方程式をリストするための既存の例には、\align 環境の方程式が含まれていません。両方の方程式タイプを同じ方程式リストに含めるにはどうすればよいですか? MWE は次のとおりです。
\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}
答え1
マクロを手動で変更するのは得策ではないと思います。これは、およびパッケージ\labelを含む、他の多くのパッケージによって使用され、再定義されるからです。方程式のリストにリストされるはずの方程式に直接ディレクティブを適用します。(および環境で使用する場合、ディレクティブは発行される必要があるようです。hyperrefcleveref\myequationsgatheralign\myequations後改行指示。理由は不明です。
ああ、そしてeqnarray環境を使用しないでください;align代わりに使用してください。
\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}
答え2
Mico がコメントしたように、 を再定義しないでください\label。ラベルを指定したくない場合は、両方を実行する簡単なコマンドを作成できます。\refここでは、ams アライメント内の正しい番号を取得するために使用しました。
\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}
