如何让自定义数学公式编号?
更详细问题的描述:如何使得数学公式默认不编号,并且不记入编号公式的计数,同时可以添加某些命令使得公式编号?
- 通过
i-figured
包,建议查看后手搓一个
typst
#import "@preview/i-figured:0.2.4"
#show math.equation: i-figured.show-equation.with(only-labeled: true)
#set heading(numbering: "1.1")
= Equation numbering
#show math.equation: i-figured.show-equation.with(only-labeled: true)
#lorem(10)
$
a+b=c
$<1>
Try to cite @eqt:1, #lorem(5)
$
1+2=3
$
#lorem(10)
$
4+5 =9
$<2>
#lorem(10)
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- 手搓 by OrangeX4, 原文地址
typst
= Equation numbering
#set math.equation(numbering: "(1)")
#show math.equation.where(block: true): it => {
if not it.has("label") {
let fields = it.fields()
fields.remove("body")
fields.numbering = none
return [#counter(math.equation).update(v => v - 1)#math.equation(..fields, it.body)<math-equation-without-label>]
}
return it
}
$ x + y $<1>
$ x + y + z $
$ x + y $<2>
Cite @1. #lorem(10)
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- 如果你还需要子公式, by obj.fake_cirno
typst
#set heading(numbering: "1.")
#let ct=counter("eq")
#set math.equation(numbering: it=>ct.display("(1-1.a)"))
#show heading.where(level: 1): it=>it+ct.step()+ct.step(level: 2)
#show math.equation.where(block: true): it=>{
it
if it.numbering !=none{
if ct.get().len()==2{
ct.step(level: 2)
}
}
}
#let eq_nonum(body)={
set math.equation(numbering: none)
body
}
#let subeqs(..args)={
for eq in args.pos(){
ct.step(level: 3)
eq
}
ct.step(level: 2)
}
= Equation numbering
$
f(x) = sin x
$
#lorem(10)
#eq_nonum(
$ x + y = z $
)
#lorem(10)
$
g(x) = cos x
$
#lorem(10)
#subeqs(
$ F &= sum $, // 编号为 (1-1.a)
$ = x $, // 编号为 (1-1.b)
$ = 1/2m v^2 $, // 编号为 (1-1.c)
)
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- 子公式另一个版本, by obj.fake_cirno
typst
#set math.equation(numbering: "(1)")
#show math.equation.where(block: true): it => {
if it.has("label") {
if "-" == str(it.label) {
counter(math.equation).update(n => n - 1)
math.equation(it.body, block: true, numbering: none)
return
} else if "::" in str(it.label) {
let (a, b) = str(it.label).split("::")
counter(math.equation).update(n => n - 2)
[#math.equation(it.body, block: true, numbering: _ => "(" + b + ")")#label(a)]
return
}
}
it
}
$ f(x) $
$ f(x) $ <eq:some::14a>
$ f(x) $ <->
$ f(x) $
@eq:some
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