答案1
可以画出这样的图吗?当然可以。但是,有许多专门的包对于特定场景特别有用。只要您不提供更多细节,您可能会错过最佳选择。根据我在您的屏幕截图中看到的内容,我做了一个猜测,以下是结果。
\documentclass[border=3.14mm,standalone]{standalone}
\usepackage{tikz-dependency}
\usetikzlibrary{fit}
\begin{document}
\begin{dependency}
\begin{deptext}[column sep=2pt]
blah2 \& blah2 \& blah2 \& blah2 \& blah2 \\[2cm]
blah \& blah \& blah \& blah \& blah \\[2cm]
blah3 \& blah3 \& blah3 \& blah3 \& blah3 \\
\end{deptext}
\begin{scope}[font=\sffamily]
\foreach \X in {1,...,5}
{\draw[thick,-latex] (\wordref{3}{\X}) -- ++ (0,-1.5) node[below]{\X};}
\path (\wordref{3}{1}) -- ++ (0,-1.5) node[midway,left=8mm] {analyze};
\path (\wordref{1}{1}) -- ++ (0,1.5) node[midway,left=8mm] {predict};
\draw[thick,-latex] (\wordref{2}{3}) -- (\wordref{3}{3})
node[midway,right]{task: analysis};
\draw[thick,-latex] (\wordref{2}{3}) -- (\wordref{1}{3})
node[midway,right]{task: prediction};
\foreach \X [count=\Z] in {subj,dobj,iobj}
{\draw[thick,-latex] (\wordref{1}{3}) -- ++ (-3+1.5*\Z,1.5) node[above]{\Z}
node[midway,left]{\X};}
\foreach \X in {1,2,3}
{\node[inner sep=0pt,fit=(\wordref{\X}{1}) (\wordref{\X}{5}),draw]{};}
\end{scope}
\end{dependency}
\end{document}