我正在尝试使用 TikZ 重新创建类似于此流程图的内容：
这是我目前的代码：
\begin{figure}[H]
\centering
\resizebox{0.99\textwidth}{!}{
\begin{tikzpicture}[node distance=2cm]
\node(first)[draw,rectangle]
{\shortstack{$\boldsymbol{M}^I = \boldsymbol{M}_s + \boldsymbol{M}_{h_0}$\\
$\boldsymbol{K}^I = \boldsymbol{K}_s + \boldsymbol{K}_h$}};
\node(second)[draw,rectangle,below of=first]
{\shortstack{$\boldsymbol{\Phi}^I$\\
$\omega_n^I$}};
\node(third)[draw,rectangle,below of=second]
{\shortstack{$\tilde{\boldsymbol{M}}^I$\\
$\tilde{\boldsymbol{K}}^I$\\
$\tilde{\boldsymbol{C}}^I = \alpha \tilde{\boldsymbol{M}}^I + \beta \tilde{\boldsymbol{K}}^I$}};
\node(fourth)[draw,rectangle,right of=first,xshift=4cm]
{\shortstack{$\boldsymbol{M}_i^{II}(\omega) = \boldsymbol{M}_{h,i}(\omega)$\\
$\boldsymbol{C}_i^{II}(\omega) = \boldsymbol{C}_{h,i}(\omega)$\\
$\boldsymbol{K}_i^{II} = \boldsymbol{K}_{h,i} - \boldsymbol{K}_{h,i} = \boldsymbol{0}$}};
\node(fifth)[draw,rectangle,below of=fourth]
{\shortstack{$\boldsymbol{M}^{II}(\omega) = \sum_{i=1}^7\boldsymbol{T}_i^T\boldsymbol{M}_i^{II}(\omega)\boldsymbol{T}_i$\\
$\boldsymbol{C}^{II}(\omega) = \sum_{i=1}^7\boldsymbol{T}_i^T\boldsymbol{C}_i^{II}(\omega)\boldsymbol{T}_i$}};
\node(sixth)[draw,rectangle,below of=fifth]
{\shortstack{$\tilde{\boldsymbol{M}}^{II}(\omega) = (\boldsymbol{\Phi}^I)^T\boldsymbol{M}^{II}(\omega)\boldsymbol{\Phi}^I$\\
$\tilde{\boldsymbol{C}}^{II}(\omega) = (\boldsymbol{\Phi}^I)^T\boldsymbol{C}^{II}(\omega)\boldsymbol{\Phi}^I$}};
\node(seventh)[draw,rectangle,below of=sixth]
{\shortstack{$\tilde{\boldsymbol{M}}(\omega) = \tilde{\boldsymbol{M}}^I + \tilde{\boldsymbol{M}}^{II}(\omega)$\\
$\tilde{\boldsymbol{C}}(\omega) = \tilde{\boldsymbol{C}}^I + \tilde{\boldsymbol{C}}^{II}(\omega)$\\
$\tilde{\boldsymbol{K}} = \tilde{\boldsymbol{K}}^I$}};
\node(eighth)[draw,rectangle,right of=fourth,xshift=5cm]
{\shortstack{$\boldsymbol{Q}_r(\omega, \theta)$}};
\node(ninth)[draw,rectangle,right of=eighth,xshift=2cm]
{\shortstack{$S_\eta(\omega,\theta) = S_\eta(\omega) D(\theta)$}};
\node(tenth)[draw,rectangle,below of=eighth,anchor=west]
{\shortstack{$\boldsymbol{S}_{p_rp_s}(\omega) = \int_\theta{\boldsymbol{Q}_r(\omega, \theta)}S_\eta(\omega, \theta)\boldsymbol{Q}_r(\omega, \theta)^H \; d\theta$}};
\node(eleventh)[draw,rectangle,below of=tenth]
{\shortstack{$\boldsymbol{S}_p(\omega) =
\begin{bmatrix}
\boldsymbol{S}_{p_1p_1}(\omega) & \cdots & \boldsymbol{S}_{p_1p_7}(\omega)\\
\vdots & \ddots & \vdots \\
\boldsymbol{S}_{p_7p_1}(\omega) & \cdots & \boldsymbol{S}_{p_7p_7}(\omega)
\end{bmatrix}$}};
