How to find the sum and average of the elements of an array in php

PHP is a commonly used server-side programming language, widely used in website development, application development and other fields. In PHP, array is a commonly used data type, often used to store a group of related data. This article will introduce how to find the sum and average of an array.

Find the sum of an array

In PHP, you can use the array_sum() function to find the sum of an array. The syntax of this function is as follows:

array_sum(array $array): float|int

Among them, $array is the array to be summed, and the return value is the sum of the array elements. The following is a sample code:

<?php $arr = array(1, 2, 3, 4, 5);
$sum = array_sum($arr);
echo &#39;数组的和为：&#39; . $sum;
?>

Run the above code, the output result is as follows:

数组的和为：15

In the above code, we define an integer array $arr, which contains elements 1, 2, 3, 4, 5. Then use the array_sum() function to find the sum of the array and assign it to the variable $sum. Finally, use the echo statement to output the sum of the array.

Find the average of an array

In PHP, you can use the count() function to get the number of elements of an array, and then divide the sum of the array by the number of elements to find it. average value. The following is a sample code:

<?php $arr = array(1, 2, 3, 4, 5);
$sum = array_sum($arr);
$count = count($arr);
$avg = $sum / $count;
echo &#39;数组的平均值为：&#39; . $avg;
?>

Run the above code, the output result is as follows:

数组的平均值为：3

In the above code, in addition to using the array_sum() function to find the sum of the array, we also use count The () function gets the number of elements of the array, and divides the sum of the array by the number of elements to find the average. Finally, use the echo statement to output the average value of the array.

For extremely large arrays, a one-time sum or average may cause memory overflow or timeout. Therefore, you can use a loop or recursion to accumulate or calculate the value of the array elements one by one.

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