CODEWITHSUNDEEP
lispcommon-lisp
WALID BELRHALMIA
WALID BELRHALMIA

Reputation: 441

lisp remove a the content of one list from another list

I have a list of string like this called F:

("hello word i'am walid" "goodbye madame") => this list contain two elements of string

and I have another list call S like this ("word" "madame") => this contain two words

now I want to remove the elements of the list S from each string of the list F and the output should be like this ("hello i'am walid" "goodbye")

i found already this function:

(defun remove-string (rem-string full-string &key from-end (test #'eql)
                      test-not (start1 0) end1 (start2 0) end2 key)
  "returns full-string with rem-string removed"
  (let ((subst-point (search rem-string full-string 
                             :from-end from-end
                             :test test :test-not test-not
                             :start1 start1 :end1 end1
                             :start2 start2 :end2 end2 :key key)))
    (if subst-point
        (concatenate 'string
                     (subseq full-string 0 subst-point)
                     (subseq full-string (+ subst-point (length rem-string))))
        full-string)))

example: (remove-string "walid" "hello i'am walid") => the output "hello i'am"

but there is a problem

example: (remove-string "wa" "hello i'am walid") => the output "hello i'am lid"

but the output should be like this "hello i'am walid" in another word i wont the remove the exact word from the string

please help me and thank's

Upvotes: 0

Views: 616

Answers (2)

Svante
Svante

Reputation: 51561

You can use the cl-ppcre library for regular expressions. Its regex flavour understands the word boundary \b.

The replacement could work like this:

(cl-ppcre:regex-replace-all "\\bwa\\b" "ba wa walid" "")

=> "ba  walid"

I guess that you want to collapse any whitespace around the removed word into one:

(cl-ppcre:regex-replace-all "\\s*\\bwa\\b\\s*" "ba wa walid" " ")

=> "ba walid"

See the documentation linked above.

UPDATE: You extended the question to punctuation. That's actually a tad more complicated, since you now have three kinds of characters: alphanumeric, punctuation, and whitespace.

I can't give a complete solution here, but the outline I envision is to create boundary definitions between all three of these kinds. You need positive/negative lookaheads/lookbehinds for that. Then you look at the replaced string, whether it starts or ends with punctuation and append or prepend the corresponding boundary to the effective expression.

For defining the boundaries in a readable manner, the parse tree syntax of cl-ppcre might prove useful.

Upvotes: 3

Throw Away Account
Throw Away Account

Reputation: 2671

The Common Lisp Cookbook provides this function:

(defun replace-all (string part replacement &key (test #'char=))
"Returns a new string in which all the occurences of the part 
is replaced with replacement."
    (with-output-to-string (out)
      (loop with part-length = (length part)
            for old-pos = 0 then (+ pos part-length)
            for pos = (search part string
                              :start2 old-pos
                              :test test)
            do (write-string string out
                             :start old-pos
                             :end (or pos (length string)))
            when pos do (write-string replacement out)
            while pos)))

Using that function:

(loop for raw-string in '("hello word i'am walid" "goodbye madame")
        collect (reduce (lambda (source-string bad-word)
                          (replace-all source-string bad-word ""))
                        '("word" "madame")
                     :initial-value raw-string))

Upvotes: 1

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