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Given a two list. Create a third list by picking an odd-index element from the first list and even index elements from the second.
For Example:
listOne = [3, 6, 9, 12, 15, 18, 21]
listTwo = [4, 8, 12, 16, 20, 24, 28]
Expected Output:
Printing Final third list
[6, 12, 18, 4, 12, 20, 28]
I think, it will be helpful for you.
<?php
$listOne = [3, 6, 9, 12, 15, 18, 21];
$listTwo = [4, 8, 12, 16, 20, 24, 28];
$NlistOne=[];
$NlistTwo=[];
//odd-index positions from list one [6, 12, 18]
foreach ($listOne as $key => $value) {
if($key%2==1){
array_push($NlistOne, $value);
}
}
//even-index positions from list two [4, 12, 20, 28]
foreach ($listTwo as $key => $value) {
if($key%2==0){
array_push($NlistTwo, $value);
}
}
//Printing Final third list [6, 12, 18, 4, 12, 20, 28]
print_r(array_merge($NlistOne,$NlistTwo));
?>
Output will be:
Array ( [0] => 6 [1] => 12 [2] => 18 [3] => 4 [4] => 12 [5] => 20 [6] => 28 )
//init result array
//loop over listOne, using for($i=1;$i<sizeof($listOne);$i=$i+2)
//and add to result for each iteration, $resultArr[] = $listOne[$i]
//do the same with listTwo, but for($i=*0*
You can merge both of arrays and then pick all odd elements
$listOne = [3, 6, 9, 12, 15, 18, 21];
$listTwo = [4, 8, 12, 16, 20, 24, 28];
$result = [];
foreach ( array_merge($listOne, $listTwo) as $value ){
if ( $key % 2 ) {
$result[] = $value;
}
}
If array length isn't fixed, say it could contain not 7 elements, then you need to check it before merging to make it have odd number of elements
$listOne = [3, 6, 9, 12, 15, 18, 21, 777];
$listTwo = [4, 8, 12, 16, 20, 24, 28];
$result = [];
if ( count($listOne) % 2 !== 1 ) {
$listOne[] = 0;
}
foreach( array_merge($listOne, $listTwo) as $value ){
if ( $key % 2 ) {
$result[] = $value;
}
}
you don't have to loop over whole array array_filter will do this for you, Constant ARRAY_FILTER_USE_KEY will check each key for odd or for even
<?php
$a1 = [3, 6, 9, 12, 15, 18, 21];
$a2 = [4, 8, 12, 16, 20, 24, 28];
function odd($var)
{
// returns whether the input integer is odd
return $var & 1;
}
function even($var)
{
// returns whether the input integer is even
return !($var & 1);
}
$result= (array_merge(array_filter($a1, 'odd', ARRAY_FILTER_USE_KEY),array_filter($a2, 'even', ARRAY_FILTER_USE_KEY)));
output you will get
Array (
[0] => 6
[1] => 12
[2] => 18
[3] => 4
[4] => 12
[5] => 20
[6] => 28 )
Iterate over first one and take only values of odds indices, and loop again through second one and take evens indices.
$listOne = [3, 6, 9, 12, 15, 18, 21];
$listTwo = [4, 8, 12, 16, 20, 24, 28];
$res = [];
for ($i=0; $i < count($listOne); $i++) {
if($i & 1) // $i is odd.
$res[] = $listOne[$i];
}
for ($j=0; $j < count($listTwo); $j++) {
if(n % 2 === 0) // $j is even.
$res[] = $listTwo[$j];
}
Result:
echo "List One :".json_encode($listOne)."<br>";
echo "List Two :".json_encode($listTwo)."<br>";
echo "Lists Merged:".json_encode($res);
Output:*
/*
List One :[3,6,9,12,15,18,21]
List Two :[4,8,12,16,20,24,28]
Lists Merged:[6,12,18,4,12,20,28]
*/
Iterate over arrays:
take odds in array starting with index One and increment it by two.
take evens in array by starting with index Zero and increment it by two.
