协调多个分割环境

Question

我想对齐多个拆分环境。到目前为止，我已经检查了这些答案，但没有一个能解决问题。如何在等号处对齐两个分割环境？，如何通过分割方程获得两个对齐点和多重比对

我希望下面的代码中的所有=符号都对齐，同时每个符号都有单独的数字。但是，我不确定我是否采取了正确的方法。提前谢谢您。

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\begin{document}
\begin{align}
\begin{split}
Y^{\text{I}}_{n} &= \alpha_{i} + \delta X_{1,n} \\
    & +X_{2,int} \times[ 
    \beta_{1} + 
    \beta_{2} z_{1} +
    \beta_{3} z_{2} +
    \beta_{4} z_{3} \\
& +\beta_{5} z_{1} z_{2} +
    \beta_{6} z_{1} z_{3} +
    \beta_{7} z_{2} z_{3} ]\\
 & +X_{3,int} \times[  
    \gamma_{1} + 
    \gamma_{2} z_{1} +
    \gamma_{3} z_{2} +
    \gamma_{4} z_{3} \\
& + \gamma_{5} z_{1} z_{2} +
    \gamma_{6} z_{1} z_{3} +
    \gamma_{7} z_{2} z_{3} ]\\
\end{split}\\
\begin{split}
Y^{\text{II}}_{int} &= \alpha_{i} +\delta X_{1,n} \\
& + X_{2,n}\times\exp[ \mu_{1}  + \sigma_{1} u_{1}] - X_{2,n}\times\exp[\eta_{1}  + \sigma_{2} u_{2}]
\end{split} \\
\begin{split}
Y^{\text{III}}_{n} &= \alpha_{i} + \delta X_{1,n} \\
& + \exp[\mu_{1} + 
\mu_{2} z_{1} +
\mu_{3} z_{2} +
\mu_{4} z_{3}  + \sigma_{1} u_{1}]\times X_{2,n}\\
& + \exp[\eta_{1} + 
\eta_{2} z_{1,n} +
\eta_{3} z_{2,n} +
\eta_{4} z_{3,n} + \sigma_{2} u_{2}]\times X_{3,n}
\end{split}         
\end{align}
\end{document}

Answer 1

split非常敏感\\，并且在第一次分裂结束时你有一个额外的。

改变

    \gamma_{7} z_{2} z_{3} ]\\

到

    \gamma_{7} z_{2} z_{3} ]

或者进行一些小调整

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\begin{document}
\begin{align}
\begin{split}
Y^{\mathrm{I}}_{n} ={}& \alpha_{i} + \delta X_{1,n} \\
    & +X_{2,int} \times[ 
    \beta_{1} + 
    \beta_{2} z_{1} +
    \beta_{3} z_{2} +
    \beta_{4} z_{3} \\
& +\beta_{5} z_{1} z_{2} +
    \beta_{6} z_{1} z_{3} +
    \beta_{7} z_{2} z_{3} ]\\
 & +X_{3,int} \times[  
    \gamma_{1} + 
    \gamma_{2} z_{1} +
    \gamma_{3} z_{2} +
    \gamma_{4} z_{3} \\
& + \gamma_{5} z_{1} z_{2} +
    \gamma_{6} z_{1} z_{3} +
    \gamma_{7} z_{2} z_{3} ]
\end{split}\\
\begin{split}
Y^{\mathrm{II}}_{\mathrm{int}} ={}& \alpha_{i} +\delta X_{1,n} \\
& + X_{2,n}\times\exp[ \mu_{1}  + \sigma_{1} u_{1}] - X_{2,n}\times\exp[\eta_{1}  + \sigma_{2} u_{2}]
\end{split} \\
\begin{split}
Y^{\mathrm{III}}_{n} ={}& \alpha_{i} + \delta X_{1,n} \\
& + \exp[\mu_{1} + 
\mu_{2} z_{1} +
\mu_{3} z_{2} +
\mu_{4} z_{3}  + \sigma_{1} u_{1}]\times X_{2,n}\\
& + \exp[\eta_{1} + 
\eta_{2} z_{1,n} +
\eta_{3} z_{2,n} +
\eta_{4} z_{3,n} + \sigma_{2} u_{2}]\times X_{3,n}
\end{split}         
\end{align}
\end{document}

Answer 2

除了 David 的回答之外，我还会这样做

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\begin{document}
\begin{align}
  \begin{split}
    Y^{\mathrm{I}}_{n} = {} & \alpha_{i} + \delta X_{1,n} \\
    & +X_{2,int} \times[ 
    \beta_{1} + 
    \beta_{2} z_{1} +
    \beta_{3} z_{2} +
    \beta_{4} z_{3} \\
    & +\beta_{5} z_{1} z_{2} +
    \beta_{6} z_{1} z_{3} +
    \beta_{7} z_{2} z_{3} ]\\
    & +X_{3,int} \times[  
    \gamma_{1} + 
    \gamma_{2} z_{1} +
    \gamma_{3} z_{2} +
    \gamma_{4} z_{3} \\
    & + \gamma_{5} z_{1} z_{2} +
    \gamma_{6} z_{1} z_{3} +
    \gamma_{7} z_{2} z_{3} ]
  \end{split}
  \\
  \begin{split}
    Y^{\mathrm{II}}_{int} = {} & \alpha_{i} +\delta X_{1,n} \\
    & + X_{2,n}\times\exp[ \mu_{1}  + \sigma_{1} u_{1}] - X_{2,n}\times\exp[\eta_{1}  + \sigma_{2} u_{2}]
  \end{split}
  \\
  \begin{split}
    Y^{\mathrm{III}}_{n} = {} & \alpha_{i} + \delta X_{1,n} \\
    & + \exp[\mu_{1} + 
    \mu_{2} z_{1} +
    \mu_{3} z_{2} +
    \mu_{4} z_{3}  + \sigma_{1} u_{1}]\times X_{2,n}\\
    & + \exp[\eta_{1} + 
    \eta_{2} z_{1,n} +
    \eta_{3} z_{2,n} +
    \eta_{4} z_{3,n} + \sigma_{2} u_{2}]\times X_{3,n}
  \end{split}         
\end{align}
\end{document}

这里我们移动+对齐方式，使其不低于 ，=因为这样看起来很糟糕。我也\text用替换了\mathrm。为什么，尝试\itshape在 之前添加\begin{align}！