Let's say we have the following data in an array:
$data1 = [3,5,7,6,8,9,13,14,17,15,16,16,16,18,22,20,21,20];
$data2 = [23,18,17,17,16,15,16,14,15,10,11,7,4,5];
As with $data1 we can say the data is increasing while in $data2 it is decreasing.
Using PHP, how do you know the data is increasing or decreasing, and is there a way on how to measure
know the rate of increasing as well as decreasing i.e in terms of percentage.
Edit
From the comments I received I got an idea and here is what I have tried.
What I want to achieve;
I want to know if the trend of the data coming in is upwards or downwards.
Want also to know the rate at which the data is rising or droping. For example $data1 = [1,3,5]; is not the same as $data2 = [1, 20, 55];. You can see $data1 rate of increase is not the same as $data2.
function increaseOrDecrease($streams = []) : array
{
$streams = [3,5,7,6,8,9,13,14,17,15,16,16,16,18,22,20,21,20]; // For the increasing
//$streams = [23,18,17,17,16,15,16,14,15,10,11,7,4,5]; // For the decreasing
$first = 0;
$diff = [];
foreach ($streams as $key => $number) {
if ($key != 0) {
$diff[] = $number - $first;
}
$first = $number;
}
$avgdifference = array_sum($diff)/count($diff); //Get the average
$side = $avgdifference > 0 ? 'UP' : 'DOWN';
$avgsum = array_sum($streams)/count($streams);
$percentage = abs($avgdifference)/$avgsum * 100;
if ($side == 'UP') {
$data = [
'up' => true,
'percent' => $percentage,
];
}else {
$data = [
'up' => false,
'percent' => $percentage,
];
}
return $data;
}
I would like some help to refactor this code or the best approach to solve the issue.
There are several ways to analyze data and extract a trend. The most classical method is called
least squares. It's a way of fitting a line
through the data. The method computes the slope and the intercept of the line. The trend is just the slope.
The formulas are given here.
A PHP implementation is the following:
function linearRegression($x, $y)
{
$x_sum = array_sum($x);
$y_sum = array_sum($y);
$xy_sum = 0;
$x2_sum = 0;
$n = count($x);
for($i=0;$i<$n;$i++)
{
$xy_sum += $x[$i] * $y[$i];
$x2_sum += $x[$i] * $x[$i];
}
$beta = ($n * $xy_sum - $x_sum * $y_sum) / ($n * $x2_sum - $x_sum * $x_sum);
$alpha = $y_sum / $n - $beta * $x_sum / $n;
return ['alpha' => $alpha, 'beta' => $beta];
}
function getTrend($data)
{
$x = range(1, count($data)); // [1, 2, 3, ...]
$fit = linearRegression($x, $data);
return $fit['beta']; // slope of fitted line
}
Examples:
echo getTrend([1, 2, 3]); // 1
echo getTrend([1, 0, -1]); // -1
echo getTrend([3,5,7,6,8,9,13,14,17,15,16,16,16,18,22,20,21,20]); // 1.065
echo getTrend([23,18,17,17,16,15,16,14,15,10,11,7,4,5]); // -1.213
You are asking for a type of data structure that can represent ascending as well as descending data. PHP got SplMinHeap and SplMaxHeap for this purpose. These built in classes make life easer when dealing with ascending or descending datasets.
A quick example ...
<?php
declare(strict_types=1);
namespace Marcel;
use SplMinHeap;
$numbers = [128, 32, 64, 8, 256];
$heap = new SplMinHeap();
foreach ($numbers as $number) {
$heap->insert($number);
}
$heap->rewind();
while($heap->valid()) {
// 8, 32, 64, 128, 256
echo $heap->current() . PHP_EOL;
$heap->next();
}
The SplMinHeap class keeps the minimum automatically on the top. So just use heaps instead of arrays that have no structure. Same goes for SplMaxHeap that keeps the highest value on the top.
Finding the differences
If you want to iterate all data and finding the differences between one to the next, you just have to iterate the heap. It 's ordered anyway.
