如何检查字符串中的每个数字？

Question

I was given the challenge of converting a string of digits into 'fake binary' on Codewars.com, and I am to convert each individual digit into a 0 or a 1, if the number is less than 5 it should become a 0, and if it's 5 or over it should become a 1. I know how to analyze the whole string's value like so:

function fakeBin(x){
if (x < 5)
return 0;
else return 1;
}

This however, analyzes the value of the whole string, how would I go about analyzing each individual digit within the string rather than the whole thing?

Note: I have already looked at the solutions on the website and don't understand them, I'm not cheating.

Answer 1

You can do it in one line with two simple global string replacement operations:

 function fakeBin(x){ return ("" + x).replace(/[0-4]/g,'0').replace(/[5-9]/g,'1'); } console.log(fakeBin(1259)) console.log(fakeBin(7815)) console.log(fakeBin("1234567890")) 

The ("" + x) part is just to ensure you have a string to work with, so the function can take numbers or strings as input (as in my example calls above).

Answer 2

Simple javascript solution to achieve expected solution

function fakeBin(x){
 x = x+'' ;
var z =[];
for(var i=0;i< x.length;i++){
  if((x[i]*1)<5){
     z[i] =0;
   }else{
    z[i]=1;
  }
}
  return z
}

console.log(fakeBin(357))

Answer 3

Split the string and apply the current function you have to each element of the string. You can accomplish this with map or with reduce :

 function fakeBin(x) { x = x.split(''); let toBin = x => { if (x < 5) return 0; else return 1 } return x.map(toBin).join(''); } console.log(fakeBin("2351")); 

refactored

 function fakeBin(x) { x = [...x]; let toBin = x => x < 5 ? 0 : 1; return x.map(toBin).join(''); } console.log(fakeBin("2351")); 

reduce

 function fakeBin(x) { let toBin = x => x < 5 ? 0 : 1; return [...x].reduce((acc,val) => acc + toBin(val), ""); } console.log(fakeBin("23519")); 

Answer 4

if you are in java you can use

charAt()

and you make a for with the word length and you can check one by one

for(int i = 0; i < text.length(); i++){
yourfunction(texto.charAt(i));
}

Answer 5

The snippet below will take a string and return a new string comprised of zeros and/or ones based on what you described.

We use a for ...of loop to traverse the input string and will add a 0 or 1 to our return array based on whether the parsed int if greater or less than 5.

Also note that we are checking and throwing an error if the character is not a number.

 const word = "1639"; const stringToBinary = function(str) { let ret = []; for (const char of word) { if (Number.isNaN(parseInt(char, 10))) { throw new Error(`${char} is not a digit!`); } else { const intVal = parseInt(char, 10); ret.push(intVal > 5 ? 1 : 0); } } return ret.join(''); }; console.log(stringToBinary(word)); 

Answer 6

You can use String.prototype.replace() with RegExp /([0-4])|([5-9])/g to match 0-4 , 5-9 , replace with 0 , 1 respectively

 let str = "8539734222673566"; let res = str.replace(/([0-4])|([5-9])/g, (_, p1, p2) => p1 ? 0 : 1); console.log(res);