How can I craft a one-liner return line that also has conditionals? For example, if I want to make a 'return the median' one:
//assuming sorted input array
return ((inputArray.length % 2) && (inputArray[int(inputArray.length/2)] + inputArray[int(inputArray.length/2)+1]) / 2) || inputArray[int(inputArray.length/2)+1];
Is there anyway to make this work?
You should use the ternary operator:
return (inputArray.length % 2 != 0) ? (inputArray[int(inputArray.length/2)] + inputArray[int(inputArray.length/2)+1]) / 2 : inputArray[int(inputArray.length/2)+1];
Which is equivalent to:
if (inputArray.length % 2 != 0) {
return (inputArray[int(inputArray.length/2)] + inputArray[int(inputArray.length/2)+1]) / 2;
} else {
return inputArray[int(inputArray.length/2)+1];
}
If you want to use only && and ||, you can use the following (which is not really a good programming style):
((inputArray.length % 2 != 0) || return inputArray[int(inputArray.length/2)+1]) && return (inputArray[int(inputArray.length/2)] + inputArray[int(inputArray.length/2)+1]);
Which is equivalent to:
(condition || return value2) && return value1;
So, thanks to the short-circuit evaluation of boolean operators:
if condition is true, return value2 is not evaluated and return value1 will be executed.
if condition is false, return value2 will be evaluated.
I would go for a more readable version without repetitions, but with some variables.
var index:int = int(inputArray.length / 2);
var item1:Number = inputArray[index];
var item2:Number = inputArray[index + 1];
var median:Number = (item1 + item2) / 2;
return (inputArray.length % 2 != 0) ? median : item2;
Please try with modified conditions:
return ((inputArray.length % 2) && (((inputArray[int(inputArray.length/2)] + inputArray[int(inputArray.length/2)+1]) / 2) || inputArray[int(inputArray.length/2)+1]));
Related
First, I am generating random integer values to get some values to check.
var A1, B2, C3:int;
A1 = Math.random() * 100 + 1;
B2 = Math.random() * 100 + 1;
C3 = Math.random() * 100 + 1;
Then I want to check if all the variables are unique from each other.
if (!(A1 == B2 || A1 == C3 || B2 == C3)){
unique = true;
}else{ // Not unique
}
If the variables are not unique to each other, I want to keep only the value for A1, and then change the two other variables B2 and C3 and then again check if they are unique.
}else{ // Not unique
if (unique = false){
do{
B2 = Math.random() * A1 + 1;
C3 = Math.random() * A1 + 1;
if (!(A1 == B2 || A1 == C3 || B2 == C3)){
unique = true;
}while (unique = true)
}
trace("Not unique");
}
My problem is that I cannot get three unique values, and any help on how I can solve this is highly appreciated.
This is a case where instead of asking to resolve the problem at hand you should probably have started by explaining what you are trying to do and what's the best way to achieve it. Since you have been posting a few times already my guess is that you are trying to extract 3 unique int from a range of 0 to 100. If that's the case then you are trying to create a complex and inefficient way to do it which is interesting by itself but is far from the point. The solution is to simply generate an array of int from 0 to 100 and then extract 3 of them randomly. In that case they will always be unique and always from 0 to 100.
var intRange:Array = [];
for(var i:int = 0; i < 100; i++)
{
intRange.push(i);
}
var index:int = Math.flor(Math.random() * intRange.length);
var a:int = intRange[index];
intRange.splice(index, 1);
index = Math.flor(Math.random() * intRange.length);
var b:int = intRange[index];
intRange.splice(index, 1);
index = Math.flor(Math.random() * intRange.length);
var c:int = intRange[index];
intRange.splice(index, 1);
//3 unique int from 0 to 100 range
Your problem is the wrong code here:
if (unique = false) { // must be: if (unique == false)
do {
...
} while (unique = true) // must be: while(unique == false)
}
also it can be reduced to
while(!unique) {
...
}
Also, there you are taking random from A1, not 100. Just too many errors. The proper code should be:
var a:int, b:int, c:int; // also, why would you call variables A1, B2, C3???
a = Math.random() * 100 + 1;
do {
b = Math.random() * 100 + 1;
while(b == a);
do {
c = Math.random() * 100 + 1;
} while(c == a || c == b);
// here you have unique
Now, the answer given by BotMaster is correct and robust but it involves filling an array and splicing it which is not always ok, especially if random range is significantly more than 100 and we need only three values from it.
I am making a game that when you click on the Monster your score gets +1. But when your score goes over 1000 I would like it like this 1,000 rather than 1000. I am not sure how to do this as I have not learnt much action script. I have embed number and punctuation into the font. Here is my code so far:
var score:Number = 0;
Multitouch.inputMode = MultitouchInputMode.TOUCH_POINT;
Monster.addEventListener(TouchEvent.TOUCH_TAP, fl_TapHandler);
function fl_TapHandler(event:TouchEvent):void
{
score = score + 1;
Taps_txt.text = (score).toString();
}
Help will greatly appreciated.
