so i have been slowly working through cs101, first question of week 5 block 4 has me stumped.
so the idea is im supposed to be writing psuedo accurate javascript to determine the number of names starting with A and B in a 2000 entry csv of baby names, im 99% sure i have written the correct code but keep getting returned "unexpected token "{" " i have tried removing all curly brackets one by one which just produces errors obviously and have no idea what i have done wrong, please help.
table = new SimpleTable("baby-2010.csv");
count1 = 0; // A count
count2 = 0; // B count
for (row: table) {
if (row.getField("name").startsWith("A") {
count1 = count1 + 1;
}
if (row.getField("name").endsWith("B") {
count2 = count2 + 1;
}
}
print("A count:", count1);
print("B count:", count2);
all good, VLAZ had the correct solution, both if statements were missing a closing ).
thanks heaps guys
use indexOf and lastIndexOf instead of startsWith and endsWith
if (row.getField("name").indexOf("A") === 0) {
count1 = count1 + 1;
}
if (row.getField("name").lastIndexOf("B") === row.getField('name').length) {
count2 = count2 + 1;
}
You could also use a regex test for /^A/ or /B$/
Related
I'm trying to return the length of the last word in a string, I am struggling with debugging at the moment. The problem is with the forEach loop I hit a breakpoint that wont update the empty array.
function lengthOfLastWord(s) {
let spaces = [];
let space = ' ';
let spaceInd = s.indexOf(space);
while (spaceInd != -1) {
spaces.push(spaceInd);
spaceInd = s.indexOf(space, spaceInd + 1);
}
let nonSpaces = [];
let wordArray = [];
for(let i = 0; i<spaces.length; i++) {
nonSpaces.push((spaces[i + 1] - spaces[i]) - 1);
}
nonSpaces.forEach(function(number) {
if (number >= 1) {
wordArray.push(number);
}
})
let words = wordArray[wordArray.length - 1];
if ((wordArray.length === 0)) {
console.log(0);
} else {
console.log(words);
}
};
lengthOfLastWord("this is my test string");
If I ignore the breakpoint I get the correct outcome, but I can't seem to figure out why the empty wordArray wont update.
(I'm sure there is an easier way to get the result using filters and mapping but i'm wondering if there is a simple fix to my current buggy code)
Right now, your code seems to return the length of the second to last word, rather than the last. The problem does not lie in the forEach, rather it is in the way you're doing it. When you take the index of spaces in the string and calculate distance between each of them, this excludes the first and the last words, as there isn't a space at the very beginning and at the very end. As a result, the second to last word's length is logged rather than the very last.
The simple fix would be to add this line right after your while loop:
spaces.push(s.length);
This will make sure that the final word length will also be calculated in the following for loop.
As you are aware, there is a much simpler way to do this.
Here is one using split:
function lengthOfLastWord(string) {
let split = string.trim().split(" ");
return split[split.length - 1].length
}
I'm making a calculator for a site project of mine where you can type your entire expression before resolving, for example: 2+3*4 would return 14, 22-4 would return 18, 20+5! would return 140, and so on.
And that works for simple expressions like the ones I showed, but when I add brackets the code breaks.
So a simple expression like (2+3)! that should return 120 actually returns 10 or 2+3!.
my original ideia to make even the basic 2+3! work was to separate the string in math simbols and the rest. so it would separate in this case it would separate it into 2, + and 3!; where it would find the symbol and resolve just that part. And that's why it solves 10 instead of not working.
But after trying to solve I couldn't make the code work except in a extremely specific situation, so I decided to redo the code and post this here in case someone could help me out.
This is the function that I'm currently using to prepare my string for evaluation:
function sepOperFat(){
//2+3! it's working
//1+(2-(2+2)+3)! want that to work in the end
var value = document.calculator.ans.value;
var operandoPos = ['0'];
var operandoInPos = [''];
var paraResolver = [];
for(i = 0; i <= value.length; i++){
//check if value[i] is equal to +, -, ×, ÷, * & /
if(value[i] == '+' || value[i] == '-' || value[i] == '×' || value[i] == '÷' || value[i] == '*' || value[i] == '/'){
operandoPos.push(i);
operandoInPos.push(value[i]);
}
}
paraResolver.push(value.slice(operandoPos[0], operandoPos[1]));
for(var total = 1; total <= operandoPos.length; total++){
paraResolver.push(value.slice(operandoPos[total] + 1, operandoPos[total + 1]));
}
document.calculator.ans.value = '';
for(var total = 0; total <= paraResolver.length - 2; total++){
if(paraResolver[total].includes('!')){
document.calculator.ans.value += "factorial(" + paraResolver[total] + ")";
}else{
document.calculator.ans.value += paraResolver[total];
}
document.calculator.ans.value += operandoInPos[total + 1];
}
}
document.calculator.ans.value is the name of the string where i have the expression.
