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方程列表未列出 \align 环境中的方程

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方程列表未列出 \align 环境中的方程

我发现的列出方程式的现有示例均未包含​​ \align 环境中的方程式。我如何才能在同一方程式列表中同时包含两种方程式类型？以下是 MWE：

    \documentclass[english]{article}
\setcounter{secnumdepth}{2}
%\setcounter{tocdepth}{1}
\usepackage{amsmath}
\usepackage{tocloft}
\usepackage{xstring}
\usepackage[unicode=true, pdfusetitle,
 bookmarks=true,bookmarksnumbered=false,bookmarksopen=false,
 breaklinks=false,pdfborder={0 0 0},backref=false,colorlinks=false]
 {hyperref}

\makeatletter
\numberwithin{equation}{section}

% we use this for our refernces as well
\AtBeginDocument{\renewcommand{\ref}[1]{\mbox{\autoref{#1}}}}

% redefinition of \equation for convenience
\let\oldequation = \equation
\let\endoldequation = \endequation
\AtBeginDocument{\let\oldlabel = \label}% \AtBeginDocument because hyperref redefines \label
\newcommand{\mynewlabel}[1]{%
  \StrBehind{#1}{eq:}[\Str]% remove "eq:" from labels
  \myequations{\Str}\oldlabel{#1}}
  \renewenvironment{equation}{%
  \oldequation
  \let\label\mynewlabel
}{\endoldequation}

% redefinition of \eqnarray for convenience
\let\oldeqnarray = \eqnarray
\let\endoldeqnarray = \endeqnarray
%\AtBeginDocument{\let\oldlabel = \label}% \AtBeginDocument because hyperref redefines \label
\newcommand{\mynewlabelarray}[1]{%
  \StrBehind{#1}{eq:}[\Str]% remove "eq:" from labels
  \myequations{\Str}\oldlabel{#1}}
  \renewenvironment{eqnarray}{%
  \oldeqnarray
  \let\label\mynewlabelarray
}{\endoldeqnarray}

\newcommand{\listequationsname}{\normalsize  List of Equations}
\newlistof{myequations}{equ}{\listequationsname}
\newcommand{\myequations}[1]{%
      \addcontentsline{equ}{myequations}{\protect\numberline{\theequation}#1}}
\setlength{\cftmyequationsnumwidth}{3em}

\makeatother

\begin{document}
%\tableofcontents
\listofmyequations

\section{Section title}
\begin{equation}
  F=q[E+(v\times B)]
  \label{eq:Force}
\end{equation}

\begin{equation}
  \tau=F\times r
  \label{eq:Torque}
\end{equation}
If the electrical force in \ref{eq:Force} is ignored,
and the remaining magnetic force is used in \ref{eq:Torque},
with the assumption that $v$ is perpendicular to $B$, we find that
\begin{equation}
      \tau=qvBrsin\theta
  \label{eq:Magnetic}
\end{equation}

\begin{align}
  \min_{u_{i}(t),y_i, i=1...N}\!\!\!\!\!\! J(u_i(t),y_i)  &:=  \sum_{i=1}^N \int_0^{T} R_i(u_{i}(t),t) dt \label{eq:objective function}\\
+&  \xi \int_{0}^{T}  \left(\theta\frac{M - I(t)}{M}K(t)  - D(t)\right)^2 dt +  \sum_{i=1}^N \gamma_i y_i  \notag\\
  + & \sum_{i=1}^N p_{i} \int_{T_i}^{T} u_i(t-T_i)dt \notag\\
 +&  h\int_{0}^T \left[\theta\frac{M - I(t)}{M}K(t) - D(t)\right]^+ dt, \notag
\end{align}

subject to
\begin{align}
  K(t)& = \sum_{i=1}^N u_i(t-T_i), & \quad t\in [0,T] \label{eq:2} \\
  u_i(t) &\le  % \theta_i S_i(t) y_i =
  \theta_i (M_i - I_i(t)) y_i,  & \ \ i = 1\ldots N \quad t\in [0,T-T_i] \label{eq:ui}\\
  u_i(t) & =  0, & i = 1\ldots N \quad t\in [T-T_i,T] \label{eq:uio}\\
  \dot{I}_i(t)& = f_i(I_i(t)), &  i = 1 \ldots N \quad t\in [0,T]& \label{eq:dotIi}\\
  \dot{I}(t)& = f(I(t)), &\quad t\in [0,T] \label{eq:dotI}\\
  u_i(t) & \ge  0, & i = 1 \ldots N \quad t\in [0,T]& \label{eq:const5}\\
  K(t)& \ge  0, &\quad t\in [0,T]& \\
  y_i &\in  \{0,1\},  & i = 1 \ldots N& \label{eq:const6}
\end{align}
%where

\begin{equation}\label{eq:capacity constraint}
    u^{j}_i(t) \leq Min_{p\in C^{j}_{i}} \sum_{k \in {K^{j}_{ip}}} u^{j+1}_k(t-T_k)
\end{equation}

