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在枚举环境中围绕表格环绕文本

tag-icon • tag-icon tables floats wrapfigure wrap
在枚举环境中围绕表格环绕文本

我想将枚举环境中的文本环绕在表格周围，这是我的实际情况（抱歉包含了太多代码，但我想准确说明枚举环境中的行数）：

\subsection{Given the following data set where “Target 2” represent the class attribute, 
compute the naive Bayesian classification for the instance $<L,white>$ and $<XS,?>$.}
\begin{table}[h!t]
\centering
\begin{tabular}{ccc}
    \toprule    
    Size        &   Color   &   Target2 \\
    \midrule
    XS      &   green   &   Yes     \\
    L       &   green   &   Yes     \\
    XS      &   white   &   No      \\
    M       &   black   &   No      \\
    XL      &   green   &   Yes     \\
    XS      &   white   &   Yes     \\
    L       &   black   &   No      \\
    M       &   green   &   Yes     \\
    \bottomrule
\end{tabular}
\end{table}
\begin{enumerate}
    \item In this case the experience $E$ is made up of $e_{1} = L$ and $e_{2} = white$, 
        which in Naive Bayes are to be considered as independent, therefore we have:
        \begin{align*}
            P(yes|E) &= P(L|yes)\cdot P(white|yes) \cdot P(yes)     \\
                     &= \sfrac{1}{5}\times \sfrac{1}{5} \times \sfrac{5}{8} \\
                     &= 0.025
        \end{align*}
        \begin{align*}
            P(no|E) &= P(L|no) \cdot P(white|no) \cdot P(no) \\
                    &= \sfrac{1}{3} \times \sfrac{1}{3} times \sfrac{3}{8}  \\
                    &= 0.041
        \end{align*}
        Then we normalize:
        \begin{align*}
            P(yes) &= \frac{0.025}{0.066}  
                    \simeq 0.38 
        \end{align*}
        \begin{align*}
            P(no) &= \frac{0.041}{0.066} 
                   \simeq 0.62
        \end{align*}
        As $P(no) > P(yes)$, we label $<L,white>$ as ``no''.
    \item Now we have to classify a sample with a missing value. During the testing phase 
        we simply omit the attribute\footnote{Classification Other Methods, slide 24}:
        \begin{align*}
            P(yes|XS) &= P(XS|yes) \cdot P(yes) = \sfrac{2}{5} \times \sfrac{5}{8}
                      = 0.25
        \end{align*}
        \begin{align*}
            P(no|XS) &= P(XS|no) \cdot P(no) 
                     = \sfrac{1}{3} \times \sfrac{3}{8}
                     = 0.125
        \end{align*}
        Let us normalize
        \begin{align*}
            P(yes) &= \sfrac{0.25}{0.375} \simeq 0.7 \\
            P(no)  &= \sfrac{0.125}{0.375} \simeq 0.3
        \end{align*}
        Bottom line this sample is classified as ``yes''.
\end{enumerate}

我尝试使用wrapfig和floatftl包，但没有成功，在这两种情况下，表格都被移到了列表的末尾。我考虑过使用两种minipage环境，但我希望文本能够真正环绕表格。

答案1

发布问题有助于完全的我猜，包括加载你需要的所有包

\usepackage{booktabs,xfrac,amsmath}

在这种情况下。我还修复了一些字体问题（针对多字母标识符和尖括号）

在 LaTeX 列表中更改边距有点棘手，但我认为这是你想要的布局

在此处输入图片描述

\documentclass{article}

\usepackage{booktabs,xfrac,amsmath}

\begin{document}

\subsection{Given the following data set where “Target 2” represent the class attribute, 
compute the naive Bayesian classification for the instance $\langle L,white\rangle$ and $\langle \mathit{XS},?\rangle$.}

\savebox0{%
\begin{tabular}{ccc}
    \toprule    
    Size        &   Color   &   Target2 \\
    \midrule
    XS      &   green   &   Yes     \\
    L       &   green   &   Yes     \\
    XS      &   white   &   No      \\
    M       &   black   &   No      \\
    XL      &   green   &   Yes     \\
    XS      &   white   &   Yes     \\
    L       &   black   &   No      \\
    M       &   green   &   Yes     \\
    \bottomrule
\end{tabular}}

\begin{enumerate}
\makeatletter
\dimen@\wd0
\advance\dimen@2em
\advance\rightmargin-\dimen@
\advance\linewidth-\dimen@
\parshape \@ne \@totalleftmargin \linewidth
\hbox to \textwidth{\hfill\vtop to \z@{\vskip1em \box\z@\vss}}
   \item

 In this case the experience $E$ is made up of $e_{1} = L$ and $e_{2} = \mathit{white}$, 
        which in Naive Bayes are to be considered as independent, therefore we have:
        \begin{align*}
            P(\mathit{yes}|E) &= P(L|\mathit{yes})\cdot P(\mathit{white}|\mathit{yes}) \cdot P(\mathit{yes})     \\
                     &= \sfrac{1}{5}\times \sfrac{1}{5} \times \sfrac{5}{8} \\
                     &= 0.025
        \end{align*}
        \begin{align*}
            P(\mathit{no}|E) &= P(L|\mathit{no}) \cdot P(\mathit{white}|\mathit{no}) \cdot P(\mathit{no}) \\
                    &= \sfrac{1}{3} \times \sfrac{1}{3} \times \sfrac{3}{8}  \\
                    &= 0.041
        \end{align*}

\advance\rightmargin\dimen@
\advance\linewidth\dimen@
\parshape \@ne \@totalleftmargin \linewidth

        Then we normalize:
        \begin{align*}
            P(\mathit{yes}) &= \frac{0.025}{0.066}  
                    \simeq 0.38 
        \end{align*}
        \begin{align*}
            P(\mathit{no}) &= \frac{0.041}{0.066} 
                   \simeq 0.62
        \end{align*}
        As $P(\mathit{no}) > P(\mathit{yes})$, we label $\langle L,\mathit{white}\rangle$ as ``no''.



    \item Now we have to classify a sample with a missing value. During the testing phase 
        we simply omit the attribute\footnote{Classification Other Methods, slide 24}:
        \begin{align*}
            P(\mathit{yes}|\mathit{XS}) &= P(\mathit{XS}|\mathit{yes}) \cdot P(\mathit{yes}) \\
&= \sfrac{2}{5} \times \sfrac{5}{8}\\
                      &= 0.25
        \end{align*}
        \begin{align*}
            P(\mathit{no}|\mathit{XS}) &= P(\mathit{XS}|\mathit{no}) \cdot P(\mathit{no})\\ 
                     &= \sfrac{1}{3} \times \sfrac{3}{8}\\
                     &= 0.125
        \end{align*}
        Let us normalize
        \begin{align*}
            P(\mathit{yes}) &= \sfrac{0.25}{0.375} \simeq 0.7 \\
            P(\mathit{no})  &= \sfrac{0.125}{0.375} \simeq 0.3
        \end{align*}
        Bottom line this sample is classified as ``yes''.
\end{enumerate}


\end{document}

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