桌子在块内损坏

Question

我在 beamer 上。我把下面的代码放在一个块内创建一个表。

\begin{block}{}
        \begin{tabular}{l|l}
            \multirow{2}{*}{$y = \begin{bmatrix}
                        w_{0}\\
                        w_{1}\\
                        \vdots \\
                        w_{D}\\
                    \end{bmatrix}^{T} *  \begin{bmatrix}
                                            x_{0}\\
                                            x_{1}\\
                                            \vdots \\
                                            x_{D}\\ 
                                            \end{bmatrix}  = W^T * X$} & $\hat{y} = W^T * X + \epsilon$ \\
                   & $\hat{y} = \sum_{j=0}^{D}w_{j}x_{j}+\epsilon$
        \end{tabular}

    \end{block}
\small where,
    \begin{itemize}
    \small\item value of $x_0$ is $1$. 
    \small\item $\epsilon$ is the residual error. 
    \small\item $W^T$ are the weight (coefficient of $x$) vector.
    \small\item $\hat{y}$ is the predicted value.
    \small\item \textcolor{red}{$D+1$} is the \textcolor{red}{dimension} of the vector $W$ and $x$.
    \end{itemize}

但我得到以下输出，

桌子为什么坏了？有没有更好的方法来解决这样的事情。

Answer 1

\multirow不知道表格会有多大，也不会告诉 TeX 它占用了多少空间。相反，用户必须告诉\multirow它允许占用多少行。然而，中的方程\multirow远高于两个行。矩阵已经使用了四行。

解决方法是提供所需的空间：

\documentclass{beamer}
\usetheme{Madrid}
\usepackage{multirow}
\begin{document}
\begin{frame}
\begin{block}{}
  \begin{tabular}{l|l}
    \multirow{2}{*}{%
      $y = \begin{bmatrix}
        w_{0}\\
        w_{1}\\
        \vdots \\
        w_{D}\\
      \end{bmatrix}^{T} *  \begin{bmatrix}
        x_{0}\\
        x_{1}\\
        \vdots \\
        x_{D}\\ 
      \end{bmatrix}  = W^T * X$}
    & $\hat{y} = W^T * X + \epsilon$ \\
    & $\hat{y} = \sum_{j=0}^{D}w_{j}x_{j}+\epsilon$\\[13mm]
  \end{tabular}
\end{block}
\small where,
  \begin{itemize}
    \item value of $x_0$ is $1$. 
    \item $\epsilon$ is the residual error. 
    \item $W^T$ are the weight (coefficient of $x$) vector.
    \item $\hat{y}$ is the predicted value.
    \item \textcolor{red}{$D+1$} is the \textcolor{red}{dimension}
      of the vector $W$ and $x$.
    \end{itemize}
\end{frame}
\end{document}

下一个示例使用一些简单的纯 TeX 框命令来设置两列并将它们对齐到公式的顶部（而不是第一行的基线）。multirow这里不需要包：

\begin{block}{}
  \mbox{%
    \vtop{%
      \kern0pt
      \hbox{%
        $y = \begin{bmatrix}
          w_{0}\\
          w_{1}\\
          \vdots \\
          w_{D}\\
        \end{bmatrix}^{T} *  \begin{bmatrix}
          x_{0}\\
          x_{1}\\
          \vdots \\
          x_{D}\\ 
        \end{bmatrix}  = W^T * X$%
      }%
    }
    \kern\tabcolsep
    \vrule
    \kern\tabcolsep
    \vtop{%
      \kern0pt
      \hbox{$%
        \begin{gathered}
          \hat{y} = W^T * X + \epsilon \\
          \hat{y} = \sum_{j=0}^{D}w_{j}x_{j}+\epsilon
        \end{gathered}%
      $}%
    }%
  }
\end{block}