\node(twelfth)[draw,rectangle,below of=eleventh]
{\shortstack{$\tilde{\boldsymbol{S}}_p(\omega) = (\boldsymbol{\Phi}^I)^T\boldsymbol{S}_p(\omega)\boldsymbol{\Phi}^I$}};
\node(thirteenth)[draw,rectangle,below of=seventh,yshift=-1cm]
{\shortstack{$\tilde{\boldsymbol{S}}_u(\omega) = \tilde{\boldsymbol{H}}(\omega)\tilde{\boldsymbol{S}}_p(\omega)\tilde{\boldsymbol{H}}(\omega)^H$}};
\node(fourteenth)[draw,rectangle,below of=thirteenth]
{\shortstack{$\boldsymbol{S}_u(\omega) = \boldsymbol{\Phi}^I\tilde{\boldsymbol{S}}_u(\omega)(\boldsymbol{\Phi}^I)^T$}};
\node[rectangle,below of=seventh,anchor=west] {\shortstack{$\tilde{\boldsymbol{H}}(\omega)$}};
\draw[->,>=stealth] (first) -- (second);
\draw[->,>=stealth] (second) -- (third);
%\draw[->,>=stealth] (third) -- (seventh);
\draw[->,>=stealth] (fourth) -- (fifth);
\draw[->,>=stealth] (fifth) -- (sixth);
\draw[->,>=stealth] (sixth) -- (seventh);
\draw[->,>=stealth] (eighth) -- (tenth);
\draw[->,>=stealth] (ninth) -- (tenth);
\draw[->,>=stealth] (tenth) -- (eleventh);
\draw[->,>=stealth] (eleventh) -- (twelfth);
\draw[->,>=stealth] (seventh) -- (thirteenth);
\draw[->,>=stealth] (thirteenth) -- (fourteenth);
\end{tikzpicture}
}
\caption{Caption}
\label{fig:my_label}
\end{figure}
结果如下：
我有两个问题：
我该如何让最右边的“列”节点与中心“列”相对于最左边“列”的水平距离相同？此外，如何让右上角的节点箭头垂直向下指向而不是倾斜？
我怎样才能获得如第一个屏幕截图所示的 90 度角箭头？
这两点是我唯一想实现的。我不想重新创建第一个屏幕截图中的颜色或编号。
答案1
正如评论中提到的，一种方法是定义节点并将它们相对于彼此移动，画出到节点特定点的线，......
问题 1 我应该如何让最右边的节点“列”与中心“列”相对于最左边的“列”具有相同的水平距离？
正如评论中提到的，\usetikzlibrary{calc}可以用于定义与合成节点之间的距离($(startnode) + (xshift,yshift)$)
\documentclass[border=5pt]{standalone}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
\node(first)[draw,rectangle] at (0,0) {\shortstack{$\boldsymbol{M}^I = \boldsymbol{M}_s + \boldsymbol{M}_{h_0}$ \\
$\boldsymbol{K}^I = \boldsymbol{K}_s + \boldsymbol{K}_h$}};
\node(fourth)[draw,rectangle,anchor=north west] at ($(first.north east) + (4,0)$)
{\shortstack{$\boldsymbol{M}_i^{II}(\omega) = \boldsymbol{M}_{h,i}(\omega)$ \\
$\boldsymbol{C}_i^{II}(\omega) = \boldsymbol{C}_{h,i}(\omega)$ \\
$\boldsymbol{K}_i^{II} = \boldsymbol{K}_{h,i} - \boldsymbol{K}_{h,i} = \boldsymbol{0}$}};
\node(eighth)[draw,rectangle,anchor=north west] at ($(fourth.north east) + (4,0)$)
{\shortstack{$\boldsymbol{Q}_r(\omega, \theta)$}};
\node(tenth)[draw,rectangle,anchor=south west] at (fourth.south east -| eighth.south west)
{\shortstack{$\boldsymbol{S}_{p_rp_s}(\omega) = \int_\theta{\boldsymbol{Q}_r(\omega, \theta)}S_\eta(\omega, \theta)\boldsymbol{Q}_r(\omega, \theta)^H \; d\theta$}};