$heap->rewind();
$smallest = $heap->current();
while($heap->valid()) {
// 8, 32, 64, 128, 256
$current = $heap->current();
echo $current . PHP_EOL;
// 0 (8 - 8), 24 (32 - 8), 32 (64 - 32), 64 (128 - 64), 128 (256 - 128)
echo "difference to the value before: " . ($current - $smallest) . PHP_EOL;
$smallest = $current;
$heap->next();
}
I would do simple things like this
$data1 = [3,5,7,6,8,9,13,14,17,15,16,16,16,18,22,20,21,20];
$data2 = [23,18,17,17,16,15,16,14,15,10,11,7,4,5];
getTrend($data1) //Returns up
getTrend($data2) // Returns down
function getTrend($arr)
{
$up = 0;
$down = 0;
$prev = "";
foreach($arr as $val)
{
if($prev != "" && $val > $prev)
{
$up = $val-$prev;
}
else if($prev != "" && $val < $prev)
{
$down = $prev-$val ;
}
$prev = $val);
}
if($up > $down)
{
return "up";
}
else if($down > $up)
{
return "down";
}
else {
return "flat";
}
}
I am trying to get year and month from the letters using established sequence. I know that the sequence is based on the following letters:
$letters = array('B','C','D','F','G','H','J','K','L','M','N','P','R','S','T','V','W','X','Y','Z');
It started with 0000BBB and when it reaches 9999 it becomes BBC, BBD etc. So I don't need the numbers in that case and only letters as I have a list of last registered sequence per year and month like this:
$plates = array(
array('2018','KHF','KHX','KJV','KKN','KLM','KML','KNK','KPD','KPR','KPT','----','----'),
array('2017','JWN','JXF','JYB','JYT','JZP','KBM','KCH','KCV','KDK','KFB','KFV','KGN'),
array('2016','JLN','JMF','JMY','JNR','JPK','JRG','JRZ','JSL','JTB','JTR','JVH','JVZ'),
array('2015','JCK','JCY','JDR','JFG','JFW','JGP','JHJ','JHT','JJH','JJW','JKK','JKZ'),
array('2014','HVN','HVZ','HWM','HXB','HXN','HYD','HTY','HZB','HZL','HZZ','JBL','JBY'),
array('2013','HNT','HPC','HPN','HPY','HRK','HRX','HSK','HSR','HSZ','HTK','HTV','HVF'),
array('2012','HJC','HJM','HKB','HKL','HKX','HLK','HLW','HMD','HML','HMT','HNC','HNK'),
array('2011','HBP','HCB','HCR','HDC','HDR','HFF','HFT','HGC','HGM','HGX','HHH','HHT'),
array('2010','GTC','GTS','GVM','GWC','GWV','GXP','GYD','GYM','GYX','GZJ','GZT','HBG'),
array('2009','GKS','GLC','GLP','GMC','GMN','GNF','GNY','GPJ','GPW','GRM','GSC','GSR'),
array('2008','FZR','GBN','GCK','GDH','GFC','GFY','GGV','GHG','GHT','GJJ','GJV','GKH'),
array('2007','FKY','FLV','FNB','FNZ','FRC','FSJ','FTP','FVJ','FWC','FXB','FXY','FYY'),
array('2006','DVW','DWT','DXZ','DYY','FBC','FCJ','FDP','FFK','FGF','FHD','FJD','FKC'),
array('2005','DFZ','DGX','DHZ','DKB','DLD','DMJ','DNP','DPK','DRG','DSC','DTB','DVB'),
array('2004','CRV','CSS','CTT','CVR','CWR','CXT','CYY','CZP','DBJ','DCH','DDG','DFF'),
array('2003','CDV','CFM','CGJ','CHF','CJC','CKB','CLD','CLV','CMM','CNK','CPF','CRC'),
array('2002','BSL','BTF','BTZ','BVW','BWT','BXP','BYP','BZF','BZV','CBP','CCH','CDC'),
array('2001','BFJ','BGF','BHG','BJC','BKB','BLC','BMF','BMW','BNL','BPG','BRB','BRT'),
array('2000','---','---','---','---','---','---','---','---','BBJ','BCD','BCY','BDR')
);
That means that array index 0 is the year and from 1 to 12 would be month. I am trying to find a match but then realize I can not search exact value and need to look for nearest value based on letters.