operandoPos is the position on the string where a symbol is at.
operandoInPos is the symbol (I maybe could have used value.charAt(operandoPos) for that too).
paraResolver is the number that I will be solving (like 3).
factorial( is the name of my function responsible for making the number factorial.
the function doesn't have a return because I still want to solve inside the document.calculator.ans.value.
to resolve the equation I'm using document.calculator.ans.value = Function('"use strict"; return '+ document.calculator.ans.value)(); that activates when I press a button.
And yeah, that's it. I just want a function capable of knowing the difference between (2+3)! and 2+(3)! so it can return factorial(2+3) instead of (2+factorial(3)).
Thank you for your help.
Your biggest problem is going to be that order of operations says parentheses need to be evaluated first. This might mean your code has to change considerably to support whatever comes out of your parentheses parsing.
I don't think you want all of that handled for you, but an approach you can take to sorting out the parenthesis part is something like this:
function parseParentheses(input) {
let openParenCount = 0;
let myOpenParenIndex = 0;
let myEndParenIndex = 0;
const result = [];
for (let i = 0; i < input.length; i++) {
if (input[i] === '(') {
if (openParenCount === 0) {
myOpenParenIndex=i;
// checking if anything exists before this set of parentheses
if (i !== myEndParenIndex) {
result.push(input.substring(myEndParenIndex, i));
}
}
openParenCount++;
}
if (input[i] === ')') {
openParenCount--;
if (openParenCount === 0) {
myEndParenIndex=i+1;
// recurse the contents of the parentheses to search for nested ones
result.push(parseParentheses(input.substring(myOpenParenIndex+1, i)));
}
}
}
// capture anything after the last parentheses
if (input.length > myEndParenIndex) {
result.push(input.substring(myEndParenIndex, input.length));
}
return result;
}
// tests
console.log(JSON.stringify(parseParentheses('1!+20'))) // ["1!+20"]
console.log(JSON.stringify(parseParentheses('1-(2+2)!'))) // ["1-",["2+2"],"!"]
console.log(JSON.stringify(parseParentheses('(1-3)*(2+5)'))) // [["1-3"],"*",["2+5"]]
console.log(JSON.stringify(parseParentheses('1+(2-(3+4))'))) // ["1+",["2-",["3+4"]]]
this will wrap your input in an array, and essentially group anything wrapped in brackets into nested arrays.
I can further explain what's happening here, but you're not likely to want this specific code so much as the general idea of how you might approach unwrapping parenthesis.
It's worth noting, the code I've provided is barely functional and has no error handling, and will behave poorly if something like 1 - (2 + 3 or 1 - )2+3( is provided.
I'm not a programmer by any means. I'm an animator trying to use JS expressions in After Effects. I'm getting an "Undefined value used in expression" error on line 1 where I define a variable.I already showed it to my friend on discord who is a cs major, and he had no clue what was wrong with it.
Here's just a paste of the code if you need it:
var count = 1;
if (framesToTime(time) % 12 == 0) {
count = count + 1
if (count % 2 == 0){
thisProperty = 95
} else {
thisProperty = 20
};
} ;
Ok I don't know why the hell this fixed it, but I changed the name of the variable from "count" to just "x" and it works now. Go figure
Try it.
var count = 1;
if (framesToTime(time) % 12 == 0) {
count = count + 1;
if (count % 2 == 0){
thisProperty = 95;
} else {
thisProperty = 20;
}
}
thisProperty;
In your code, thisProperty has become an ordinary variable. If you write its name at the end of the code, then its value will be assigned to the property.
In AE, if there is nothing inside an if statement or the if statement contains malformed/error code you will receive this error. Put a temp value inside the curly braces or something to process and ensure nothing inside will throw an error.
I also received this error with this:
pastTime = timeToFrames(time)-1;
curPos = transform.xPosition;
pastPos = transform.xPosition.valueAtTime(framesToTime(pastTime));
if (curPos-pastPos[0] != 0) {
// Here is the problem in my case. added a value here 99 to fix until finished testing.
}
else {
effect("Angle Control")("Angle")
}
if/else statements are strict
The syntax for if/else statements is strict in the JavaScript engine
and need to be written for standardized JavaScript.
https://helpx.adobe.com/after-effects/using/expression-language-reference.html*
I got this error because there was a missing semicolon.
