The flat() method creates a new array with all sub-array elements concatenated into it recursively up to the specified depth.JS Tutorial JS HOME JS Introduction JS Where To JS Output JS Statements JS Syntax JS Comments JS Variables JS Let JS Const JS Operators JS Arithmetic JS Assignment JS Data Types JS Functions JS Objects JS Events JS Strings JS String Methods JS String Search JS String Templates JS Numbers JS BigInt JS Number Methods JS Number Properties JS Arrays JS Array Methods JS Array Sort JS Array Iteration JS Array Const JS Dates JS Date Formats JS Date Get Methods JS Date Set Methods JS Math JS Random JS Booleans JS Comparisons JS If Else JS Switch JS Loop For JS Loop For In JS Loop For Of JS Loop While JS Break JS Iterables JS Sets JS Maps JS Typeof JS Type Conversion JS Bitwise JS RegExp JS Precedence JS Errors JS Scope JS Hoisting JS Strict Mode JS this Keyword JS Arrow Function JS Classes JS Modules JS JSON JS Debugging JS Style Guide JS Best Practices JS Mistakes JS Performance JS Reserved Words

At the basis of all this reasoning is the () method. Finally I convert that index into a pair of two-dimensional coordinates. Then I use indexOf('🦌') to find Rudolf's position.

If so then I can transform the two dimensional array into a one dimensional array. The basic assumption is that the forest is made up of rows all of the same size. I’d like to be able to avoid both forEach() and some(). Use flat() to find an element in an arrayĮverything is very nice but I am not satisfied yet. Then I loop for each element in the forest and break it when r.indexOf('🦌') > -1, which is when I have found the value '🦌' in the array. But I also save an additional variable, found, to be used as a reference to find out whether or not I was able to find Rudolf. I save the coordinate value in two variables, col and row.

Javascript find element in array