\end{document}

答案1

我认为手动修改宏不是一个好主意\label，因为它被很多其他包使用和重新定义，包括hyperref和cleveref包。我只需将\myequations指令直接应用于应该在方程列表中列出的那些方程即可。（当与gather和align环境一起使用时，似乎\myequations已经发出了指令后换行指令。不知道为什么。）

哦，还有请不要使用eqnarray环境；改用align。

在此处输入图片描述

\documentclass[english]{article}
\setcounter{secnumdepth}{2}
\usepackage{mathtools} % for \coloneqq and \mathclap macros
\usepackage{tocloft}
\usepackage[unicode=true, pdfusetitle, bookmarks=true, 
    bookmarksnumbered=false, bookmarksopen=false, 
    breaklinks=false, backref=false, 
    colorlinks=true,allcolors=black]
   {hyperref}
\usepackage[noabbrev]{cleveref}

\counterwithin{equation}{section}

\newcommand{\listequationsname}{\normalsize List of Equations}
\newlistof{myequations}{equ}{\listequationsname}
\newcommand{\myequations}[1]{%
      \addcontentsline{equ}{myequations}%
      {\protect\numberline{\theequation}#1}}     
\setlength{\cftmyequationsnumwidth}{2.5em}

\begin{document}
%\tableofcontents
\listofmyequations

\allowdisplaybreaks

\section{Section title}
\begin{gather}
   F=q[E+(v\times B)]
   \label{eq:Force}  \\ \myequations{Force}
   \tau=F\times r
   \label{eq:Torque} 
\end{gather} \myequations{Torque}
If the electrical force in \cref{eq:Force} is ignored, and 
if the remaining magnetic force is used in \cref{eq:Torque}, 
with the assumption that $v$ is perpendicular to~$B$, we 
find that
\begin{equation}
   \tau=qvBr\sin\theta
   \label{eq:Magnetic} 
   \myequations{Magnetic}
\end{equation}

\begin{equation} 
   \label{eq:objective function} 
   \myequations{Objective function}
\begin{aligned}[t] 
\smash[b]{\min_{\mathclap{%
     \substack{u_{i}(t),y_i,\\ i=1,\dots,N}}}}
     \,J(u_i(t),y_i)  
&\coloneqq \sum_{i=1}^N \int_0^{T} R_i(u_{i}(t),t)\, dt 
    \\
&\quad+ \xi \int_{0}^{T} \left(\theta\frac{M - I(t)}{M}K(t) - D(t)\right)^{\!2} dt 
      + \sum_{i=1}^N \gamma_i y_i \\
&\quad+ \sum_{i=1}^N p_{i} \int_{T_i}^{T} u_i(t-T_i)\,dt\\
&\quad+  h\int_{0}^T \left[\theta\frac{M - I(t)}{M}K(t) - D(t)\right]^+ dt,
\end{aligned}
\end{equation}
subject to
\begin{align}
  K(t) &= \sum\nolimits_{i=1}^N u_i(t-T_i), 
    && t\in [0,T] 
       \label{eq:2} \\
  u_i(t) &\le  % \theta_i S_i(t) y_i =
  \theta_i (M_i - I_i(t)) y_i,  
    && i = 1,\ldots, N, 
       \quad t\in [0,T-T_i] 
       \label{eq:ui} \\ \myequations{ui}
  u_i(t) &=  0, 
    && i = 1,\ldots, N, 
       \quad t\in [T-T_i,T] 
       \label{eq:uio} \\ \myequations{uio}
  \dot{I}_i(t) &= f_i(I_i(t)), 
    && i = 1,\ldots,N, 
       \quad t\in [0,T] 
       \label{eq:dotIi} \\ \myequations{dotIi}
  \dot{I}(t) &= f(I(t)), 
    && t\in [0,T] 
       \label{eq:dotI}\\ \myequations{dotI}
  u_i(t) &\ge  0, 
    && i = 1,\ldots,N, 
       \quad t\in [0,T]
       \label{eq:const5} \\ \myequations{const5}
  K(t) &\ge  0, 
    && t\in [0,T] \\
  y_i &\in  \{0,1\},  
    && i = 1,\ldots,N 
      \label{eq:const6}
\end{align} \myequations{const6}
where
\begin{equation}
  \label{eq:capacity constraint}
  \myequations{Capacity constraint}
    u^{j}_i(t) \leq \min_{p\in C^{j}_{i}} \sum_{k \in {K^{j}_{ip}}} u^{j+1}_k(t-T_k)
\end{equation}

