多字节安全计数字符串中的不同字符

Question

我不想找到一种智能且有效的方法来计算一个字符串中有多少个不同的字母字符。示例：

$str = "APPLE";
echo char_count($str) // should return 4, because APPLE has 4 different chars 'A', 'P', 'L' and 'E'

$str = "BOB AND BOB"; // should return 5 ('B', 'O', 'A', 'N', 'D'). 

$str = 'PLÁTANO'; // should return 7 ('P', 'L', 'Á', 'T', 'A', 'N', 'O')

它应该支持 UTF-8 字符串！

Answer 1

如果您正在处理 UTF-8（您确实应该考虑这一点，恕我直言），则所有发布的解决方案（使用 strlen、str_split 或 count_chars）都不起作用，因为它们都将一个字节视为一个字符（对于显然是 UTF-8）。

<?php

$treat_spaces_as_chars = true;
// contains hälöwrd and a space, being 8 distinct characters (7 without the space)
$string = "hällö wörld"; 
// remove spaces if we don't want to count them
if (!$treat_spaces_as_chars) {
  $string = preg_replace('/\s+/u', '', $string);
}
// split into characters (not bytes, like explode() or str_split() would)
$characters = preg_split('//u', $string, -1, PREG_SPLIT_NO_EMPTY);
// throw out the duplicates
$unique_characters = array_unique($characters);
// count what's left
$numer_of_characters = count($unique_characters);

如果你想扔掉所有非单词字符：

<?php

$ignore_non_word_characters = true;
// contains hälöwrd and PIE, as this is treated as a word character (Greek)
$string = "h,ä*+l•π‘°’lö wörld"; 
// remove spaces if we don't want to count them
if ($ignore_non_word_characters) {
  $string = preg_replace('/\W+/u', '', $string);
}
// split into characters (not bytes, like explode() or str_split() would)
$characters = preg_split('//u', $string, -1, PREG_SPLIT_NO_EMPTY);
// throw out the duplicates
$unique_characters = array_unique($characters);
// count what's left
$numer_of_characters = count($unique_characters);

var_dump($characters, $unique_characters, $numer_of_characters);

Answer 2

只需使用 count_chars：

echo count(array_filter(count_chars($str)));

从 count_chars()< 返回的数组/code> 还会告诉您字符串中每个字符有多少个。

Answer 3

count_chars 返回所有 ASCII 字符的映射，告诉您字符串中每个字符的数量。这是您自己实施的起点。

function countchars($str, $ignoreSpaces) {
  $map = array();
  $len = strlen($str);
  for ($i=0; $i < $len; $i++) {
    if (!isset($map[$str{$i}])) {
      $map[$str{$i}] = 1;
    } else {
      $map[$str{$i}]++;
    }    
  }

  if ($ignoreSpaces) {
    unset($map[' ']);
  }

  return $map;
}

print_r(countchars('Hello World'));

Answer 4

这是一个使用关联数组的魔力来完成此操作的函数。以线性时间工作。 （大O = log(n)）

function uniques($string){
   $arr = array();
   $parts = str_split($string);
   foreach($parts as $part)
      $arr["$part"] = "yup";
   return count($arr);
}

$str = "APPLE";
echo uniques($str);  // outputs 4

Answer 5

我的看法是，

$chars = array_count_values(str_split($input));

这将为您提供一个唯一字母的关联数组作为键，出现次数作为值。

如果你对出现的次数不感兴趣，

$chars = array_unique(str_split($input));
$numChars = count($chars);