我需要找到大于特定浮点值的记录数，并找到重复次数最多的数据组。例如，我有下面的数据，我需要找到有多少条目的值>4。
1.5
1.7
4.5
4.7
4.8
1.4
4.5
4.9
在上述数据中，大于4的值的最长连续重复为4.5、4.7、4.8。因此，我希望返回的总数应该是3。正如你所看到的，这个模式在4.8之后中断，因为这个数字是1.4以上。有没有办法识别这种模式？
试试这个，我在这里使用了一个数组：
$arr = array(
0 => '1.5',
1 => '1.7',
2 => '4.5',
3 => '4.7',
4 => '4.8',
5 => '1.4',
6 => '4.5',
7 => '4.9'
);
$chk_val = 4; // value which is checking
$cnt = 0;$inc = 0;
foreach ($arr as $i => $val) {
if ($val > $chk_val) {
$inc++;
if ($inc > $cnt) { $cnt = $inc;}
} else {
$inc = 0;
}
}
echo $cnt;
尝试这个
$n = 4; // number to check
$count = 0;
$max = 0;
$ele = array(1.5, 1.7, 4.5, 4.7, 4.8, 1.4, 4.5, 4.9);
for ($i = 0; $i < count($ele); $i++) {
if ($ele[$i] >= $n) { // check for greater element than given number
$count++; // increase consecutive counter variable
$arr[$max] = $count; //save continues max counter to array
} else {
$count = 0; //reset consecutive counter
$max++;
}
}
echo max($arr);
快速且肮脏。。。
function findNums($nums, $min = 4) {
$groups = array();
$groupcounts = array();
$groupindex = 0;
foreach($nums as $num) {
if($num > $min) {
$groups[$groupindex][] = $num;
if(array_key_exists($groupindex, $groupcounts)) {
$groupcounts[$groupindex]++;
} else {
$groupcounts[$groupindex] = 1;
}
} else {
$groupindex++;
}
}
return array($groupcounts, $groups);
}
// $your_numbers is your list
$nums = array_map('trim', explode("'n", $your_numbers));
$result = findNums($nums);
$counts = $result[0];
$maxcount = max($counts);
$groups = $result[1];
echo "max count is ".$maxcount." with values:'n";
$key = array_search($maxcount, $counts);
var_dump($groups[$key]);