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我是 Python 的新手。我想我会尝试编写一个脚本,使用 etree 从 SVG 图标文件中检索 glyph-name
和 unicode
以从中创建一个 LESS 文件。
我编写了一个解决方案,在该解决方案中,我遍历 SVG 元素以获取元素属性,将它们放入列表中,然后根据列表中的属性编写我的 LESS 样式。它看起来像这样:
<?xml version="1.0" standalone="no"?><!DOCTYPE svg PUBLIC "-//W3C//DTD SVG 1.1//EN" "http://www.w3.org/Graphics/SVG/1.1/DTD/svg11.dtd"><!--
2019-3-13: Created with FontForge (http://fontforge.org)
-->
<svg xmlns="http://www.w3.org/2000/svg"
xmlns:xlink="http://www.w3.org/1999/xlink" version="1.1">
<metadata>
Created by FontForge 20170731 at Wed Mar 13 07:05:16 2019 By root
</metadata>
<defs>
<font id="Flaticon" horiz-adv-x="512">
<font-face font-family="Flaticon" font-weight="400" font-stretch="normal" units-per-em="512" panose-1="2 0 5 3 0 0 0 0 0 0" ascent="448" descent="-64" bbox="-0.00048785 -64.001 512 448" underline-thickness="25.6" underline-position="-51.2" unicode-range="U+0020-F131" />
<missing-glyph />
<glyph glyph-name="space" unicode=" " horiz-adv-x="200" />
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</font>
</defs>
</svg>
from lxml import etree
filename = "font.svg"
tree = etree.parse(open(filename, 'r'))
# Brute Force Method: Iterate through everything and then iterate through attributes.
for element in tree.iter():
attributes = []
for name, value in element.attrib.items():
if (name == "glyph-name") or (name == "unicode"):
attributes.insert(0, value)
if type(attributes) != 'bool' and len(attributes) == 2:
lessPreStyle += '''\
.icon-{name} {{
&:before {{
content: {unicode};
}}
}}
'''.format(name=attributes[0], unicode=repr(attributes[1]).replace('\\u', '\\'))
现在我试图通过仅遍历字形而不是每个元素并直接从对象获取属性而不是遍历属性来检查它是否被称为 glyph-name
或 unicode
。到目前为止,我有这样的事情:
# Go directly to the parent of the glyphs
font = tree.getroot()[1][0]
# Filter only glyphs
for glyph in font.iter('glyph'):
print (glyph.attrib['unicode'])
print (glyph.attrib['glyph-name'])
我当前的问题是 for glyph in font.iter('glyph')
不返回任何内容。根据Python 3.6.8 Documentation我应该能够通过在 iter()
函数中输入我想要的子元素的名称来过滤子元素。但是,我什么也没得到。如果我删除过滤器并只打印子项的标签名称,我将获得所有标签名称。我不想像上次那样迭代和检查单个元素。我在这里错过了什么吗?或者是否有更好的方法来实现相同的目标?