Answer 2

基于Micos 的回答，但使用 beamers 自己的柱机制：

\documentclass{beamer}
\usetheme{Madrid}
\begin{document}
\begin{frame}
\begin{block}{}
    \vskip-\baselineskip
    \begin{columns}[T]
        \begin{column}{.55\textwidth}
            \vskip-0.2cm
            \[
            y = \begin{bmatrix}
                  w_{0}\\
                  w_{1}\\
                  \vdots \\
                  w_{D}\\
            \end{bmatrix}^{T}
            \begin{bmatrix}
                  x_{0}\\
                  x_{1}\\
                  \vdots \\
                  x_{D}\\ 
            \end{bmatrix} + \epsilon = W^T X+\epsilon
            \]
        \end{column}
            \vrule
        \begin{column}{.35\textwidth}
            \begin{align*}
            \hat{y} &= W^T X \\
                    &= \sum_{j=0}^{D} w_{j}x_{j}
            \end{align*}
        \end{column}
    \end{columns}
\end{block}
\bigskip
where
\begin{itemize}
      \item $X$ is the vector of observed or predicted factors. $x_0\equiv1$ 
      \item $W$ is the coefficient vector
      \item $\epsilon$ is the error term (a scalar)
      \item $\hat{y}$ is the fitted or predicted value
      \item \alert{$D+1$} is the \alert{dimension} of the vectors $W$ and $X$
\end{itemize}
\end{frame}
\end{document}

Answer 3

这是一个在环境内使用两个并排minipage环境的解决方案block。（没有垂直规则是故意的，因为我认为它既不需要也没什么用。）

请注意，我也试图简化一些术语。

\documentclass{beamer}
\usetheme{Madrid} % just for this example
\begin{document}
\begin{frame}
\begin{block}{}
\vskip-\baselineskip % per @samcarter's suggestion :-)
\begin{minipage}{0.6\textwidth}
\[
y = \begin{bmatrix}
      w_{0}\\
      w_{1}\\
      \vdots \\
      w_{D}\\
\end{bmatrix}^{T}
\begin{bmatrix}
      x_{0}\\
      x_{1}\\
      \vdots \\
      x_{D}\\ 
\end{bmatrix} + \epsilon = W^T X+\epsilon
\]
\end{minipage}%
\begin{minipage}{0.3\textwidth}
\begin{align*}
\hat{y} &= W^T X \\
        &= \sum_{j=0}^{D} w_{j}x_{j}
\end{align*}
\end{minipage}
\end{block}

\bigskip
where
\begin{itemize}
      \item $X$ is the vector of observed or predicted factors. $x_0\equiv1$ 
      \item $W$ is the coefficient vector
      \item $\epsilon$ is the error term (a scalar)
      \item $\hat{y}$ is the fitted or predicted value
      \item \alert{$D+1$} is the \alert{dimension} of the vectors $W$ and $X$
\end{itemize}

\end{frame}
\end{document}

Answer 4

环境\multirow中等式的右边没有和：aligned

\documentclass{beamer}
\usetheme{Madrid}
\usepackage{amsmath}

\begin{document}
\begin{frame}
\begin{block}{}
\setlength\tabcolsep{12pt}
  \begin{tabular}{l|l}
      $y = \begin{bmatrix}
        w_{0}\\
        w_{1}\\
        \vdots \\
        w_{D}\\
      \end{bmatrix}^{T} \ast  \begin{bmatrix}
        x_{0}\\
        x_{1}\\
        \vdots \\
        x_{D}\\
      \end{bmatrix}  = W^T \ast X$
    & $\begin{aligned}\hat{y}   & = W^T \ast X + \epsilon \\
                                & = \sum_{j=0}^{D}w_{j}x_{j}+\epsilon
        \end{aligned}$
  \end{tabular}
\end{block}

\small
where,
  \begin{itemize}
    \item value of $x_0$ is $1$.
    \item $\epsilon$ is the residual error.
    \item $W^T$ are the weight (coefficient of $x$) vector.
    \item $\hat{y}$ is the predicted value.
    \item \textcolor{red}{$D+1$} is the \textcolor{red}{dimension}
      of the vector $W$ and $x$.
    \end{itemize}
\end{frame}
\end{document}