HTML to LaTeX

hbghlyj · 发表于 2023-3-2 09:28

Some old websites use HTML to display math.
We can use ChatGPT to convert HTML to LaTeX:
Screenshot 2023-03-02 at 01-26-26 ChatGPT.png

Pandoc converts HTML to LaTeX, but didn't recognize HTML markup as math formula.
Screenshot 2023-03-02 at 01-29-13 Try pandoc!.png

Is there any other tools to convert HTML to LaTeX

hbghlyj · 发表于 2023-3-6 09:17

本帖最后由 hbghlyj 于 2023-3-11 11:24 编辑 Here's a Python code replacing  tags with math delimiters and removing the  tags:

import os
import re
def html_to_latex(html):
# Remove tags
html = re.sub(r'<\/?p>', '', html)
# Treat tags as math
html = re.sub(r'(.*?)<\/em>', r'\\(\1\\)', html)
# Treat tags as operator
html = re.sub(r'(sin|cos|tan)<\/strong>', r'\\(\\\1\\)', html)
html = re.sub(r'(.*?)<\/strong>', r'\\(\\textbf{\1}\\)', html)
# Replace and tags with LaTeX commands
html = re.sub(r'(.*?)', r'\\(_{\1}\\)', html).replace('>',r'\(>\)').replace('<',r'\(<\)')
# Replace and tags with LaTeX commands
html= re.sub(r'(.*?)', r'\\(^{\1}\\)', html)
# Wrap math symbols with delimiter
html = re.sub(r'(\\\(.*?\\\))|([Α-Ωα-ω\d=·\+′≌△⇒∠°－])', lambda m:r'\('+m.group(2)+r'\)' if m.group(2) else m.group(1), html)
# Merge adjacent formulas if the only characters between them are comma or ( ) or whitespace
html= re.sub(r'\\\)([\s,\(\)]*)\\\(', '\g<1>', html).replace('′',"'")
return html
# Find all HTML files in the current directory
html_files = [file for file in os.listdir() if file.endswith('.html')]
# Loop through each HTML file
for file_name in html_files:
with open(file_name, 'r') as f:
contents = f.read()
# Use regular expressions to remove all <img> tags
contents = html_to_latex(contents)
with open('converted/'+file_name, 'w') as f:
f.write(contents)

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Example

html = r'''△ABC和△PQR满足如下条件：A和P分别是线段QR和BC的中点，QR和BC是∠BAC和∠QPR的内角平分线．求证AB + AC = PQ + PR．
证：设X为线段BC、PQ中垂线的交点，Q′、R′是Q、R关于直线XP的对称点，B′、C′是B、C关于直线XZ的对称点．
∴ ∠Q′PB = ∠QPC = ∠RPC ⇒ Q′、P、R三点共线．
同理Q、P、R′；B′、A、C；B、A、C′分别三点共线．
又 XR = XR′, XQ = XQ′, XR = XQ ⇒ X为Q、Q′、R、R′共圆的圆心．
同理X为B、B′、C、C′共圆的圆心．
设Y点是XP与QR的交点，BY = YC, ∠BAY = ∠CAY，
∴ B、A、C、Y共圆，∠RPQ = ∠PQQ′ + ∠PQ′Q = 2∠PQ′Q = ∠RXQ．
∴ P、X、R、Q共圆．同理X、A、C、B共圆．
∴ X、A、C、B、Y五点共圆，∠XCY = ∠XAT = 90° = ∠XPC．
∴ YC 2 = YP·YX = TR·YQ = YX 2－XR 2．
又 YC 2 = YX 2－XC 2 ⇒ XR = XC．
∴ B、B′、C、C′、R、R′、Q、Q′共圆，圆心为Y，YB、YC为该圆的切线．
∴ ∠BXC′ = ∠YBC′ = ∠YXA = 90°－∠Q′YQ = ∠YQQ′ = ∠QXR′．
∴ △BXC′≌△QXR′ ⇒ BC′ = QR′．
又 BC′ = AB + AC′ = AB + AC, QR′ = QP + PR′ = PQ + PR．
∴ AB + AC = PQ + PR，命题得证．
评注：此题看似不易下手，关键在于由内角平分到外角平分的过渡，一旦看到点X的好性质，问题便迎刃而解了．'''
html_to_latex(html)

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This will output:

\(△ABC\)和\(△PQR\)满足如下条件：\(A\)和\(P\)分别是线段\(QR\)和\(BC\)的中点，\(QR\)和\(BC\)是\(∠BAC\)和\(∠QPR\)的内角平分线．求证\(AB+AC=PQ+PR\)．
证：设\(X\)为线段\(BC\)、\(PQ\)中垂线的交点，\(Q'\)、\(R'\)是\(Q\)、\(R\)关于直线\(XP\)的对称点，\(B'\)、\(C'\)是\(B\)、\(C\)关于直线\(XZ\)的对称点．
∴ \(∠Q'PB=∠QPC=∠RPC⇒Q'\)、\(P\)、\(R\)三点共线．
同理\(Q\)、\(P\)、\(R'\)；\(B'\)、\(A\)、\(C\)；\(B\)、\(A\)、\(C'\)分别三点共线．
又 \(XR=XR'XQ=XQ'XR=XQ⇒X\)为\(Q\)、\(Q'\)、\(R\)、\(R'\)共圆的圆心．
同理\(X\)为\(B\)、\(B'\)、\(C\)、\(C'\)共圆的圆心．
设\(Y\)点是\(XP\)与\(QR\)的交点，\(BY=YC∠BAY=∠CAY\)，
∴ \(B\)、\(A\)、\(C\)、\(Y\)共圆，\(∠RPQ=∠PQQ'+∠PQ'Q=2∠PQ'Q=∠RXQ\)．
∴ \(P\)、\(X\)、\(R\)、\(Q\)共圆．同理\(X\)、\(A\)、\(C\)、\(B\)共圆．
∴ \(X\)、\(A\)、\(C\)、\(B\)、\(Y\)五点共圆，\(∠XCY=∠XAT=90°=∠XPC\)．
∴ \(YC^{2}=YP·YX=TR·YQ=YX^{2}－XR^{2}\)．
又 \(YC^{2}=YX^{2}－XC^{2}⇒XR=XC\)．
∴ \(B\)、\(B'\)、\(C\)、\(C'\)、\(R\)、\(R'\)、\(Q\)、\(Q'\)共圆，圆心为\(Y\)，\(YB\)、\(YC\)为该圆的切线．
∴ \(∠BXC'=∠YBC'=∠YXA=90°－∠Q'YQ=∠YQQ'=∠QXR'\)．
∴ \(△BXC'≌△QXR'⇒BC'=QR'\)．
又 \(BC'=AB+AC'=AB+ACQR'=QP+PR'=PQ+PR\)．
∴ \(AB+AC=PQ+PR\)，命题得证．
评注：此题看似不易下手，关键在于由内角平分到外角平分的过渡，一旦看到点\(X\)的好性质，问题便迎刃而解了．

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