最佳答案
您的 SVG 有一个默认命名空间 (http://www.w3.org/2000/svg
)。
要在 iter()
中指定命名空间,您必须使用 Clark Notation。这基本上是命名空间 uri 和本地名称的组合:{http://www.w3.org/2000/svg}glyph
您还可以使用 *
而不是实际的命名空间 uri 来匹配任何命名空间中的元素:{*}glyph
例子:
from lxml import etree
tree = etree.parse("font.svg")
# Filter only glyphs
for glyph in tree.iter('{http://www.w3.org/2000/svg}glyph'):
print(glyph.attrib['unicode'])
print(glyph.attrib['glyph-name'])
关于python - 当我运行 `for glyph in font.iter(' glyph')` 时,为什么 etree 没有从我的 SVG 中返回任何内容?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55385936/
我正在处理一组标记为 160 个组的 173k 点。我想通过合并最接近的(到 9 或 10 个组)来减少组/集群的数量。我搜索过 sklearn 或类似的库,但没有成功。 我猜它只是通过 knn 聚类
我有一个扁平数字列表,这些数字逻辑上以 3 为一组,其中每个三元组是 (number, __ignored, flag[0 or 1]),例如: [7,56,1, 8,0,0, 2,0,0, 6,1,
我正在使用 pipenv 来管理我的包。我想编写一个 python 脚本来调用另一个使用不同虚拟环境(VE)的 python 脚本。 如何运行使用 VE1 的 python 脚本 1 并调用另一个 p
假设我有一个文件 script.py 位于 path = "foo/bar/script.py"。我正在寻找一种在 Python 中通过函数 execute_script() 从我的主要 Python
这听起来像是谜语或笑话,但实际上我还没有找到这个问题的答案。 问题到底是什么? 我想运行 2 个脚本。在第一个脚本中,我调用另一个脚本,但我希望它们继续并行,而不是在两个单独的线程中。主要是我不希望第
我有一个带有 python 2.5.5 的软件。我想发送一个命令,该命令将在 python 2.7.5 中启动一个脚本,然后继续执行该脚本。 我试过用 #!python2.7.5 和http://re
我在 python 命令行(使用 python 2.7)中,并尝试运行 Python 脚本。我的操作系统是 Windows 7。我已将我的目录设置为包含我所有脚本的文件夹,使用: os.chdir("
剧透:部分解决(见最后)。 以下是使用 Python 嵌入的代码示例: #include int main(int argc, char** argv) { Py_SetPythonHome
假设我有以下列表,对应于及时的股票价格: prices = [1, 3, 7, 10, 9, 8, 5, 3, 6, 8, 12, 9, 6, 10, 13, 8, 4, 11] 我想确定以下总体上最
所以我试图在选择某个单选按钮时更改此框架的背景。 我的框架位于一个类中,并且单选按钮的功能位于该类之外。 (这样我就可以在所有其他框架上调用它们。) 问题是每当我选择单选按钮时都会出现以下错误: co
我正在尝试将字符串与 python 中的正则表达式进行比较,如下所示, #!/usr/bin/env python3 import re str1 = "Expecting property name
考虑以下原型(prototype) Boost.Python 模块,该模块从单独的 C++ 头文件中引入类“D”。 /* file: a/b.cpp */ BOOST_PYTHON_MODULE(c)
如何编写一个程序来“识别函数调用的行号?” python 检查模块提供了定位行号的选项,但是, def di(): return inspect.currentframe().f_back.f_l
我已经使用 macports 安装了 Python 2.7,并且由于我的 $PATH 变量,这就是我输入 $ python 时得到的变量。然而,virtualenv 默认使用 Python 2.6,除
我只想问如何加快 python 上的 re.search 速度。 我有一个很长的字符串行,长度为 176861(即带有一些符号的字母数字字符),我使用此函数测试了该行以进行研究: def getExe
list1= [u'%app%%General%%Council%', u'%people%', u'%people%%Regional%%Council%%Mandate%', u'%ppp%%Ge
这个问题在这里已经有了答案: Is it Pythonic to use list comprehensions for just side effects? (7 个答案) 关闭 4 个月前。 告
我想用 Python 将两个列表组合成一个列表,方法如下: a = [1,1,1,2,2,2,3,3,3,3] b= ["Sun", "is", "bright", "June","and" ,"Ju
我正在运行带有最新 Boost 发行版 (1.55.0) 的 Mac OS X 10.8.4 (Darwin 12.4.0)。我正在按照说明 here构建包含在我的发行版中的教程 Boost-Pyth
学习 Python,我正在尝试制作一个没有任何第 3 方库的网络抓取工具,这样过程对我来说并没有简化,而且我知道我在做什么。我浏览了一些在线资源,但所有这些都让我对某些事情感到困惑。 html 看起来
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