You can do like that:
function affScore(n:Number, d:int):String {
return n.toFixed(d).replace(/(\d)(?=(\d{3})+\b)/g,'$1,');
}
trace(affScore(12345678, 0)); // 12,345,678
This may not be the most elegant approach, but I wrote a function that will return the string formatted with commas;
public function formatNum(str:String):String {
var strArray:Array = str.split("");
if (strArray.length >= 4) {
var count:uint = 0;
for (var i:uint = strArray.length; i > 0; i--) {
if (count == 3) {
strArray.splice(i, 0, ",");
count = 0;
}
count++;
}
return strArray.join("");
}
else {
return str;
}
}
I tested it on some pretty large numbers, and it seems to work just fine. There's no upper limit on the size of the number, so;
trace (formatNum("10000000000000000000"));
Will output:
10,000,000,000,000,000,000
So in your example, you could use it thusly;
Taps_txt.text = formatNum(String(score));
(This is casting the type implicitly rather than explicitly using toString();, but either method is fine. Casting just looks a little neater in function calls)
Use the NumberFormatter class:
import flash.globalization.NumberFormatter;
var nf:NumberFormatter = new NumberFormatter("en_US");
var numberString:String = nf.formatNumber(1234567.89);
trace("Formatted Number:" + numberString);
// Formatted Number:1,234,567.89
To show the score with comma, you can do like this : ( comments are inside the code )
var score:Number = 0;
var score_str:String;
var score_str_len:int;
Multitouch.inputMode = MultitouchInputMode.TOUCH_POINT;
Monster.addEventListener(TouchEvent.TOUCH_TAP, fl_TapHandler);
function fl_TapHandler(event:TouchEvent):void
{
score = score + 1;
score_str = score.toString();
score_str_len = score_str.length;
// here you can use score > 999 instead of score_str_len > 3
Taps_txt.text =
score_str_len > 3
// example : 1780
// score_str_len = 4
// score_str.substr(0, 4 - 3) = 1 : get thousands
// score_str.substr(4 - 3) = 780 : get the rest of our number : hundreds, tens and units
// => 1 + ',' + 780 = 1,780 : concatenate thousands + comma + (hundreds, tens and units)
? score_str.substr(0, score_str_len-3) + ',' + score_str.substr(score_str_len-3)
: score_str
;
// gives :
// score == 300 => 300
// score == 1285 => 1,285
// score == 87903 => 87,903
}
EDIT :
To support numbers greater than 999.999, you can do like this :
function fl_TapHandler(event:MouseEvent):void
{
score = score + 1;
score_str = score.toString();
Taps_txt.text = add_commas(score_str);
}
function add_commas(nb_str:String):String {
var tmp_str:String = '';
nb_str = nb_str.split('').reverse().join('');
for(var i = 0; i < nb_str.length; i++){
if(i > 0 && i % 3 == 0) tmp_str += ',';
tmp_str += nb_str.charAt(i);
}
return tmp_str.split('').reverse().join('');
/*
gives :
1234 => 1,234
12345 => 12,345
123456 => 123,456
1234567 => 1,234,567
12345678 => 12,345,678
123456789 => 123,456,789
1234567890 => 1,234,567,890
*/
}
Hope that can help you.
I have this expression:
!(1 && !(0 || 1))
The output returns 1 true. And that's ok. When I read the expression I came to the same conclusion before checking the output. But I would really appreciate if someone can explain to me why the returning value is true, that way, I will have a better understanding of boolean logic and how to implement better evaluators in my code.
Key observation here: ! is not, && is the "And" operator, and || is the "Inclusive Or" Operator.
What are you really asking when you say "why it's true?".
0 = false
1 = true
AND && table
0 0 -> 0
0 1 -> 0
1 0 -> 0
1 1 -> 1
OR || table
0 0 -> 0
0 1 -> 1
1 0 -> 1
1 1 -> 1
NOT ! table
0 -> 1
1 -> 0
With parentheses implying "do this first", the statement reduces using the tables above:
!(1 && !(0 || 1))
!(1 && !1)
!(1 && 0)
!0
1
But I don't know "why" it's true. Because that's what an AND operation is, what an OR operation is, and what a NOT operation is, and how reducing a statement works. With those definitions, it can't be another answer, so it's that answer. But you already know that, because you did it yourself and got the same answer ... so what does the question mean?
The innermost expression (0 || 1) is always true.
So !(0 || 1) is always false.
That leaves 1 && 0, which is always false.
So !(false) is always true.
Please forgive my freely intermixing 0/false and 1/true.
The human evaluator (:-).
Working through the expression, following order of operation:
!(1 && !(0 || 1))
= !(1 && !(1))
= !(1 && 0)
= !(0)
= 1
Step by step explanation:
1 = true
0 = false
Starting point: !(1 && !(0 || 1))
Lets start with the inner most expression: !(0 || 1)
Var1 || Var2 =
Var1 or Var2 =
If Var1 or Var2 is 1 or both are 1, the result is 1.