\node(ninth)[draw,rectangle,anchor=east] at (eighth.east -| tenth.east)
{\shortstack{$S_\eta(\omega,\theta) = S_\eta(\omega) D(\theta)$}};
\end{tikzpicture}
\end{document}
第二季度 另外，如何让右上角的节点箭头垂直向下而不是倾斜？
由于我们不知道端点的锚点，我们可以简单地使用以下方法判断箭头的长度--++(xshift,yshift)
\documentclass[border=5pt]{standalone}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
% nodes
\node(eighth)[draw,rectangle,anchor=north west] at (0,0)
{\shortstack{$\boldsymbol{Q}_r(\omega, \theta)$}};
\node(tenth)[draw,rectangle,anchor=north west] at ($(eighth.south west) - (0,1)$)
{\shortstack{$\boldsymbol{S}_{p_rp_s}(\omega) = \int_\theta{\boldsymbol{Q}_r(\omega, \theta)}S_\eta(\omega, \theta)\boldsymbol{Q}_r(\omega, \theta)^H \; d\theta$}};
\node(ninth)[draw,rectangle,anchor=east] at (eighth.east -| tenth.east)
{\shortstack{$S_\eta(\omega,\theta) = S_\eta(\omega) D(\theta)$}};
% arrows
\draw[-latex] (eighth.south) --++ (0,-1);
\draw[-latex] (ninth.south) --++ (0,-1);
\end{tikzpicture}
\end{document}
第三季度 我怎样才能获得如第一个屏幕截图所示的 90 度角箭头？
当知道起点和终点时，可以轻松完成此操作\draw[->] (startnode) |- (endnode)。
\documentclass[border=5pt]{standalone}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
\begin{tikzpicture}
% nodes
\node(third)[draw,rectangle] at (0,0)
{\shortstack{$\tilde{\boldsymbol{M}}^I$\\
$\tilde{\boldsymbol{K}}^I$\\
$\tilde{\boldsymbol{C}}^I = \alpha \tilde{\boldsymbol{M}}^I + \beta \tilde{\boldsymbol{K}}^I$}};
\node(seventh)[draw,rectangle,anchor=north west] at ($(third.south east) + (4,-4)$)
{\shortstack{$\tilde{\boldsymbol{M}}(\omega) = \tilde{\boldsymbol{M}}^I + \tilde{\boldsymbol{M}}^{II}(\omega)$\\
$\tilde{\boldsymbol{C}}(\omega) = \tilde{\boldsymbol{C}}^I + \tilde{\boldsymbol{C}}^{II}(\omega)$\\
$\tilde{\boldsymbol{K}} = \tilde{\boldsymbol{K}}^I$}};
% arrows
\draw[-latex] (third.south) |- ($(seventh.west) + (0,0.5)$);
\draw[-latex] (third.south) |- (seventh.west);
\draw[-latex] (third.south) |- ($(seventh.west) - (0,0.5)$);
\end{tikzpicture}
\end{document}
编辑
最终结果如下：
tikzstyle您可以尝试一下，也许为不同的节点定义自己的节点，使它们具有相同的宽度，从而实现更好的对齐。
\documentclass[border=5pt]{standalone}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{calc,fit,backgrounds}
\begin{document}
\tikzset{%
every node/.style={minimum height=0.75cm},
}
\tikzstyle{node1} = [draw,fill=white,minimum width=3cm]
\tikzstyle{node2} = [draw,fill=white,minimum width=5cm]
\tikzstyle{node3} = [draw,fill=white,minimum width=7cm]
\tikzstyle{circ} = [draw,fill=white,circle]
\begin{tikzpicture}
% first column
\node(1)[node1] at (0,0)
{\shortstack{
$\boldsymbol{M}^I = \boldsymbol{M}_s + \boldsymbol{M}_{h_0}$ \\
$\boldsymbol{K}^I = \boldsymbol{K}_s + \boldsymbol{K}_h$}
