PHP替換字符串的隨機詞

Question

我想替換一個隨機單詞，其中一個字符串中包含多個單詞。

假設字符串是

$str = 'I like blue, blue is my favorite colour because blue is very nice and blue is pretty';

假設我想用紅色替換藍色一詞，但在隨機位置僅2次。

所以完成一個功能后，輸出可能像

I like red, blue is my favorite colour because red is very nice and blue is pretty

另一個可能是

I like blue, red is my favorite colour because blue is very nice and red is pretty

因此，我想多次替換相同的單詞，但每次都在不同的位置。

我想到使用preg_match，但是沒有選擇peing替換的單詞的位置也是隨機的。

有人知道如何實現這一目標嗎？

Answer 1

就像我不喜歡使用regex來處理表面上很簡單的事情一樣，為了保證正好n個替換，我認為它可以在這里有所幫助，因為它可以輕松使用array_rand() ，它可以完成您的工作想要-從不確定長度的列表中選擇n個隨機項（已改進 ）。

<?php

    function replace_n_occurences ($str, $search, $replace, $n) {

        // Get all occurences of $search and their offsets within the string
        $count = preg_match_all('/\b'.preg_quote($search, '/').'\b/', $str, $matches, PREG_OFFSET_CAPTURE);

        // Get string length information so we can account for replacement strings that are of a different length to the search string
        $searchLen = strlen($search);
        $diff = strlen($replace) - $searchLen;
        $offset = 0;

        // Loop $n random matches and replace them, if $n < 1 || $n > $count, replace all matches
        $toReplace = ($n < 1 || $n > $count) ? array_keys($matches[0]) : (array) array_rand($matches[0], $n);
        foreach ($toReplace as $match) {
            $str = substr($str, 0, $matches[0][$match][1] + $offset).$replace.substr($str, $matches[0][$match][1] + $searchLen + $offset);
            $offset += $diff;
        }

        return $str;

    }

    $str = 'I like blue, blue is my favorite colour because blue is very nice and blue is pretty';

    $search = 'blue';
    $replace = 'red';
    $replaceCount = 2;

    echo replace_n_occurences($str, $search, $replace, $replaceCount);

看到它正常工作

Answer 2

echo preg_replace_callback('/blue/', function($match) { return rand(0,100) > 50 ? $match[0] : 'red'; }, $str);

Answer 3

好吧，您可以使用以下算法：

1. 計算您要替換字符串的隨機次數
2. 將字符串分解為數組
3. 對於該數組，僅當1到100之間的隨機值是％3（用於等距）時，才替換字符串出現
4. 減少在第1點計算的數字。
5. 重復直到數字達到0。

Answer 4

<?php
$amount_to_replace = 2;
$word_to_replace = 'blue';
$new_word = 'red';

$str = 'I like blue, blue is my favorite colour because blue is very nice and blue is pretty';

$words = explode(' ', $str); //convert string to array of words
$blue_keys = array_keys($words, $word_to_replace); //get index of all $word_to_replace

if(count($blue_keys) <= $amount_to_replace) { //if there are less to replace, we don't need to randomly choose.  just replace them all
    $keys_to_replace = array_keys($blue_keys);
}
else {
    $keys_to_replace = array();
    while(count($keys_to_replace) < $amount_to_replace) { //while we have more to choose
        $replacement_key = rand(0, count($blue_keys) -1);
        if(in_array($replacement_key, $keys_to_replace)) continue; //we have already chosen to replace this word, don't add it again
        else {
            $keys_to_replace[] = $replacement_key;
        }
    }
}

foreach($keys_to_replace as $replacement_key) {
    $words[$blue_keys[$replacement_key]] = $new_word;
}

$new_str = implode(' ', $words); //convert array of words back into string
echo $new_str."\n";
?>

注意，我剛剛意識到這不會替換第一個藍色，因為它以單詞“ blue”輸入到單詞數組中，因此在array_keys調用中不匹配。