madbernard: a long angled pier (Default)
Here's one I thought would be simple, which wasn't. The problem is, "return a count of all the vowels (not y) in a string." I thought I'd use regex.exec to capture an array of vowels and find the length of the array... But it only ever returned length 2. I see in the Mozilla Developer Network info page that RexExp.length is 2, but I don't know why I'm getting that here. Can someone troubleshoot?
function VowelCount(str) { 
  var re = /([aeiou])/ig;
  var arr = re.exec(str);
  return arr.length; 

So I pivoted to for loops and came up with this,
function VowelCount(str) { 
  var re = /[aeiou]/i;
  var count = 0;
  for (var i = 0; i < str.length; i++) {
    if (re.test(str.charAt(i))) {
      count += 1;
  return count; 
This also took some troubleshooting, since the count was wrong when I was using the 'g' flag in the regex. I'm also not sure why that was. Other people I looked at also had mangled looking answers, though of course Matt Larsh has something good,
function VowelCount(str) { 
  var vowels = str.match(/[aeiou]/g);
  return vowels.length;
Looks like I need to bone up on str.match.
madbernard: a long angled pier (Default)
Last evening I had a moment and tried the Coderbyte challenge "when given a string of single letters, plusses, and equals signs, return true only if every letter has a plus on both sides". I came up with a four part if statement reliant on regexes.
function SimpleSymbols(str) {
  if (/(\+\w\+)+/g.test(str) && !/^\w/g.test(str) && !/=\w/g.test(str) && !/\w(?!\+)/g.test(str)) 
  {return true;}
  else {return false;}
In English, that's if there is a plus on either side of a letter, and if there is no letter at the start of the string, and if there is no letter with an equals sign immediately in front of it, and if there is no letter that is not followed by a plus sign... Then true.

Ben, who does JS for a living, gave me the learning that if there's more than one regex, it's probably not ideal... "There's so many edge cases". Makes sense. So, later, the Coderbyte challenge appears, "when given a string of lowercase letters, if there is an a followed by a b three spots later, return true". And I'm like, "I'll use a for loop, like we were talking about last night!"Read more... )
madbernard: a long angled pier (Default)
Challenge 5 was, given a number between 1 and 1000, add up all the numbers inclusive in that set. Obviously I could repurpose Matt Larsh's recursive function to do this, so I did so, and it worked first try. Woo! Though, no one else seems to be doing +=. I wonder if this would be super wrong if I wasn't recursing and thus seeing it only once each function call? Or do they just like num + SimpleAdding(stuff)?
function SimpleAdding(num) {
  if (num === 1) {
    return 1;  
  else {
    return num += SimpleAdding(num - 1);
Challenge 6 was one I hadn't yet solved in Free Code Camp: Capitalize every word in a string. I abandoned a couple different paths while trying for speed, and eventually came up with this answer that seems cludgy:
function LetterCapitalize(str) { 
  var foo = str.split('');
  var bar = [];
  for(var i = 1; i < foo.length; i++) {
    if (foo[i - 1] === ' ') {
    else {
  return bar.join('');
LetterCapitalize('to see how to enter arguments in JavaScript');
I tried replacing the space in the if statement with /\s/ and that didn't work, nor did /\s/g; I'm not sure why not.

One of my abandoned paths was to do a str.replace(regex, function) but I wasn't getting the function to work... So I was super curious to see what Matt Larsh did, and sure enough he did something cool:
function LetterCapitalize(str) { 
  return str.replace(/\b[a-z]/g,function(c){return c.toUpperCase()});
That "\b" was something I thought should exist, but didn't find: the signal for "match a word boundary". I just tried /\b\w/g and that works, too; I suppose his way is better, though, since it limits the capitalization to only things that we absolutely know need capitalizing. Unless if \w includes accented letters and other slightly less usual things? My RegEx skillz are not yet l33t and m4d.
madbernard: a long angled pier (Default)
So, 4 was trivial, which was great: make a function that factorializes numbers. Like, 5! is 1 * 2 * 3 * 4 * 5 = 120.

function factorialize(num) {
for (var i = (num - 1); i > 0; i--) {
num *= i;
return num;


But 5... I feel like I shouldn't have to make my own function that takes an array and jams it all into a single string without commas separating the bits. But the concat stuff I was looking at in the Mozilla Developer Network wasn't working that way (even though their example shows it should. So baffled).

And I feel like there must be some way to not declare like 7 variables... But I couldn't figure out how to chain these methods/functions. Anyway: here it is, a function that checks if a statement is a palindrome (ignoring all the non-letter things! The RegEx stuff was honestly fun learning).

function reverseString(str) {
var reverseStr = '';
for (var i = str.length - 1; i > -1; i--){
reverseStr += str[i];
return reverseStr;

function arrToRevStr(arr) {
var str = '';
for (var i = arr.length - 1; i > -1; i--){
str += arr[i];
return str;

function palindrome(str) {
var cleanStrArray = /(\w)/g.exec(str);
var cleanStr = arrToRevStr(cleanStrArray);
var cleanStrLow = cleanStr.toLowerCase();

var reverseStr = reverseString(str);
var cleanReverseStrArray = /(\w)/g.exec(reverseStr);
var cleanReverseStr = arrToRevStr(cleanReverseStrArray);
var cleanReverseStrLow = cleanReverseStr.toLowerCase();

if (cleanStrLow == cleanReverseStrLow) {return true;}
else {return false;}


May 2016

8910 11121314


RSS Atom

Most Popular Tags

Style Credit

Expand Cut Tags

No cut tags
Page generated Sep. 22nd, 2017 06:58 pm
Powered by Dreamwidth Studios