我正在尝试做一些基于句子中字符数量的动态规划。英语字母表中哪个字母在屏幕上占像素最多?


当前回答

或者如果你想要一个宽度的映射,包含不止上面描述的alpha(数字)字符(就像我在非浏览器环境中需要的那样)

const chars = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "!", "\"", "#", "$", "%", "'", "(", ")", "*", "+", ",", "-", ".", "/", ":", ";", "=", "?", "@", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z", "[", "\\", "]", "^", "_", "`", "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "{", "|", "}", "~", " ", "&", ">", "<"] const test = document.createElement('div') test.id = "Test" document.body.appendChild(test) test.style.fontSize = 12 const result = {} chars.forEach(char => { let newStr = "" for (let i = 0; i < 10; i++) { if (char === " ") { newStr += "&nbsp;" } else { newStr += char } } test.innerHTML = newStr const width = (test.clientWidth) result[char] = width / 10 }) console.log('RESULT:', result) #Test { position: absolute; /* visibility: hidden; */ height: auto; width: auto; white-space: nowrap; /* Thanks to Herb Caudill comment */ }

其他回答

这段代码将获取数组中所有字符的宽度:

const alphabet = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"; var widths = []; var div = document.createElement('div'); div.style.float = 'left'; document.body.appendChild(div); var highestObservedWidth = 0; // widest characters (not just one) var answer = ''; for (var i = 0; i < alphabet.length; i++) { div.innerText = alphabet[i]; var computedWidthString = window.getComputedStyle(div, null).getPropertyValue("width"); var computedWidth = parseFloat(computedWidthString.slice(0, -2)); // console.log(typeof(computedWidth)); widths[i] = computedWidth; if(highestObservedWidth == computedWidth) { answer = answer + ', ' + div.innerText; } if(highestObservedWidth < computedWidth) { highestObservedWidth = computedWidth; answer = div.innerText; } } if (answer.length == 1) { div.innerHTML = ' Winner: ' + answer + '.'; } else { div.innerHTML = ' Winners: ' + answer + '.'; } div.innerHTML = div.innerHTML ; // console.log(widths); // console.log(widths.sort((a, b) => a - b));

这取决于字体。我会用你最熟悉的编程语言创建一个小程序,把字母表中的每个字母画成n乘以m的位图。用白色初始化每个像素。然后,在你画完每个字母后,数一数白色像素的数量,并保存这个数字。你找到的最大的数字就是你要找的数字。

编辑:如果你实际上只是对哪个占据了最大的矩形感兴趣(但看起来你真的是在追求它,而不是像素),你可以使用各种API调用来找到大小,但这取决于你的编程语言。例如,在Java中,您将使用FontMetrics类。

嗯,让我想想:

咔嚓咔嚓

cccccccccccccccccccccccccccccccccccccccc

DDDD

eeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

ffffffffffffffffffffffffffffffffffffffff

gggggggggggggggggggggggggggggggggggggggg

hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

唔��

mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

购买力平价

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

rrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrr

ssssssssssssssssssssssssssssssssssssssss

啧��

uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

wwwwww

xxxx

yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy

zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB

CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC

DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF

GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG

HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII

JJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJJ

KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK

LLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL

MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM

NNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNN

公私合营

QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ

RRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR

SSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSSS

TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTT

UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU

VVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVVV

www

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

W获胜。

当然,这是一个愚蠢的经验实验。哪个字母最宽没有唯一的答案。这取决于字体。所以你必须做一个类似的经验实验来找出你所处环境的答案。但事实是,大多数字体遵循相同的惯例,大写W将是最宽的。

主旨与这些字符宽度的比率形式(W = 100)在这里捕获使用特定的示例字体:

https://gist.github.com/imaurer/d330e68e70180c985b380f25e195b90c

这也取决于字体。我在1或2年前用Processing和Helvetica做过这个,它是ILJTYFVCPAXUZKHSEDORGNBQMW,按增加像素的顺序。这个想法是用你正在看的字体在画布上绘制文本,计算像素,然后用HashMap或Dictionary排序。

当然,这可能与您的使用没有直接关系,因为它计算像素面积而不仅仅是宽度。可能也有点过头了。

void setup() { 
 size(30,30);
 HashMap hm = new HashMap();
 fill(255);
 PFont font = loadFont("Helvetica-20.vlw");
 textFont(font,20);
 textAlign(CENTER);

 for (int i=65; i<91; i++) {
    background(0);
    text(char(i),width/2,height-(textDescent()+textAscent())/2); 
    loadPixels();
    int white=0;
    for (int k=0; k<pixels.length; k++) {
       white+=red(pixels[k]);
    }
    hm.put(char(i),white);
  }

  HashMap sorted = getSortedMap(hm);

  String asciiString = new String();

  for (Iterator<Map.Entry> i = sorted.entrySet().iterator(); i.hasNext();) { 
    Map.Entry me = (Map.Entry)i.next();
    asciiString += me.getKey();
  }

  println(asciiString); //the string in ascending pixel order

}

public HashMap getSortedMap(HashMap hmap) {
  HashMap map = new LinkedHashMap();
  List mapKeys = new ArrayList(hmap.keySet());
  List mapValues = new ArrayList(hmap.values());

  TreeSet sortedSet = new TreeSet(mapValues);
  Object[] sortedArray = sortedSet.toArray();
  int size = sortedArray.length;

  // a) Ascending sort

  for (int i=0; i<size; i++) {
    map.put(mapKeys.get(mapValues.indexOf(sortedArray[i])), sortedArray[i]);
  }
  return map;
}

Arial 30px在Chrome - W获胜。