I have 2 different possible values I use in an equation. I want to select whichever one exists and is greater using the least amount of code. It's possible neither variable exists, in which case year = 0, but one or both might exist. I.e:
if(isset($this->average['year'] || isset($this->Listings['year']) {
$year = whichever is greater of the above.
} else {
$year = 0;
}
It seems like there must be a shorter/ less messy way to do this than:
if (isset($this->average['year']) && ($this->average['year'] > $this->Listings['year']) {
$year = $this->average['year'];
} elseif( isset($this->Listings['year'])) {
$year = $this->Listings['year'];
} else {
$year = 0;
}
Thanks
Using max and the ternary operator to do isset checks on both variables you can shorten it to this:
$year = max(array(
isset($this->average['year']) ? $this->average['year'] : 0,
isset($this->Listings['year']) ? $this->Listings['year'] : 0
));
Related
I need to find the value of x where the variance of two results (which take x into account) is the closest to 0. The problem is, the only way to do this is to cycle through all possible values of x. The equation uses currency, so I have to check in increments of 1 cent.
This might make it easier:
$previous_var = null;
$high_amount = 50;
for ($i = 0.01; $i <= $high_amount; $i += 0.01) {
$val1 = find_out_1($i);
$val2 = find_out_2();
$var = variance($val1, $val2);
if ($previous_var == null) {
$previous_var = $var;
}
// If this variance is larger, it means the previous one was the closest to
// 0 as the variance has now started increasing
if ($var > $previous_var) {
$l_s -= 0.01;
break;
}
}
$optimal_monetary_value = $i;
I feel like there is a mathematical formula that would make the "cycling through every cent" more optimal? It works fine for small values, but if you start using 1000's as the $high_amount it takes quite a few seconds to calculate.
Based on the comment in your code, it sounds like you want something similar to bisection search, but a little bit different:
function calculate_variance($i) {
$val1 = find_out_1($i);
$val2 = find_out_2();
return variance($val1, $val2);
}
function search($lo, $loVar, $hi, $hiVar) {
// find the midpoint between the hi and lo values
$mid = round($lo + ($hi - $lo) / 2, 2);
if ($mid == $hi || $mid == $lo) {
// we have converged, so pick the better value and be done
return ($hiVar > $loVar) ? $lo : $hi;
}
$midVar = calculate_variance($mid);
if ($midVar >= $loVar) {
// the optimal point must be in the lower interval
return search($lo, $loVar, $mid, $midVar);
} elseif ($midVar >= $hiVar) {
// the optimal point must be in the higher interval
return search($mid, $midVar, $hi, $hiVar);
} else {
// we don't know where the optimal point is for sure, so check
// the lower interval first
$loBest = search($lo, $loVar, $mid, $midVar);
if ($loBest == $mid) {
// we can't be sure this is the best answer, so check the hi
// interval to be sure
return search($mid, $midVar, $hi, $hiVar);
} else {
// we know this is the best answer
return $loBest;
}
}
}
$optimal_monetary_value = search(0.01, calculate_variance(0.01), 50.0, calculate_variance(50.0));
This assumes that the variance is monotonically increasing when moving away from the optimal point. In other words, if the optimal value is O, then for all X < Y < O, calculate_variance(X) >= calculate_variance(Y) >= calculate_variance(O) (and the same with all > and < flipped). The comment in your code and the way have you have it written make it seem like this is true. If this isn't true, then you can't really do much better than what you have.
Be aware that this is not as good as bisection search. There are some pathological inputs that will make it take linear time instead of logarithmic time (e.g., if the variance is the same for all values). If you can improve the requirement that calculate_variance(X) >= calculate_variance(Y) >= calculate_variance(O) to be calculate_variance(X) > calculate_variance(Y) > calculate_variance(O), you can improve this to be logarithmic in all cases by checking to see how the variance for $mid compares the the variance for $mid + 0.01 and using that to decide which interval to check.
Also, you may want to be careful about doing math with currency. You probably either want to use integers (i.e., do all math in cents instead of dollars) or use exact precision numbers.
If you known nothing at all about the behavior of the objective function, there is no other way than trying all possible values.
On the opposite if you have a guarantee that the minimum is unique, the Golden section method will converge very quickly. This is a variant of the Fibonacci search, which is known to be optimal (require the minimum number of function evaluations).
