var str = "ajdisoiureenvmcnmvm"
var arr1 = ["e","t","a","i","n","o","h","r","d","q","l","c","u","m","w","f","s","g","y","p","b","v","k","j","x","z"]
var arr2 = ["m","i","e","n","v","r","d","j","s","o","u","c","b","a","f","g","h","k","l","p","q","t","w","x","z","y"]
What is the most efficient way of replacing the string such that if it contains an "m" replace it with "e", "i" with "t" and so on.
If you're fine with map:
function encode (string) {
return string.split("").map(function (letter) {
return arr1[arr2.indexOf(letter)];
}).join("");
}
This should be self explanatory and relatively efficient.
str = str.split("")
for(var i = 0; i < str.length; i++) {
var c = str[i];
var j = arr1.indexOf(c);
str[i] = arr2[j];
}
str = str.join("");
I'd use a mapping for this kind of thing though.
str[i] = arr2[j];
str[i] = arr2[j];.
var str = "ajdisoiureenvmcnmvm"
var arr1 = ["e","t","a","i","n","o","h","r","d","q","l","c","u","m","w","f","s","g","y","p","b","v","k","j","x","z"]
var arr2 = ["m","i","e","n","v","r","d","j","s","o","u","c","b","a","f","g","h","k","l","p","q","t","w","x","z","y"]
newStr = str.split("");
for (var i=0, len=newStr.length; i<len; i++)
{
index = arr2.indexOf(newStr[i]);
if(index>=0)
newStr[i] = arr1[index];
}
str = newStr.join("");
console.log(str);
Probably a more straight forward approach is to put the original string into an array first so you can manipulate it more easily character by character:
var arr1 = ["e","t","a","i","n","o","h","r","d","q","l","c","u","m","w","f","s","g","y","p","b","v","k","j","x","z"];
var arr2 = ["m","i","e","n","v","r","d","j","s","o","u","c","b","a","f","g","h","k","l","p","q","t","w","x","z","y"];
var str = "ajdisoiureenvmcnmvm";
var data = str.split(""), index;
for (var i = 0; i < arr2.length; i++) {
index = str.indexOf(arr2[i]);
if (index >= 0) {
data[index] = arr1[i];
}
}
// build the string back again
str = data.join("");
Working demo: http://jsfiddle.net/jfriend00/fAQTe/
str.split("")It depends on your definition of "efficient." With that structure, it's quite hard to be efficient. But something like this is functional and short:
var rex = new RegExp("[" + arr1.join("") + "]", "g");
var result = str.replace(rex, function(letter) {
var index = arr1.indexOf(letter);
return index === -1 ? letter : arr2[index];
});
There, we create a regular expression using a character class to match each of the source characters, then do a replacement operation using a function to look up the replacement (if any) from the arrays.
Now, if you can change the structure, there are better ways, like using a map rather than arr1.indexOf. E.g.:
var map = {
"e": "m",
"t": "i",
"a": "e",
// ...
};
Then (this uses ES5's Object.keys, which can be shimmed if needed):
var rex = new RegExp("[" + Object.keys(map).join("") + "]", "g");
var result = str.replace(rex, function(letter) {
return map[letter] || letter;
});
I knew you don't ask for the way to do by shell script, but it will give you the feeling how easily it can be done by tr command in this special request.
echo "ajdisoiureenvmcnmvm" |tr "etainohrdqlcumwfsgypbvkjxz" "mienvrdjsoucbafghklpqtwxzy"
exsnhrnbjmmvtacvata