I am sorry in advance if my title is somehow misleading and I am really sorry for my English if you wouldn't understand me, it's just not my native language!
I will try to explain as better as I can about what I try to achieve. I try to do this for past two entire days and I really need your help!
Let's say I have array with the following numbers:
2 4 6 8 10 1 3 5 2 4
I am trying to count how many even and odd numbers are here in a row, and when even/odd changes - count it again. So my answer from the array above should be:
5 (5 even numbers in a row) 3 (3 odd lines in a row) (2 even lines in a row)
Also when the counting isn't stopped it should post "<br>" instead of counted evens/odds, so it could show me results one time near to each line.
Check this example image:
I have this script which is counting, but it has a few issues: when number is even, it shows counting twice. Next, I can't figure it out how to add <br> to these lines where counting and add result only at the last line of counting. Also my counting result should be at the top, so the script should count from the end as I guess, and when I try i-- it starts the infinite loop...
var digits = ["2, 4, 6, 8, 10, 1, 3, 5, 2, 4"]
var evenCount=1, oddCount=1;
for(var i =0; i < digits.length; i++){
if(digits[i] % 2 ==0){
var oddCount=1;
$("#res").append(evenCount + " (l) <br>");
evenCount++;
}
else
var evenCount=1;
$("#res").append(oddCount + " (n) <br>");
oddCount++;
}
Check my fiddle to see it in action:
https://jsfiddle.net/xk861vf9/8/
First, I think your code show counting twice because you misses two '{' after "for loop" and "else". After I fix the code format, I don't see it counting twice anymore.
$(document).ready(function() {
$("#sub").bind("click", function() {
$("#res").html("");
var digits = $('#content').find("span").map(function() {
return $(this).text();
});
var evenCount = 1;
var oddCount = 1;
for(var i =0; i < digits.length; i++) {
if (digits[i] % 2 ==0) {
oddCount = 1;
$("#res").append(evenCount + " (l) <br>");
evenCount++;
} else {
evenCount=1;
$("#res").append(oddCount + " (n) <br>");
oddCount++;
}
}
});
});
Second, they are many ways to implement that. Take a look at this jsfiddle code as an example.
https://jsfiddle.net/xk861vf9/11/
The concept is to print the counted number after even/odd number changes. Then use for loop to print <br> x times (counted number - 1) so if counted number is 4, there will be 3 <br> tags followed.We also have to check if current number is the last number in array and print the counted number or else the last counted number will be skipped.
Hope this help! :)
Ps. Sorry for my bad English, not my native language too.
I'm starting to learn javascript and I basically needed a countup that adds an x value to a number(which is 0) every 1 second. I adapted a few codes I found on the web and came up with this:
var d=0;
var delay=1000;
var y=750;
function countup() {
document.getElementById('burgers').firstChild.nodeValue=y+d;
d+=y;
setTimeout(function(){countup()},delay);
}
if(window.addEventListener){
window.addEventListener('load',countup,false);
}
else {
if(window.attachEvent){
window.attachEvent('onload',countup);
}
}
There's probably residual code there but it works as intended.
Now my next step was to divide the resultant string every 3 digits using a "," - basically 1050503 would become 1,050,503.
This is what I found and adapted from my research:
"number".match(/.{1,3}(?=(.{3})+(?!.))|.{1,3}$/g).join(",");
I just can't find a way to incorporate this code into the other. What should I use to replace the "number" part of this code?
The answer might be obvious but I've tried everything I knew without sucess.
Thanks in advance!
To use your match statement, you need to convert your number to a String.
Let's say you have 1234567.
var a = 1234567;
a = a + ""; //converts to string
alert(a.match(/.{1,3}(?=(.{3})+(?!.))|.{1,3}$/g).join(","));
If you wish, you can wrap this into a function:
function baz(a) {
a = a + "";
return a.match(/.{1,3}(?=(.{3})+(?!.))|.{1,3}$/g).join(",");
}
Usage is baz(1234); and will return a string for y our.
While I do commend you for using a pattern matching algorithm, this would probably be easier to, practically speaking, implement using a basic string parsing function, as it doesn't look anywhere as intimidating from just looking at the match statement.
function foo(bar) {
charbar = (""+bar).split(""); //convert to a String
output = "";
for(x = 0; x < charbar.length; x++) { //work backwards from end of string
i = charbar.length - 1 - x; //our index
output = charbar[i] + output; //pre-pend the character to the output
if(x%3 == 2 && i > 0) { //every 3rd, we stick in a comma, except if it is not the leftmost digit
output = ',' + output;
}
}
return output;
}
Usage is basically foo(1234); which yields 1,234.