如何从字符串中获取奇数和偶数位置字符？

Question

I'm trying to figure out how to remove every second character (starting from the first one) from a string in Javascript. For example, the string "This is a test!" should become "hsi etTi sats!" I also want to save every deleted character into another array.

I have tried using replace method and splice method, but wasn't able to get them to work properly. Mostly because replace only replaces the first character.

function encrypt(text, n) {
  if (text === "NULL") return n;
  if (n <= 0) return text;
  var encArr = [];
  var newString = text.split("");
  var j = 0;
  for (var i = 0; i < text.length; i += 2) {
    encArr[j++] = text[i];
    newString.splice(i, 1); // this line doesn't work properly
  }
}

Answer 1

You could reduce the characters of the string and group them to separate arrays using the % operator. Use destructuring to get the 2D array returned to separate variables

 let str = "This is a test!"; const [even, odd] = [...str].reduce((r,char,i) => (r[i%2].push(char), r), [[],[]]) console.log(odd.join('')) console.log(even.join('')) 

Using a for loop:

 let str = "This is a test!", odd = [], even = []; for (var i = 0; i < str.length; i++) { i % 2 === 0 ? even.push(str[i]) : odd.push(str[i]) } console.log(odd.join('')) console.log(even.join('')) 

Answer 2

It would probably be easier to use a regular expression and .replace : capture two characters in separate capturing groups, add the first character to a string, and replace with the second character. Then, you'll have first half of the output you need in one string, and the second in another: just concatenate them together and return:

 function encrypt(text) { let removedText = ''; const replacedText1 = text.replace(/(.)(.)?/g, (_, firstChar, secondChar) => { // in case the match was at the end of the string, // and the string has an odd number of characters: if (!secondChar) secondChar = ''; // remove the firstChar from the string, while adding it to removedText: removedText += firstChar; return secondChar; }); return replacedText1 + removedText; } console.log(encrypt('This is a test!')); 

Answer 3

You could take an array and splice and push each second item to the end of the array.

 function encrypt(string) { var array = [...string], i = 0, l = array.length >> 1; while (i <= l) array.push(array.splice(i++, 1)[0]); return array.join(''); } console.log(encrypt("This is a test!")); 

Answer 4

 function encrypt(text) { text = text.split(""); var removed = [] var encrypted = text.filter((letter, index) => { if(index % 2 == 0){ removed.push(letter) return false; } return true }).join("") return { full: encrypted + removed.join(""), encrypted: encrypted, removed: removed } } console.log(encrypt("This is a test!")) 

Splice does not work, because if you remove an element from an array in for loop indexes most probably will be wrong when removing another element.

Answer 5

Pretty simple with .reduce() to create the two arrays you seem to want.

 function encrypt(text) { return text.split("") .reduce(({odd, even}, c, i) => i % 2 ? {odd: [...odd, c], even} : {odd, even: [...even, c]} , {odd: [], even: []}) } console.log(encrypt("This is a test!")); 

They can be converted to strings by using .join("") if you desire.

Answer 6

I think you were on the right track. What you missed is replace is using either a string or RegExp.

The replace() method returns a new string with some or all matches of a pattern replaced by a replacement. The pattern can be a string or a RegExp, and the replacement can be a string or a function to be called for each match. If pattern is a string, only the first occurrence will be replaced.

Source: String.prototype.replace()

If you are replacing a value (and not a regular expression), only the first instance of the value will be replaced. To replace all occurrences of a specified value, use the global (g) modifier

Source: JavaScript String replace() Method

So my suggestion would be to continue still with replace and pass the right RegExp to the function, I guess you can figure out from this example - this removes every second occurrence for char 't':

 let count = 0; let testString = 'test test test test'; console.log('original', testString); // global modifier in RegExp let result = testString.replace(/t/g, function (match) { count++; return (count % 2 === 0) ? '' : match; }); console.log('removed', result); 

Answer 7

like this?

 var text = "This is a test!" var result = "" var rest = "" for(var i = 0; i < text.length; i++){ if( (i%2) != 0 ){ result += text[i] } else{ rest += text[i] } } console.log(result+rest) 

Answer 8

Maybe with split, filter and join:

const remaining = myString.split('').filter((char, i) => i % 2 !== 0).join('');
const deleted = myString.split('').filter((char, i) => i % 2 === 0).join('');

Answer 9

I don't know how much you care about performance, but using regex is not very efficient. Simple test for quite a long string shows that using filter function is on average about 3 times faster, which can make quite a difference when performed on very long strings or on many, many shorts ones.

 function test(func, n){ var text = ""; for(var i = 0; i < n; ++i){ text += "a"; } var start = new Date().getTime(); func(text); var end = new Date().getTime(); var time = (end-start) / 1000.0; console.log(func.name, " took ", time, " seconds") return time; } function encryptREGEX(text) { let removedText = ''; const replacedText1 = text.replace(/(.)(.)?/g, (_, firstChar, secondChar) => { // in case the match was at the end of the string, // and the string has an odd number of characters: if (!secondChar) secondChar = ''; // remove the firstChar from the string, while adding it to removedText: removedText += firstChar; return secondChar; }); return replacedText1 + removedText; } function encrypt(text) { text = text.split(""); var removed = ""; var encrypted = text.filter((letter, index) => { if(index % 2 == 0){ removed += letter; return false; } return true }).join("") return encrypted + removed } var timeREGEX = test(encryptREGEX, 10000000); var timeFilter = test(encrypt, 10000000); console.log("Using filter is faster ", timeREGEX/timeFilter, " times") 

Using actually an array for storing removed letters and then joining them is much more efficient, than using a string and concatenating letters to it.
I changed an array to string in filter solution to make it the same like in regex solution, so they are more comparable.