to do in Python. The method I chose was to put the first string ("hello") into an array then iterate over the second string ("h%lo") comparing the elements to those in the array. If the element was in the array i.e. 'h', 'l', 'o' then I would pop it from the array. The resulting array is then ['e','l']. The special character can be found as it is the element which does not exist in the initial array.
One can then substitute the special character for the joined array "el" in the string and compare with the first string.
In the first case this will give "hello" == "hello" -> True
In the second case this will give "hello" == "helol" -> False
I hope this helps and makes sense.
scala has a standard way of splitting a string in StringOps.split
it's behaviour somewhat surprised me though.
To demonstrate, using the quick convenience function
def sp(str: String) = str.split('.').toList
the following expressions all evaluate to true
(sp("") == List("")) //expected
(sp(".") == List()) //I would have expected List("", "")
(sp("a.b") == List("a", "b")) //expected
(sp(".b") == List("", "b")) //expected
(sp("a.") == List("a")) //I would have expected List("a", "")
(sp("..") == List()) // I would have expected List("", "", "")
(sp(".a.") == List("", "a")) // I would have expected List("", "a", "")
so I expected that split would return an array with (the number a separator occurrences) + 1 elements, but that's clearly not the case.
It is almost the above, but remove all trailing empty strings, but that's not true for splitting the empty string.
I'm failing to identify the pattern here. What rules does StringOps.split follow?
For bonus points, is there a good way (without too much copying/string appending) to get the split I'm expecting?
For curious you can find the code here.https://github.com/scala/scala/blob/v2.12.0-M1/src/library/scala/collection/immutable/StringLike.scala
See the split function with the character as an argument(line 206).
I think, the general pattern going on over here is, all the trailing empty splits results are getting ignored.
Except for the first one, for which "if no separator char is found then just send the whole string" logic is getting applied.
I am trying to find if there is any design documentation around these.
Also, if you use string instead of char for separator it will fall back to java regex split. As mentioned by #LRLucena, if you provide the limit parameter with a value more than size, you will get your trailing empty results. see http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#split(java.lang.String,%20int)
You can use split with a regular expression. I´m not sure, but I guess that the second parameter is the largest size of the resulting array.
def sp(str: String) = str.split("\\.", str.length+1).toList
Seems to be consistent with these three rules:
1) Trailing empty substrings are dropped.
2) An empty substring is considered trailing before it is considered leading, if applicable.
3) First case, with no separators is an exception.
split follows the behaviour of http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#split(java.lang.String)
That is split "around" the separator character, with the following exceptions:
Regardless of anything else, splitting the empty string will always give Array("")
Any trailing empty substrings are removed
Surrogate characters only match if the matched character is not part of a surrogate pair.