英语字母表中哪个字母占像素最多?

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?

115102 次浏览
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大写的“M”通常是最宽的。

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我认为W是最宽的。

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这取决于字体。例如，交叉0所占的空间要比普通0大得多。

但如果要猜的话，我会选X或B。

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最佳答案

嗯，让我想想:

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gggggggggggggggggggggggggggggggggggggggg

hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

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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

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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。

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Arial 30px在Chrome - W获胜。

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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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接下来是Ned Batchelder非常实用的回答，因为我来这里是想知道数字:

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2222222222222222222222222222222222222222

3333333333333333333333333333333333333333

4444444444444444444444444444444444444444

5555555555555555555555555555555555555555

6666666666666666666666666666666666666666

7777777777777777777777777777777777777777

8888888888888888888888888888888888888888

9999999999999999999999999999999999999999

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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;
}
小开

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)。

原来的帖子: 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()
小开

那么程序化的解决方案呢?

. 
var capsIndex = 65;
var smallIndex = 97
var div = document.createElement('div');
div.style.float = 'left';
document.body.appendChild(div);
var highestWidth = 0;
var elem;


for(var i = capsIndex; i < capsIndex + 26; i++) {
div.innerText = String.fromCharCode(i);
var computedWidth = window.getComputedStyle(div, null).getPropertyValue("width");
if(highestWidth < parseFloat(computedWidth)) {
highestWidth = parseFloat(computedWidth);
elem = String.fromCharCode(i);
}
}
for(var i = smallIndex; i < smallIndex + 26; i++) {
div.innerText = String.fromCharCode(i);
var computedWidth = window.getComputedStyle(div, null).getPropertyValue("width");
if(highestWidth < parseFloat(computedWidth)) {
highestWidth = parseFloat(computedWidth);
elem = String.fromCharCode(i);
}
}
div.innerHTML = '<b>' + elem + '</b>' + ' won';

小开

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占据了最多的空间。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>

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想知道真正的最长的字形，而不仅仅是猜测?
我说的不仅仅是字母、数字和常用符号(!， @等等)。我指的是UTF-16的32,834个字符中最长的字形。
所以我从@NK的答案开始，它有一个程序化的解决方案，并进行了修改:

var capsIndex = 65;
var smallIndex = 97;
var div = document.createElement('div');
div.style.float = 'left';
document.body.appendChild(div);
var highestWidth = 0;
var elem;


for(var i = capsIndex; i < 32834; i++) {
div.innerText = String.fromCharCode(i);
var computedWidth = window.getComputedStyle(div, null).getPropertyValue("width");
if(highestWidth < parseFloat(computedWidth)) {
highestWidth = parseFloat(computedWidth);
elem = String.fromCharCode(i);
}
}
div.innerHTML = '<b>' + elem + '</b>' + ' won';

在运行并等待(和等待)之后，它给出输出ௌ won。
这就是UTF-32中最长的字符! 请注意，在大多数字体上，最长的字形是﷽，但有些字体(特别是单行字体)与字符重叠，就像运行程序时使用的字体一样

小开

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: ';
} else {
div.innerHTML = ' Winners: ';
}
div.innerHTML = div.innerHTML + answer + '.';
console.log(widths.sort((a, b) => a - b));

小开

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 */
}


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