Your function may have different properties which call for other algorithms.
Why not implementing binary search ?
<?php
$high_amount = 50;
// computed val2 is placed outside the loop
// no need te recalculate it each time
$val2 = find_out_2();
$previous_var = variance(find_out_1(0.01), $val2);
$start = 0;
$end = $high_amount * 100;
$closest_variance = NULL;
while ($start <= $end) {
$section = intval(($start + $end)/2);
$cursor = $section / 100;
$val1 = find_out_1($cursor);
$variance = variance($val1, $val2);
if ($variance <= $previous_var) {
$start = $section;
}
else {
$closest_variance = $cursor;
$end = $section;
}
}
if (!is_null($closest_variance)) {
$closest_variance -= 0.01;
}
how do i add an if statement to a Php value, this is what i tried
&n = 1;
$number = if(10 < 1)
{
$n = 0;
}
is there any way to fix this or something i didn't ad that made it not to work?
because it kept showing an error message that the code is not right
You have a typo and some syntax error.
&n = 1;
That should be:
$n = 1;
Also, you can't really assign the value of an if statement to your variable.
Try simply using a ternary instead.
$n = (10 < 1) ? 1 : 0;
In PHP, if is a statement, but the = operator expects an expression, so that is invalid syntax. If you want to assign something to $number when the condition is true, then you can use something like:
if (10 < 1)
{
$number = ...;
$n = 0;
}
I have checked a bunch of posts on stackoverflow and on articles on google but none of them were able to answer my question. Here is my code (i've simplified it instead of posting my code)
$first = 10;
$second = 0; //comes from db row count
$total = !is_int($first/$second) ? 0 : $first/$second;
problem is when i do this I keep getting the Division by zero error. I have a bunch and $second isnt always 0, it can be any number. But it does come out to 0 since the row counts for whatever query it comes out as 0. Is there a safe way of checking to see if $first can be divided by $second without giving an error? I have tried # before the !is_int and that just breaks all other statements.
Try this:
$total = ($second == 0) ? 0 : $first / $second;
You can't divide by 0 it is undefined. If you want to handle division by 0 just check if the divisor isn't equals to 0. Or a safer way, chack if it is a positive integer:
$first = 10;
$dbRowCount = dbFunction();
if ($dbRowCount > 0) {
$total = $first / $dbRowCount;
} else {
//Error handling
}
The ternary structure can accept more than one condition. and it will work just as any other if condition, and won't try the second condition if the first fails.
So, just add it
$total = ($first!==0 && $second!==0 && !is_int($first/$second)) ? 0 : $first/$second;
You might want to try checking if your $Second variable is 0.
Something like:
$First = 10;
$Second = $row['table_column'];
if ($Second == 0) {
echo "Oops this will be an error";
}
else
$First/$second = $me;
There are a few instances where I've come across this and I just need to ask to see if there is something better out there. I have a value that can come through the pike that has a number assigned to it. The number means something, so to output it to the user. For example, I need to change 1 to something that means something to the user. Here's a code example:
switch($data['priority']) {
case 1:
$priority = 'Low';
break;
case 2:
$priority = 'Medium';
break;
case 3:
$priority = 'High';
break;
default:
$priority = 'None';
break;
}
Is there a way to assign the value to priority with less code? I know that in JavaScript you can do multiple ternary operators (var data = type == 1 ? 'Good' : type == 2 ? 'Bad' : type == 3 ? 'Unknown' : '';) but PHP doesn't allow this. I know you could also do multiple if, else if, else if, else statements, but this just seems like a lot of extra code. Is there a better way to do what I'm asking?
One possibility is to use an array as map:
$map = array(1 => 'Low', 2 => 'Mid', ...);
$prioName = isset($map[$prio]) ? $map[$prio] : 'None';
Try an array map:
$priorities = array('None', 'Low', 'Medium', 'High');
// you can use an easy "if"
if ($data['priority'] < 1 || $data['priority'] > 3)
$data['priority'] = 0;
// or you can use a ternary assignment:
$data['priority'] = $data['priority'] < 1 || $data['priority'] > 3 ? 0 : $data['priority'];
$priority = $priorities[$data['priority']];
I am trying to to mark some trends, so I have 1 as the lowest and 5 as the biggest value.