I would deeply appreciate if anyone could direct me in right direction what would be the best method of doing this.
This is a test so far but this will just return an exact match, I would have to search any possible letters such as KHW as an example that would have to match as nearest value to KHX
foreach ($plates as $key => $val) {
$search = array_search('KHX', $plates[$key]);
if($search){
echo $search."\n";
echo $plates[$key][0];
break;
}
}
You can solve it with O(log n) with a binary search. But in a more straightforward solution, you can solve it with O(n).
You can calculate the difference between each word with the below algorithm.
<?php
function strToInt($str)
{
$result = 0;
for ($i = 0; $i < strlen($str); $i++) {
$result = $result * 100 + ord($str[$i]);
}
return $result;
}
function find($searchStr)
{
$plates = [
['2018','KHF','KHX','KJV','KKN','KLM','KML','KNK','KPD','KPR','KPT','----','----'],
['2017','JWN','JXF','JYB','JYT','JZP','KBM','KCH','KCV','KDK','KFB','KFV','KGN'],
['2016','JLN','JMF','JMY','JNR','JPK','JRG','JRZ','JSL','JTB','JTR','JVH','JVZ'],
['2015','JCK','JCY','JDR','JFG','JFW','JGP','JHJ','JHT','JJH','JJW','JKK','JKZ'],
['2014','HVN','HVZ','HWM','HXB','HXN','HYD','HTY','HZB','HZL','HZZ','JBL','JBY'],
['2013','HNT','HPC','HPN','HPY','HRK','HRX','HSK','HSR','HSZ','HTK','HTV','HVF'],
['2012','HJC','HJM','HKB','HKL','HKX','HLK','HLW','HMD','HML','HMT','HNC','HNK'],
['2011','HBP','HCB','HCR','HDC','HDR','HFF','HFT','HGC','HGM','HGX','HHH','HHT'],
['2010','GTC','GTS','GVM','GWC','GWV','GXP','GYD','GYM','GYX','GZJ','GZT','HBG'],
['2009','GKS','GLC','GLP','GMC','GMN','GNF','GNY','GPJ','GPW','GRM','GSC','GSR'],
['2008','FZR','GBN','GCK','GDH','GFC','GFY','GGV','GHG','GHT','GJJ','GJV','GKH'],
['2007','FKY','FLV','FNB','FNZ','FRC','FSJ','FTP','FVJ','FWC','FXB','FXY','FYY'],
['2006','DVW','DWT','DXZ','DYY','FBC','FCJ','FDP','FFK','FGF','FHD','FJD','FKC'],
['2005','DFZ','DGX','DHZ','DKB','DLD','DMJ','DNP','DPK','DRG','DSC','DTB','DVB'],
['2004','CRV','CSS','CTT','CVR','CWR','CXT','CYY','CZP','DBJ','DCH','DDG','DFF'],
['2003','CDV','CFM','CGJ','CHF','CJC','CKB','CLD','CLV','CMM','CNK','CPF','CRC'],
['2002','BSL','BTF','BTZ','BVW','BWT','BXP','BYP','BZF','BZV','CBP','CCH','CDC'],
['2001','BFJ','BGF','BHG','BJC','BKB','BLC','BMF','BMW','BNL','BPG','BRB','BRT'],
['2000','---','---','---','---','---','---','---','---','BBJ','BCD','BCY','BDR']
];
$minYear = null;
$minKey = null;
$minDiff = strToInt('ZZZ');
$searchInt = strToInt($searchStr);
for ($i = 0; $i < count($plates); $i++) {
for ($j = 1; $j < 13; $j++) {
if(abs($searchInt - strToInt($plates[$i][$j])) < $minDiff) {
$minDiff = abs($searchInt - strToInt($plates[$i][$j]));
$minYear = $plates[$i][0];
$minKey = $plates[$i][$j];
}
}
}
return [$minYear, $minKey];
}
print_r(find('KHW'));
The code down below is by no means optimized, but it's rather a concept of how you might solve your problem.
