Asked  7 Months ago    Answers:  5   Viewed   44 times

i have an array:

Array 
(
[0] => Array
    (
        [setid] => 2
        [income] => 100
    )

[1] => Array
    (
        [setid] => 2
        [income] => 120
    )

[2] => Array
    (
        [setid] => 3
        [income] => 700
    )
)

i need to find entrys with the same setid, sum their income up and delete duplicate entrys - in the end it should look like this:

Array
(
[0] => Array
    (
        [setid] => 2
        [income] => 220
    )

[1] => Array
    (
        [setid] => 3
        [income] => 700
    )
)

does someone know a sophosticated solution for my problem or do i have to take the long road and do every step manually?

thanks and greetings

 Answers

38

Just create a new array which you make fast adressable by using the setid as key. And reindex the array at the end.

$result = array();
foreach ($array as $val) {
    if (!isset($result[$val['setid']]))
        $result[$val['setid']] = $val;
    else
        $result[$val['setid']]['income'] += $val['income'];
}
$result = array_values($result); // reindex array
Wednesday, March 31, 2021
 
fhonics
answered 7 Months ago
59

Try below code:

<?php

$arr = array(
        array('city' => 'NewYork', 'cash' => '1000'),
        array('city' => 'Philadelphia', 'cash' => '2300'),
        array('city' => 'NewYork', 'cash' => '2000'),
    );

$newarray = array();
foreach($arr as $ar)
{
    foreach($ar as $k => $v)
    {
        if(array_key_exists($v, $newarray))
            $newarray[$v]['cash'] = $newarray[$v]['cash'] + $ar['cash'];
        else if($k == 'city')
            $newarray[$v] = $ar;
    }
}

print_r($newarray);


Output:

Array
(
    [NewYork] => Array
        (
            [city] => NewYork
            [cash] => 3000
        )

    [Philadelphia] => Array
        (
            [city] => Philadelphia
            [cash] => 2300
        )

)


Demo:
http://3v4l.org/D8PME

Wednesday, March 31, 2021
 
RenegadeAndy
answered 7 Months ago
42
function set_val(array &$arr, $path,$val)
{
   $loc = &$arr;
   foreach(explode('.', $path) as $step)
   {
     $loc = &$loc[$step];
   }
   return $loc = $val;
}
Wednesday, March 31, 2021
 
laurent
answered 7 Months ago
23

I am not sure if this will be useful. I have noticed that the ArrayObject class is 'interesting'...

I am not sure that this is even an 'answer'. It is more an observation about this class.

It handles the 'multidimensional array' stuff correctly as standard.

You may be able to add methods to make it do more of what you wish?

<?php //

class Config extends ArrayObject
{

//    private $data = array();

    public function __construct(array $data = array())
    {
        parent::__construct($data);
    }
}

$conf = new Config(array('a' => 'foo', 'b' => 'bar', 'c' => array('sub' => 'baz')));
$conf['c']['sub'] = 'notbaz';
$conf['c']['sub2'] = 'notbaz2';

var_dump($conf, $conf['c'], $conf['c']['sub']);

unset($conf['c']['sub']);

var_dump('isset?: ', isset($conf['c']['sub']));

var_dump($conf, $conf['c'], $conf['c']['sub2']);

Output:

object(Config)[1]
  public 'a' => string 'foo' (length=3)
  public 'b' => string 'bar' (length=3)
  public 'c' => 
    array
      'sub' => string 'notbaz' (length=6)
      'sub2' => string 'notbaz2' (length=7)

array
  'sub' => string 'notbaz' (length=6)
  'sub2' => string 'notbaz2' (length=7)

string 'notbaz' (length=6)

string 'isset?: ' (length=8)

boolean false

object(Config)[1]
  public 'a' => string 'foo' (length=3)
  public 'b' => string 'bar' (length=3)
  public 'c' => 
    array
      'sub2' => string 'notbaz2' (length=7)

array
  'sub2' => string 'notbaz2' (length=7)

string 'notbaz2' (length=7)
Saturday, May 29, 2021
 
Jubair
answered 5 Months ago
65

Solution using data.table:

require(data.table)
df <- structure(list(year = c(2015, 2015), ID = c(200, 200), Lats = c(30.5417, 
            30.5417), Longs = c(-20.5254, -20.5254), N = c(150, 90), n = c(30, 
            50), c_id = c(4142, 4142)), .Names = c("year", "ID", "Lats", 
            "Longs", "N", "n", "c_id"), row.names = c(NA, -2L), 
            class = "data.frame")
dt <- data.table(df)
dt[, lapply(.SD, sum), by="c_id,year,ID,Lats,Longs"]

   c_id year  ID    Lats    Longs   N  n
1: 4142 2015 200 30.5417 -20.5254  240 80

Solution using plyr:

require(plyr)
ddply(df, .(c_id, year, ID, Lats, Longs), function(x) c(N=sum(x$N), n=sum(x$n)))

  c_id year  ID    Lats    Longs   N  n
1 4142 2015 200 30.5417 -20.5254 240 80
Wednesday, September 22, 2021
 
abkds
answered 1 Month ago
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