(0 || 1) = 0 or 1 -> the second variable is 1 so the expression is 1.
Insert the result (0 || 1) = 1 into Startingpoint: !(1 && !(1))
! = not (inverts the value of what is behinde)
!1 = 0
!0 = 1
!(0 || 1) = !(1) = 0
Insert the result !(1) = 0 into Startingpoint: !(1 && 0)
So we have !(1 && 0)
Var1 && Var2 = And =
the opossite of or =
If Var1 AND Var2 are both 1, the result is 1. Else it is 0 =
If Var1 or Var2 is 0, the result is zero
1 && 1 = 1
1 && 0 = 0
everything else: 0
So this is left: !(0)
Reminder: ! = not = inverts the expression behind it. So !0 = 1 (and !1 = 0)
This is 1. Or in your case: true
A good book for Beginner C programmers and people who want to learn about programming and logic in an easy, understandable way:
C for Dummies by Dan Godkins
!(1 && !(0 || 1))
Since, you have used parenthesis, evaluation takes place according to them.
First, evaluate innermost parenthesis.
0 || 1 => always true.
!(0 || 1) => !(true) => always false.
1 && !(0 || 1) => 1 && false => always false.
!(1 && !(0 || 1)) => !false => always true.
In a case where at least two out of three booleans are true, this is the easiest way to find out:
BOOL a, b, c;
-(BOOL)checkAtLeastTwo
{
return a && (b || c) || (b && c);
}
What will be the optimal solution if there is ten booleans and at least two of them needs to be true? Thanks in advance.
Your original implementation is sub-optimal - you can just sum true values:
return (int)a + (int)b + (int)c >= 2;
Obviously you can extend this to 10 variables:
return (int)a + (int)b + (int)c + (int)d + (int)e +
(int)f + (int)g + (int)h + (int)i + (int)j >= 2;
In C you can just check sum of your variables
return a + b + .... + n >= 2;
If implicit conversion from boolean to integer doesn't in your language, you can simply convert your variables to integer and check sum of converted values.
Basically I'm looking for something like
SELECT ordinal(my_number) FROM my_table
which would return
1st
11th
1071st
...
etc
but preferrably without the use of a stored procedure
I don't know of a built-in function but it's pretty easy to write:
SELECT
CONCAT(my_number, CASE
WHEN my_number%100 BETWEEN 11 AND 13 THEN "th"
WHEN my_number%10 = 1 THEN "st"
WHEN my_number%10 = 2 THEN "nd"
WHEN my_number%10 = 3 THEN "rd"
ELSE "th"
END)
FROM my_table;
mysql doesn't have support for this. You'll have to handle the strings in whichever language you are getting the mysql data from.
Based on Ken's code, a custom MySQL function would be as follows:
DELIMITER $$
CREATE FUNCTION ordinal(number BIGINT)
RETURNS VARCHAR(64)
DETERMINISTIC
BEGIN
DECLARE ord VARCHAR(64);
SET ord = (SELECT CONCAT(number, CASE
WHEN number%100 BETWEEN 11 AND 13 THEN "th"
WHEN number%10 = 1 THEN "st"
WHEN number%10 = 2 THEN "nd"
WHEN number%10 = 3 THEN "rd"
ELSE "th"
END));
RETURN ord;
END$$
DELIMITER ;
Then it can be used as:
SELECT ordinal(1) -- 1st
SELECT ordinal(11) -- 11th
SELECT ordinal(21) -- 21st
SELECT ordinal(my_number) FROM my_table
It is possible in MySQL using the string functions but it gets messy real fast. You'd better just do the suffix in the language you're using. For example, in PHP you could do something like this:
function ordSuffix($num) {
if(empty($num) || !is_numeric($num) || $num == 0) return $num;
$lastNum = substr($num, -1);
$suffix = 'th';
if($lastNum == 1 && $num != 11) { $suffix = 'st'; }
elseif($lastNum == 2 && $num != 12) { $suffix = 'nd'; }
elseif($lastNum == 3 && $num != 13) { $suffix = 'rd'; }
return $num.$suffix;
}
echo ordSuffix(4); // 4th
echo ordSuffix(1); // 1st
echo ordSuffix(12); // 12th
echo ordSuffix(1052); // 1052nd
I found a way that works for me but its a bit of a hack
DATE_FORMAT(CONCAT('2010-01-', my_number), '%D')
That works because currently the number I'm looking at never gets above 25. But it doesn't generalize well so someone might be entertained by this:
CONCAT(
IF(my_number % 100 BETWEEN 11 AND 13,
FLOOR(my_number / 100),
FLOOR(my_number / 10)),
DATE_FORMAT(
CONCAT('2010-01-',
IF(my_number % 100 BETWEEN 11 AND 13
my_number % 100,
my_number % 10)),
'%D'))
But that's a lot of work just to get at the DATE_FORMAT functionality when Ken's code is simpler.