$listOne = [3, 6, 9, 12, 15, 18, 21]; $listTwo = [4, 8, 12, 16, 20, 24, 28];
$res = [];
for ($i=1; $i < count($listOne); $i+=2) {
$res[] = $listOne[$i];
}
for ($j=0; $j < count($listTwo); $j+=2) {
$res[] = $listTwo[$j];
}
print(json_encode($res)); // [6,12,18,4,12,20,28]
Given two arrays $A1 and $A2, sort $A1 in such a way that the relative order among the elements will be same as those in $A2. For the elements not present in $A2, move them to the back of the array in ascending order.
$A1 = [2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8];
$A2 = [2, 1, 8, 3];
Desired output:
[2, 2, 1, 1, 8, 8, 3, 5, 6, 7, 9]
Coding attempt:
$sorted = array();
foreach($a1 as $key => $value) {
if(in_array($value, $a2)) {
$sorted[array_search($value, $a1)] = $value;
}
}
This can be done via for each loop :
$arr1 = array(2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8); // array to be sorted
$arr2 = array(2, 1, 8, 3); // refrence array for sort logic
// Output: A1[] = {2, 2, 1, 1, 8, 8, 3, 5, 6, 7, 9}
$sortarr = array(); // array to store final sorted values
foreach ($arr2 as $a) {
foreach ($arr1 as $k => $b) {
if($b==$a) {
$sortarr[]=$b;
unset($arr1[$k]);
}
}
}
$finalarr = array_merge($sortarr, $arr1);
print_r($finalarr);
You can use usort like this:
$k = array_flip($a2); // Create an associative array for the second array
usort($a1, function($a, $b) use ($k) {
return isset($k[$a]) ? (isset($k[$b]) ? $k[$a]-$k[$b] : -1) : (isset($k[$b]) ? 1 : $a-$b);
});
Other solutions that use a nested loop result in a time complexity of O(n²), while this has a time complexity of O(nlogn).
i tried this code and works:
<?php
/*
Input: A1[] = {2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8}
A2[] = {2, 1, 8, 3}
Output: A1[] = {2, 2, 1, 1, 8, 8, 3, 5, 6, 7, 9}
*/
$a1=array(2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8);
$a2=array(2, 1, 8, 3);
$a3=array();
sort($a1);//order array
//order array a3 with a2 value....
for($i=0;$i<sizeof($a2);$i++){
for($j=0;$j<sizeof($a1);$j++){
if($a1[$j]==$a2[$i]){
array_push($a3,$a2[$i]);
//if exsist value i change value in a1 in x
$a1[$j]="x";
}
}
}
//write in a3 the next number not present in a2
for($i=0;$i<sizeof($a1);$i++){
if($a1[$i]<>"x"){
array_push($a3, $a1[$i]);
}
}
print_r($a3);
//out:Array ( [0] => 2 [1] => 2 [2] => 1 [3] => 1 [4] => 8 [5] => 8 [6] => 3 [7] => 5 [8] => 6 [9] => 7 [10] => 9 )
?>
Hope this helps
I do not endorse the use of nested loops because they will be doing too many unnecessary cycles. #trincot's approach moreso represents what I would use in a professional application, but I will demonstrate a more modern and concise style with no iterated function calls.
Code: (Demo)
$arr1 = [2, 1, 2, 5, 7, 1, 9, 3, 6, 8, 8];
$arr2 = [2, 1, 8, 3];
$priority = array_flip($arr2);
$fallback = count($arr2);
usort(
$arr1,
fn($a, $b) =>
[$priority[$a] ?? $fallback, $a]
<=>
[$priority[$b] ?? $fallback, $b]
);
var_export($arr1);
// [2, 2, 1, 1, 8, 8, 3, 5, 6, 7, 9]
$arr2 is flipped into a lookup array, and count() is used when a value is not found in the lookup array. Whenever the first rule results in a tie, the second rule orders the numbers numerically in an ascending direction.