//Flatten out array (one dimension without years and ----)
$flatten = array();
foreach($plates as $platevalues) {
foreach($platevalues as $pv) {
if ($pv != '---' && $pv != '----' && intval($pv) == 0) {
//Create a string only if valid letters included in the $letters-array
//This string is then added to the new array that is flattened out
$pv2 = '';
for($i=0;$i<strlen($pv);$i++) {
$letter = substr($pv,$i,1);
if (in_array($letter, $letters) !== false) {
$pv2 .= $letter;
}
}
$flatten[] = $pv2;
}
}
}
//Do the search
$search = 'GWN';
$search_result = '';
//Create a new search string based on first found in flattened
//plates array (first G, then GW, then GWN)
for($i=0;$i<strlen($search);$i++) {
foreach($flatten as $key=>$f) {
if (substr($search,0,$i+1) == substr($f,0,$i+1)) {
$search_result .= substr($search,$i,1);
break;
}
}
}
/*
$search_result is: GW
*/
//Create a new array where all items that begins with GW are included
$result = [];
foreach($flatten as $key=>$item) {
if (substr($search_result,0,strlen($search_result)) ==
substr($item,0,strlen($search_result))) {
$result[] = $item;
}
}
/*
$result =
array (size=2)
0 => string 'GWC' (length=3)
1 => string 'GWV' (length=3)
*/
//Create an array with total ASCII-value for each item
//in the $result array above
$result_o = [];
foreach($result as $item) {
$o = 0;
for($i=0;$i<strlen($item);$i++) {
$o += ord(substr($item,$i,1));
}
$result_o[]= $o;
}
/*
$result_o =
array (size=2)
0 => int 225
1 => int 244
*/
//Get the total ASCII-value for the original search string
$search_o = 0;
for($i=0;$i<strlen($search);$i++) {
$search_o += ord(substr($search,$i,1));
}
/*
$search_ o = 236
*/
//Find closest value in the $result_o (ASCII) - array compared (225,244)
//to the original $search_o ASCII value above (236)
$closest = 0;
$use_key = 0;
foreach($result_o as $key=>$item) {
if ($closest == 0 || abs($search_o - $closest) > abs($item - $search_o)) {
$closest = $item;
$use_key = $key;
}
}
/*
$closest = 244 (it's closer to 236 than 225 is)
$use_key = 1
*/
To get the result you have:
/*
$result =
array (size=2)
0 => string 'GWC' (length=3)
1 => string 'GWV' (length=3)
*/
//This should print out GWV
echo 'result=' . $result[$use_key];
This code is working fine when the array length is 8 or 10 only. When we are checking this same code for more than 10 array length.it get loading not showing the results.
How do reduce my code. If you have algorithm please share. Please help me.
This program working flow:
$allowed_per_room_accommodation =[2,3,6,5,3,5,2,5,4];
$allowed_per_room_price =[10,30,60,40,30,50,20,60,80];
$search_accommodation = 10;
i am get subsets = [5,5],[5,3,2],[6,4],[6,2,2],[5,2,3],[3,2,5]
Show lowest price room and then equal of 10 accommodation; output like as [5,3,2];
<?php
$dp=array(array());
$GLOBALS['final']=[];
$GLOBALS['room_key']=[];
function display($v,$room_key)
{
$GLOBALS['final'][] = $v;
$GLOBALS['room_key'][] = $room_key;
}
function printSubsetsRec($arr, $i, $sum, $p,$dp,$room_key='')
{
// If we reached end and sum is non-zero. We print
// p[] only if arr[0] is equal to sun OR dp[0][sum]
// is true.
if ($i == 0 && $sum != 0 && $dp[0][$sum]) {
array_push($p,$arr[$i]);
array_push($room_key,$i);
display($p,$room_key);
return $p;
}
// If $sum becomes 0
if ($i == 0 && $sum == 0) {
display($p,$room_key);
return $p;
}
// If given sum can be achieved after ignoring
// current element.