So for example,
I may have the following case:
5,4,5,5 (UP)
3,4, (UP)
4,3,3 (DOWN)
4,4,4,4, (FLAT - this is OK for all same numbers)
I am planning to have unlimited number of ordered values as input, an as an output I will just show an (UP), (DOWN), or (FLAT) image.
Any ideas on how I can achieve this?
Sorry if I am not descriptive enough.
Thank you all for you time.
Use least square fit to calculate the "slope" of the values.
function leastSquareFit(array $values) {
$x_sum = array_sum(array_keys($values));
$y_sum = array_sum($values);
$meanX = $x_sum / count($values);
$meanY = $y_sum / count($values);
// calculate sums
$mBase = $mDivisor = 0.0;
foreach($values as $i => $value) {
$mBase += ($i - $meanX) * ($value - $meanY);
$mDivisor += ($i - $meanX) * ($i - $meanX);
}
// calculate slope
$slope = $mBase / $mDivisor;
return $slope;
} // function leastSquareFit()
$trend = leastSquareFit(array(5,4,5,5));
(Untested)
If the slope is positive, the trend is upwards; if negative, it's downwards. Use your own judgement to decide what margin (positive or negative) is considered flat.
A little bit hard to answer based on the limited info you provide, but assuming that:
if there's no movement at all the trend is FLAT,
otherwise, the trend is the last direction of movement,
then this code should work:
$input = array();
$previousValue = false;
$trend = 'FLAT';
foreach( $input as $currentValue ) {
if( $previousValue !== false ) {
if( $currentValue > $previousValue ) {
$trend = 'UP';
} elseif( $currentValue < $previousValue ) {
$trend = 'DOWN';
}
}
$previousValue = $currentValue;
}
For your examples :
Calculate longest increasing subsequence, A
Calulate longest decreasing subsequence , B
Going by your logic, if length of A is larger than B , its an UP , else DOWN.
You will also need to keep track of all equals using one boolean variable to mark FLAT trend.
Query :
What trend would be :
3,4,5,4,3 ?
3,4,4,4,3 ?
1,2,3,4,4,3,2,2,1 ?
Then the logic might need some alterations depending upon what your requirements are .
I'm not sure if i understand your problem totally but I would put the values in an array and use a code like this (written in pseudocode):
int i = 0;
String trend = "FLAT":
while(i<length(array)) {
if(array(i)<array(i+1)) {
trend = "UP";
}
else if(array(i)>array(i+1) {
trend = "DOWN";
}
i++;
}
EDIT: this would obviously only display the trend of the latest alteration
one would also may count the number of times the trend is up or down and determine the overall trend by that values
echo foo(array(5,4,5,5)); // UP
echo foo(array(3,4)); // UP
echo foo(array(4,3,3)); // DOWN
echo foo(array(4,4,4,4)); // FLAT
function foo($seq)
{
if (count(array_unique($seq)) === 1)
return 'FLAT';
$trend = NULL;
$count = count($seq);
$prev = $seq[0];
for ($i = 1; $i < $count; $i++)
{
if ($prev < $seq[$i])
{
$trend = 'UP';
}
if ($prev > $seq[$i])
{
$trend = 'DOWN';
}
$prev = $seq[$i];
}
return $trend;
}
I used the code from #liquorvicar to determine Google search page rank trends, but added some extra trend values to make it more accurate:
nochange - no change
better (higher google position = lower number)
worse (lower google position = higher number)
I also added extra checks when the last value had no change, but taking in account the previous changes i.e.
worsenochange (no change, previouse was worse - lower number)
betternochange (no change, previouse was better - lower number)
I used these values to display a range of trend icons:
$_trendIndicator="<img title="trend" width="16" src="/include/main/images/trend-'. $this->getTrend($_positions). '-icon.png">";
private function getTrend($_positions)
{
// calculate trend based on last value
//
$_previousValue = false;
$_trend = 'nochange';
foreach( $_positions as $_currentValue ) {
if( $_previousValue !== false ) {
if( $_currentValue > $_previousValue ) {
$_trend = 'better';
} elseif( $_currentValue < $_previousValue ) {
$_trend = 'worse';
}
if ($_trend==='worse' && ($_previousValue == $_currentValue)) {$_trend = 'worsenochange';}
if ($_trend==='better' && ($_previousValue == $_currentValue)) {$_trend = 'betternochange';}
}
$_previousValue = $_currentValue;
}
return $_trend;
}