The meaning of custom function's syntax is:
[left value rule1, left value rule2] compared to [right value rule1, right value rule2]
The spaceship operator (<=>) will compare the corresponding element one at a time "rule1 vs rule1" then "rule2 vs rule2" as needed.
I have this array [1,1,2,2,2,3,4,4,5,6,6,6,7], may I group the array according to the range value, so get the final result:
'1-3' = [1,1,2,2,2,3]; // Count is 6
'4-5' = [4,4,5]; // Count is 3
'6-7' = [6,6,6,7]; // Count is 4
What you need I believe is:
function array_get_range($array, $min, $max) {
return array_filter($array, function($element) use ($min, $max) {
return $element >= $min && $element <= $max;
});
}
$array = [1,1,2,2,2,3,4,4,5,6,6,6,7];
$range13 = array_get_range($array, 1, 3); // [1, 1, 2, 2, 2, 3]
$range45 = array_get_range($array, 4, 5); // [4, 4, 5]
$range67 = array_get_range($array, 6, 7); // [6, 6, 6, 7]
Create a new array with your ranges, then iterate through the values and through the ranges inside. If the current value is inside the range, add the record to the current range:
<?php
$numbers = [1,1,2,2,2,3,4,4,5,6,6,6,7];
$counts = [];
$counts[] = ["values"=> [1, 3], "records" => []]; // first value in "values" is min, second is max
$counts[] = ["values"=> [4, 5], "records" => []];
$counts[] = ["values"=> [6, 7], "records" => []];
foreach ($numbers as $number) {
foreach ($counts as $key => &$value) {
if ($number >= $value["values"][0] && $number <= $value["values"][1]) { // if between the range, add it to the records
$value["records"][] = $number;
}
}
}
echo "<pre>";
foreach ($counts as $count) {
echo $count["values"][0]." - ".$count["values"][1]." = ".count($count["records"])." elements<br>";
}
Demo
I think array_intersect() and range() with sizeof()/count() does a cleaner job for this task. It eliminates the double-conditional.
Code: (Demo)
function count_range($array, $min, $max) {
return sizeof(array_intersect($array, range($min, $max)));
}
$array = [1, 1, 2, 2, 2, 3, 4, 4, 5, 6, 6, 6, 7];
echo count_range($array, 1, 3); // 6 from [1, 1, 2, 2, 2, 3]
echo count_range($array, 4, 4); // 2 from [4, 4]
echo count_range($array, 2, 7); // 11 from [2, 2, 2, 3, 4, 4, 5, 6, 6, 6, 7]
range() creates a whitelist array of one or more consecutive integers.
array_intersect() compares the input array and the whitelist array all at once.
sizeof() is just an alias of count() to determine how many elements were retained.
am trying to get output of following array in one format. its not getting
<?php
$distance_covered = array( '1_JAN_2017' => array('DRIVER_1' => array(2, 5, 3),'DRIVER_2' => array(3, 2, 6, 9)),
'2_JAN_2017' => array('DRIVER_1' => array(3, 9), 'DRIVER_3' => array(1, 4, 8)),
'3_JAN_2017' => array('DRIVER_4' => array(9), 'DRIVER_1' => array(2, 7, 5, 2)),
'4_JAN_2017' => array('DRIVER_1' => array(5, 3, 3, 2), 'DRIVER_4' => array(4, 9, 8, 5)),
'5_JAN_2017' => array('DRIVER_2' => array(8, 5), 'DRIVER_5' => array(3, 9, 7)),
'6_JAN_2017' => array('DRIVER_5' => array(2, 1, 7, 5), 'DRIVER_4' => array(1, 9, 6)),
'7_JAN_2017' => array('DRIVER_4' => array(5, 2, 9), 'DRIVER_3' => array(4, 1, 6)), );
The above is my array
i want output in the following format
Output: Array ( [DRIVER_1] => 51, [DRIVER_2] => 33, [DRIVER_3] => 24, [DRIVER_4] => 67, [DRIVER_5] => 34 )
this is the sum of distance travelled by each driver in all trips
i tried code like this,anybody knows please help
$res = array();
foreach($distance_covered as $value) {
foreach($value as $key => $number) {
(!isset($res[$key])) ?
