![分割和对齐内的对齐](https://linux22.com/image/362158/%E5%88%86%E5%89%B2%E5%92%8C%E5%AF%B9%E9%BD%90%E5%86%85%E7%9A%84%E5%AF%B9%E9%BD%90.png)
我希望(Kollo Skewness)
与右侧对齐(Row Sums)
,同时(Orthogonality)
保持(4.16)
在第三行和第四行之间。
这是我的代码
\begin{align}
&\mathbf{L}_{mn}'\mathbf{L}_{mn} = m \mathbf{I}_n & \text{(Orthogonality)}\\
&\mathbf{1}_m^{'} \mathbf{L}_{mn} = \mathbf{0}'_n & \text{(Row Sums)} \\
\begin{split} \label{eq:kollo:b}
&\mathbf{b} = m^{-1} \left[ s_1 \mathbf{c}_1 + \dots + s_m \mathbf{c}_m \right] \text{(Kollo skewness)}\\
&\text{where} \quad s_k = \left(c_{1k} + \dots + c_{nk} \right)^2 = \left( \mathbf{1}' \mathbf{c}_k \right)^2
\end{split}
\end{align}
这个问题可能在其他地方也得到解答了。如果是这样,请告诉我那里。我找不到它。
这是一个最小示例
\documentclass[12pt,a4paper]{report}
\usepackage{amsmath,amsfonts,amsthm}
\setlength{\topmargin}{1.5cm}
\setlength{\headheight}{15pt}
\setlength{\headsep}{20pt}
\setlength{\topskip}{12pt}
\setlength{\evensidemargin}{0pt}
\setlength{\oddsidemargin}{0pt}
\setlength{\textheight}{240mm}
\setlength{\textwidth}{160mm}
\setlength{\voffset}{-2cm}
\setlength{\parindent}{0pt}
\setlength{\parskip}{6pt}
\begin{document}
\begin{align}
&\mathbf{L}_{mn}'\mathbf{L}_{mn} = m \mathbf{I}_n & \text{(Orthogonality)}\\
&\mathbf{1}_m^{'} \mathbf{L}_{mn} = \mathbf{0}'_n & \text{(Row Sums)} \\
\begin{split} \label{eq:kollo:b}
&\mathbf{b} = m^{-1} \left[ s_1 \mathbf{c}_1 + \dots + s_m \mathbf{c}_m \right] \text{(Kollo skewness)}\\
&\text{where} \quad s_k = \left(c_{1k} + \dots + c_{nk} \right)^2 = \left( \mathbf{1}' \mathbf{c}_k \right)^2
\end{split}
\end{align}
\end{document}
答案1
这里有两种不同的可能对齐方式,用aligned
而不是split
:
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{fourier}
\usepackage{mathtools}
\usepackage{cleveref}
\begin{document}
\begin{align}
&\mathbf{L}_{mn}'\mathbf{L}_{mn} = m \mathbf{I}_n & \text{(Orthogonality)}\\
&\mathbf{1}_m^{'} \mathbf{L}_{mn} = \mathbf{0}'_n & \text{(Row Sums)} \\
& \begin{aligned}[t] \label{eq:kollo:b}
&\mathbf{b} = m^{-1} \left[ s_1 \mathbf{c}_1 + \dots + s_m \mathbf{c}_m \right]\\
&\quad\text{where} \quad s_k = \left(c_{1k} + \dots + c_{nk} \right)^2 = \left( \mathbf{1}' \mathbf{c}_k \right)^2
\end{aligned} & \text{(Kollo skewness)}
\end{align}
\Cref{eq:kollo:b}
\begin{align}
&\mathbf{L}_{mn}'\mathbf{L}_{mn} = m \mathbf{I}_n & & \text{(Orthogonality)}\\
&\mathbf{1}_m^{'} \mathbf{L}_{mn} = \mathbf{0}'_n & & \text{(Row Sums)} \\
& \begin{aligned}[t] \label{eq:kollo:c}
&\mathbf{b} = m^{-1} \left[ s_1 \mathbf{c}_1 + \dots + s_m \mathbf{c}_m \right]\\
&\quad\text{where} \quad s_k = \left(c_{1k} + \dots + c_{nk} \right)^2 = \left( \mathbf{1}' \mathbf{c}_k \right)^2
\end{aligned} & & \text{(Kollo skewness)}
\end{align}
\end{document}
答案2
从 Bernard 的回答中,我看到他如何将 & 放在嵌套之外,这aligned
对我也有同样的效果split
。这样我的第二个要求,即在第 3 行和第 4 行之间进行编号仍然得到满足。
\begin{align}
&\mathbf{L}_{mn}'\mathbf{L}_{mn} = m \mathbf{I}_n & \text{(Orthogonoalitly)}\\
&\mathbf{1}_m^{'} \mathbf{L}_{mn} = \mathbf{0}'_n & \text{(Row Sums)} \\
\begin{split} \label{eq:kollo:b}
&\mathbf{b} = m^{-1} \left[ s_1 \mathbf{c}_1 + \dots + s_m \mathbf{c}_m \right] \\
&\quad \text{where} \quad s_k = \left(c_{1k} + \dots + c_{nk} \right)^2 = \left( \mathbf{1}' \mathbf{c}_k \right)^2
\end{split} & \text{(Kollo skewness)}
\end{align}