if (isset($dp[$i-1][$sum])) {
// Create a new vector to store path
// if(!is_array(#$b))
// $b = array();
$b = $p;
printSubsetsRec($arr, $i-1, $sum, $b,$dp,$room_key);
}
// If given $sum can be achieved after considering
// current element.
if ($sum >= $arr[$i] && isset($dp[$i-1][$sum-$arr[$i]]))
{
if(!is_array($p))
$p = array();
if(!is_array($room_key))
$room_key = array();
array_push($p,$arr[$i]);
array_push($room_key,$i);
printSubsetsRec($arr, $i-1, $sum-$arr[$i], $p,$dp,$room_key);
}
}
// Prints all subsets of arr[0..n-1] with sum 0.
function printAllSubsets($arr, $n, $sum,$get=[])
{
if ($n == 0 || $sum < 0)
return;
// Sum 0 can always be achieved with 0 elements
// $dp = new bool*[$n];
$dp = array();
for ($i=0; $i<$n; ++$i)
{
// $dp[$i][$sum + 1]=true;
$dp[$i][0] = true;
}
// Sum arr[0] can be achieved with single element
if ($arr[0] <= $sum)
$dp[0][$arr[0]] = true;
// Fill rest of the entries in dp[][]
for ($i = 1; $i < $n; ++$i) {
for ($j = 0; $j < $sum + 1; ++$j) {
// echo $i.'d'.$j.'.ds';
$dp[$i][$j] = ($arr[$i] <= $j) ? (isset($dp[$i-1][$j])?$dp[$i-1][$j]:false) | (isset($dp[$i-1][$j-$arr[$i]])?($dp[$i-1][$j-$arr[$i]]):false) : (isset($dp[$i - 1][$j])?($dp[$i - 1][$j]):false);
}
}
if (isset($dp[$n-1][$sum]) == false) {
return "There are no subsets with";
}
$p;
printSubsetsRec($arr, $n-1, $sum, $p='',$dp);
}
$blockSize = array('2','3','6','5','3','5','2','5','4');
$blockvalue = array('10','30','60','40','30','50','20','60','80');
$blockname = array("map","compass","water","sandwich","glucose","tin","banana","apple","cheese");
$processSize = 10;
$m = count($blockSize);
$n = count($processSize);
// sum of sets in array
printAllSubsets($blockSize, $m, $processSize);
$final_subset_room = '';
$final_set_room_keys = '';
$final_set_room =[];
if($GLOBALS['room_key']){
foreach ($GLOBALS['room_key'] as $set_rooms_key => $set_rooms) {
$tot = 0;
foreach ($set_rooms as $set_rooms) {
$tot += $blockvalue[$set_rooms];
}
$final_set_room[$set_rooms_key] = $tot;
}
asort($final_set_room);
$final_set_room_first_key = key($final_set_room);
$final_all_room['set_room_keys'] = $GLOBALS['room_key'][$final_set_room_first_key];
$final_all_room_price['set_room_price'] = $final_set_room[$final_set_room_first_key];
}
if(isset($final_all_room_price)){
asort($final_all_room_price);
$final_all_room_first_key = key($final_all_room_price);
foreach ($final_all_room['set_room_keys'] as $key_room) {
echo $blockname[$key_room].'---'. $blockvalue[$key_room];
echo '<br>';
}
}
else
echo 'No Results';
?>
I'm assuming your task is, given a list rooms, each with the amount of people it can accommodate and the price, to accommodate 10 people (or any other quantity).
This problem is similar to 0-1 knapsack problem which is solvable in polynomial time. In knapsack problem one aims to maximize the price, here we aim to minimize it. Another thing that is different from classic knapsack problem is that full room cost is charged even if the room is not completely occupied. It may reduce the effectiveness of the algorithm proposed at Wikipedia. Anyway, the implementation isn't going to be straightforward if you have never worked with dynamic programming before.
If you want to know more, CLRS book on algorithms discusses dynamic programming in Chapter 15, and knapsack problem in Chapter 16. In the latter chapter they also prove that 0-1 knapsack problem doesn't have trivial greedy solution.