$res[$key] = $number :
$res[$key] += $number;
}
}
print_r($res);
?>
This one works for me
$res = array();
foreach($distance_covered as $value)//the array which you have given us
{
foreach($value as $key => $number) //loop over array of date
{
if(!isset($res[$key]))//check if the key exist in over defined array if no then run this
{
$res[$key] = array_sum($number);// Sum all distances of that driver
continue;//set the key and continue the foreach...
}
$res[$key] += array_sum($number);// Sum all distances of that driver
}
}
print_r($res);
die;
And the Output is
Array
(
[DRIVER_1] => 51
[DRIVER_2] => 33
[DRIVER_3] => 24
[DRIVER_4] => 67
[DRIVER_5] => 34
)
This should work:
$res = array();
foreach($distance_covered as $value) {
foreach($value as $key => $number) {
foreach ($number as $n) {
if (isset($res[$key])) {
$res[$key] += $n;
} else {
$res[$key] = $n;
}
}
}
}
print_r($res);
Just traverse through array of arrays.
$distance_covered = array(
'1_JAN_2017' => array('DRIVER_1' => array(2, 5, 3),'DRIVER_2' => array(3, 2, 6, 9)),
'2_JAN_2017' => array('DRIVER_1' => array(3, 9), 'DRIVER_3' => array(1, 4, 8)),
'3_JAN_2017' => array('DRIVER_4' => array(9), 'DRIVER_1' => array(2, 7, 5, 2)),
'4_JAN_2017' => array('DRIVER_1' => array(5, 3, 3, 2), 'DRIVER_4' => array(4, 9, 8, 5)),
'5_JAN_2017' => array('DRIVER_2' => array(8, 5), 'DRIVER_5' => array(3, 9, 7)),
'6_JAN_2017' => array('DRIVER_5' => array(2, 1, 7, 5), 'DRIVER_4' => array(1, 9, 6)),
'7_JAN_2017' => array('DRIVER_4' => array(5, 2, 9), 'DRIVER_3' => array(4, 1, 6)), );
// Counting.
$merged = [];
foreach ($distance_covered as $day => $stats) {
foreach ($stats as $driver => $distances) {
if (!isset($merged[$driver])) {
$merged[$driver] = 0;
}
$merged[$driver] += array_sum($distances);
}
}
// Display.
echo "<pre>";
print_r($merged);
echo "</pre>";
You are close, but...
$res = array ();
foreach ( $distance_covered as $value ) {
foreach ( $value as $key=> $driver ) {
if ( isset($res[$key]) == false ){
$res[$key]=0;
}
$res[$key] += array_sum($driver);
}
}
print_r($res);
The first foreach simply splits the data down to the days.
The second one returns elements like $key = 'DRIVER_1' and $driver = array(3, 9).
If this is the first time you've encountered this driver, then you need to ensure that the element in $res exists, so set it to 0.
Once you know there is an element there, you can add in the sum of the values ( 3 & 9 in this case ) using the += array_sum($driver) bit. The += is simply adding to rather than having to say a=a+b, you can say a+=b.
[sarcastic voice] I can't believe everybody overlooked this convoluted function-based one-liner...
Code: (Demo)
var_export(array_map('array_sum', array_merge_recursive(...array_values($distance_covered))));
Output:
array (
'DRIVER_1' => 51,
'DRIVER_2' => 33,
'DRIVER_3' => 24,
'DRIVER_4' => 67,
'DRIVER_5' => 34,
)
*this is virtually guaranteed to process slower than any other posted answer.
