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;
}

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 */ }

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;
}

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。

Alex Michael在他的博客上发布了一个计算字体宽度的解决方案，有点像xxx发布的解决方案(有趣的是，他在这里链接了我)。

简介:

对于Helvetica，前三个字母是:M(2493像素)，W(2414像素)和B(1909像素)。 对于他的Mac附带的一组字体，结果大致相同:M(2217.51±945.19)，W(2139.06±945.29)和B(1841.38±685.26)。

原文:http://alexmic.net/letter-pixel-count/

代码:

# -*- coding: utf-8 -*-
from __future__ import division
import os
from collections import defaultdict
from math import sqrt
from PIL import Image, ImageDraw, ImageFont


# Make a lowercase + uppercase alphabet.
alphabet = 'abcdefghijklmnopqrstuvwxyz'
alphabet += ''.join(map(str.upper, alphabet))


def draw_letter(letter, font, save=True):
    img = Image.new('RGB', (100, 100), 'white')

    draw = ImageDraw.Draw(img)
    draw.text((0,0), letter, font=font, fill='#000000')

    if save:
        img.save("imgs/{}.png".format(letter), 'PNG')

    return img


def count_black_pixels(img):
    pixels = list(img.getdata())
    return len(filter(lambda rgb: sum(rgb) == 0, pixels))


def available_fonts():
    fontdir = '/Users/alex/Desktop/English'
    for root, dirs, filenames in os.walk(fontdir):
        for name in filenames:
            path = os.path.join(root, name)
            try:
                yield ImageFont.truetype(path, 100)
            except IOError:
                pass


def letter_statistics(counts):
    for letter, counts in sorted(counts.iteritems()):
        n = len(counts)
        mean = sum(counts) / n
        sd = sqrt(sum((x - mean) ** 2 for x in counts) / n)
        yield letter, mean, sd


def main():
    counts = defaultdict(list)

    for letter in alphabet:
        for font in available_fonts():
            img = draw_letter(letter, font, save=False)
            count = count_black_pixels(img)
            counts[letter].append(count)

        for letter, mean, sd in letter_statistics(counts):
            print u"{0}: {1:.2f} ± {2:.2f}".format(letter, mean, sd)


    if __name__ == '__main__':
        main()

根据您的平台，可能有一种方法从字符串或DrawText()函数中“getWidth”，以某种方式使用width属性。

我会做一个简单的算法时间，利用所需的字体，然后通过alfabet运行，并将其存储在一个小配置或只是计算它在初始化作为一个循环从a到Z并不难。