\end{document}

答案2

正如 Mico 所评论的，你不应该重新定义\label。如果你想避免提供标签，你可以制作一个可以同时执行这两项操作的简单命令。我\ref在这里用它来在 ams 对齐中选择正确的数字。

\documentclass[english]{article}
\setcounter{secnumdepth}{2}
\usepackage{mathtools} % for \coloneqq and \mathclap macros
\usepackage{tocloft}
\usepackage[unicode=true, pdfusetitle, bookmarks=true, 
    bookmarksnumbered=false, bookmarksopen=false, 
    breaklinks=false, backref=false, 
    colorlinks=true,allcolors=black]
   {hyperref}
\usepackage[noabbrev]{cleveref}

\counterwithin{equation}{section}

\newcommand{\listequationsname}{\normalsize List of Equations}
\newlistof{myequations}{equ}{\listequationsname}
\newcommand{\myequations}[1]{%
      \addcontentsline{equ}{myequations}%
      {\protect\numberline{\ref{eq:#1}}#1}}     
\setlength{\cftmyequationsnumwidth}{2.5em}

\newcommand\myeq[1]{\label{eq:#1}\myequations{#1}}

\begin{document}
%\tableofcontents
\listofmyequations

\allowdisplaybreaks

\section{Section title}
\begin{gather}
   F=q[E+(v\times B)]
   \myeq{Force}  \\
   \tau=F\times r
   \myeq{Torque} 
\end{gather}
If the electrical force in \cref{eq:Force} is ignored, and 
if the remaining magnetic force is used in \cref{eq:Torque}, 
with the assumption that $v$ is perpendicular to~$B$, we 
find that
\begin{equation}
   \tau=qvBr\sin\theta
   \myeq{Magnetic} 
\end{equation}

\begin{equation} 
   \myeq{objective function} 
\begin{aligned}[t] 
\smash[b]{\min_{\mathclap{%
     \substack{u_{i}(t),y_i,\\ i=1,\dots,N}}}}
     \,J(u_i(t),y_i)  
&\coloneqq \sum_{i=1}^N \int_0^{T} R_i(u_{i}(t),t)\, dt 
    \\
&\quad+ \xi \int_{0}^{T} \left(\theta\frac{M - I(t)}{M}K(t) - D(t)\right)^{\!2} dt 
      + \sum_{i=1}^N \gamma_i y_i \\
&\quad+ \sum_{i=1}^N p_{i} \int_{T_i}^{T} u_i(t-T_i)\,dt\\
&\quad+  h\int_{0}^T \left[\theta\frac{M - I(t)}{M}K(t) - D(t)\right]^+ dt,
\end{aligned}
\end{equation}
subject to
\begin{align}
  K(t) &= \sum\nolimits_{i=1}^N u_i(t-T_i), 
    && t\in [0,T] 
       \label{eq:2} \\
  u_i(t) &\le  % \theta_i S_i(t) y_i =
  \theta_i (M_i - I_i(t)) y_i,  
    && i = 1,\ldots, N, 
       \quad t\in [0,T-T_i] 
       \myeq{ui} \\
  u_i(t) &=  0, 
    && i = 1,\ldots, N, 
       \quad t\in [T-T_i,T] 
       \myeq{uio} \\
  \dot{I}_i(t) &= f_i(I_i(t)), 
    && i = 1,\ldots,N, 
       \quad t\in [0,T] 
       \myeq{dotIi} \\
  \dot{I}(t) &= f(I(t)), 
    && t\in [0,T] 
       \myeq{dotI}\\
  u_i(t) &\ge  0, 
    && i = 1,\ldots,N, 
       \quad t\in [0,T]
       \myeq{const5} \\
  K(t) &\ge  0, 
    && t\in [0,T] \\
  y_i &\in  \{0,1\},  
    && i = 1,\ldots,N 
      \myeq{const6}
\end{align}
where
\begin{equation}
  \myeq{capacity constraint}
    u^{j}_i(t) \leq \min_{p\in C^{j}_{i}} \sum_{k \in {K^{j}_{ip}}} u^{j+1}_k(t-T_k)
\end{equation}

\end{document}

在此处输入图片描述

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