我知道公认的答案是W, W代表胜利。

但是，在本例中，W也表示宽度。案例研究使用了一个简单的宽度测试来检查像素，但它只是宽度，而不是总像素数。作为一个简单的反例，公认的答案假设O和Q占用相同数量的像素，但它们只占用相同数量的空间。

因此，W占的空间最大。但是，这是所有的像素，它被吹捧吗?

让我们来看看一些经验数据。我从下面的B, M和w中创建了imgur图像，然后分析了它们的像素数(见下文)，结果如下:

B: 114像素

M: 150像素

女:157像素

以下是我如何将它们放入画布并分析图像中的原始像素数据。

var imgs = { B : "//i.imgur.com/YOuEPOn.png", M : "//i.imgur.com/Aev3ZKQ.png", W : "//i.imgur.com/xSUwE7w.png" }; window.onload = function(){ for(var key in imgs){(function(img,key){ var Out = document.querySelector("#"+key+"Out"); img.crossOrigin = "Anonymous"; img.src=imgs[key]; img.onload = function() { var canvas = document.querySelector('#'+key); (canvas.width = img.width,canvas.height = img.height); var context = canvas.getContext('2d'); context.drawImage(img, 0, 0); var data = context.getImageData(0, 0, img.width, img.height).data; Out.innerHTML = "Total Pixels: " + data.length/4 + "<br>"; var pixelObject = {}; for(var i = 0; i < data.length; i += 4){ var rgba = "rgba("+data[i]+","+data[i+1]+","+data[i+2]+","+data[i+3]+")"; pixelObject[rgba] = pixelObject[rgba] ? pixelObject[rgba]+1 : 1; } Out.innerHTML += "Total Whitespace: " + pixelObject["rgba(255,255,255,255)"] + "<br>"; Out.innerHTML += "Total Pixels In "+ key +": " + ((data.length/4)-pixelObject["rgba(255,255,255,255)"]) + "<br>"; }; })(new Image(),key)} }; <table> <tr> <td> <canvas id="B" width="100%" height="100%"></canvas> </td> <td id="BOut"> </td> </tr> <tr> <td> <canvas id="M" width="100%" height="100%"></canvas> </td> <td id="MOut"> </td> </tr> <tr> <td> <canvas id="W" width="100%" height="100%"></canvas> </td> <td id="WOut"> </td> </tr> </table>

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。

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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

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()

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。