Checking and deleting multiple occurrences of the same value in an array using javascript

Question

This is what I've been trying, but something is wrong with the last part, can sy please tell me whats wrong, or show me an other method of doing this. Thanks in advance.

function removeSpaces(string) {
  var str = string.replace(/^\s*|\s*$/g,'')
  str = str.replace(/[^a-z|A-z|\r\n]/g,'');
  while (str.indexOf("\r\n\r\n") >= 0) {
    str = str.replace(/\r\n\r\n/g, "\r\n");
  }
  words = str.split("\r\n");
  var i = 0;
  var j = 0;
  for (i = 0; i < words.length; i++) {
    for (j = i; j < words.length; j++) {
      if ((i != j) && (words[i] == words[j])) {
        words.splice(j,1);
        j--;
      }
    }
  }
  str = words.join("\r\n");
  return str;
}

This question is similar to this one: stackoverflow.com/questions/6940103/how-do-i-make-unique-array/… — F-A
– F-A, Commented Aug 10, 2011 at 12:38

F-A · Accepted Answer · 2011-08-10 13:01:47Z

1

You could use the filter function. The filter function of an array is returning an array containing elements which pass to a certain filter.

var isLastElement = function(element, index, array){
    return (index == array.lastIndexOf(element));
};
words = words.filter(isLastElement);

In this example, the filter function goes through the elements of the initial array ad add this element to the new array only if isLastElement(element, elementIndex, initArray) return true.
The isLastElement function returns true only if the element is not present a second time in the end of the array.

edited Aug 10, 2011 at 13:01

answered Aug 10, 2011 at 12:37

F-A

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4 Comments

mgabor Over a year ago

Im sorry, I don't really see what this does. could you please explain?

F-A Over a year ago

@Gabor Magyar: I edited my answer, to explain what is the function doing.

Felix Kling Over a year ago

Note that IE (don't know which versions) does not support indexOf and lastIndexOf for arrays.

F-A Over a year ago

@Gabor Magyar: You're welcome. Glad it helped you. However as Felix Kling said, it might not work on all browser, you can have a look at the link I posted as a comment to your question for other solutions.

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