Remove the first level associative keys (date strings) with array_values()
Unpack the array of arrays with the "splat operator" (...) and feed to array_merge_recursive() to group values
Sum the subarray values by calling array_sum() on each subarray with array_map()
(This is merely an exercise of thinking outside the box.)
Beyond that no one suggested using a null coalescing operator, so I'll post what that can look like:
$driver_totals = [];
foreach ($distance_covered as $daily_log) {
foreach ($daily_log as $driver => $distances) {
$driver_totals[$driver] = ($driver_totals[$driver] ?? 0) + array_sum($distances);
}
}
var_export($driver_totals);
And if you have a special scenario where you only need to know the distance for a single specific driver, you can call upon array_column() like this:
$target_driver = 'DRIVER_4';
$total_distance = 0;
foreach (array_column($distance_covered, $target_driver) as $distances) {
$total_distance += array_sum($distances);
}
echo "{$target_driver} drove for a distance of {$total_distance}";
*Notice that the order of the drivers within each date array is inconsequential because array_column() is smart enough to find the desired distance subarray.
Finally, if you declare a whitelist array of all possible drivers, you can:
control the order of the drivers in the output
avoid the iterated isset() conditions
ensure that drivers without any distance records are included in the output
Code:
$roster = ['DRIVER_6', 'DRIVER_5', 'DRIVER_4', 'DRIVER_3', 'DRIVER_2', 'DRIVER_1'];
$driver_totals = array_fill_keys($roster, 0);
foreach ($distance_covered as $daily_log) {
foreach ($daily_log as $driver => $distances) {
$driver_totals[$driver] += array_sum($distances);
}
}
var_export($driver_totals);
I have multiple arrays, that get generated when a for loop completes. Now I need to get the total of all these arrays together, but in order, for instance:
array1(2, 4, 6, 8, 10)
array2(2, 4, 6, 8, 10)
I need to add the 2's together, the 4's etc etc
Then put them into another array.
Some combination of array_combine and array_sum seems to be the solution but I can't figure it out.
Any help is greatly appreciated.
You could do this without any php function except count,
$limit = count($array1);
for ($i = 0; $i < $limit; $i++) {
$array_res[$i] = $array1[$i] + $array2[$i];
should work. This will combine your two arrays and add them in the same order. But be sure you used sort or ksort on both of your arrays before if you're not sure they are in the same order.
1st stlye
$a1 = array(
"a" => 2
,"b" => 0
,"c" => 5
);
$a2 = array(
"a" => 3
,"b" => 9
,"c" => 7
,"d" => 10
);
$a3 = array_merge($a1, $a2);
print_r($a3);
2nd style
$a3 = $a1;
foreach($a2 as $k=>$v) {
if(array_key_exists($k,$a3))
$a3[$k] + = $v;
else
$a3[$k] = $v;
}
You should use array_map():
$arr1 = array(2, 4, 6, 8, 10);
$arr2 = array(1, 2, 3, 4, 5);
$arr3 = array_map(
function($foo, $bar){
return $foo+$bar;
},
$arr1, $arr2
);
print_r($arr3); // outputs:
Array
(
[0] => 3
[1] => 6
[2] => 9
[3] => 12
[4] => 15
)
You can do something like this
$a = array(2, 4, 6, 8, 10);
$b = array(2, 4, 6, 8, 10);
$c = array_map("test", $a, $b);
function test($a, $b) {
return $a + $b;
}
print_r($c);
As far as I understand your question You want sum of all the similar digits in both the array:
<?php
$array1= array(2, 4, 6, 8, 10);
$array2=array(2, 4, 6, 8, 10);
$array_sum=array();
foreach($array1 as $key1=>$val1)
{
$sum=0;
foreach($array2 as $key2 => $val2)
{
if($val1==$val2)
{
$sum=$sum+$val1;
}
}
$array_sum[]=$sum;
}
print_r($array_sum);
?>