};
\node(2)[node1,anchor=north] at ($(1.south) + (0,-1)$)
{\shortstack{
$\boldsymbol{\Phi}^I$ \\
$\omega_n^I$}
};
\node(3)[node1,anchor=north] at ($(2.south) + (0,-1)$)
{\shortstack{
$\tilde{\boldsymbol{M}}^I$\\
$\tilde{\boldsymbol{K}}^I$\\
$\tilde{\boldsymbol{C}}^I = \alpha \tilde{\boldsymbol{M}}^I + \beta \tilde{\boldsymbol{K}}^I$}
};
% second column
\node(4)[node2,anchor=north west] at ($(1.north east) + (2,0)$)
{\shortstack{
$\boldsymbol{M}_i^{II}(\omega) = \boldsymbol{M}_{h,i}(\omega)$\\
$\boldsymbol{C}_i^{II}(\omega) = \boldsymbol{C}_{h,i}(\omega)$\\
$\boldsymbol{K}_i^{II} = \boldsymbol{K}_{h,i} - \boldsymbol{K}_{h,i} = \boldsymbol{0}$}
};
\node(5)[node2,anchor=north] at ($(4.south) + (0,-1)$)
{\shortstack{$\boldsymbol{M}^{II}(\omega) = \sum_{i=1}^7\boldsymbol{T}_i^T\boldsymbol{M}_i^{II}(\omega)\boldsymbol{T}_i$\\
$\boldsymbol{C}^{II}(\omega) = \sum_{i=1}^7\boldsymbol{T}_i^T\boldsymbol{C}_i^{II}(\omega)\boldsymbol{T}_i$}};
\node(6)[node2,anchor=north] at ($(5.south) + (0,-1)$)
{\shortstack{
$\tilde{\boldsymbol{M}}^{II}(\omega) = (\boldsymbol{\Phi}^I)^T\boldsymbol{M}^{II}(\omega)\boldsymbol{\Phi}^I$\\
$\tilde{\boldsymbol{C}}^{II}(\omega) = (\boldsymbol{\Phi}^I)^T\boldsymbol{C}^{II}(\omega)\boldsymbol{\Phi}^I$}
};
\node(7)[node2,anchor=north] at ($(6.south) + (0,-1)$)
{\shortstack{
$\tilde{\boldsymbol{M}}(\omega) = \tilde{\boldsymbol{M}}^I + \tilde{\boldsymbol{M}}^{II}(\omega)$\\
$\tilde{\boldsymbol{C}}(\omega) = \tilde{\boldsymbol{C}}^I + \tilde{\boldsymbol{C}}^{II}(\omega)$\\
$\tilde{\boldsymbol{K}} = \tilde{\boldsymbol{K}}^I$}
};
% third column
\node(8)[draw,fill=white,anchor=north west] at ($(4.north east) + (2,0)$)
{\shortstack{
$\boldsymbol{Q}_r(\omega, \theta)$}
};
\node(10)[node3,anchor=north west] at ($(8.south west) + (0,-1)$)
{\shortstack{
$\boldsymbol{S}_{p_rp_s}(\omega) = \int_\theta{\boldsymbol{Q}_r(\omega, \theta)}S_\eta(\omega, \theta)\boldsymbol{Q}_r(\omega, \theta)^H \; d\theta$}
};
\node(9)[draw,fill=white,anchor=east] at (10.east |- 8.east)
{\shortstack{
$S_\eta(\omega,\theta) = S_\eta(\omega) D(\theta)$}
};
\node(11)[node3,anchor=north] at ($(10.south) + (0,-1)$)
{\shortstack{
$\boldsymbol{S}_p(\omega) =
\begin{bmatrix}
\boldsymbol{S}_{p_1p_1}(\omega) & \cdots & \boldsymbol{S}_{p_1p_7}(\omega)\\
\vdots & \ddots & \vdots \\
\boldsymbol{S}_{p_7p_1}(\omega) & \cdots & \boldsymbol{S}_{p_7p_7}(\omega)
\end{bmatrix}$}
};
\node(12)[node3,anchor=north] at ($(11.south) + (0,-1)$)
{\shortstack{
$\tilde{\boldsymbol{S}}_p(\omega) = (\boldsymbol{\Phi}^I)^T\boldsymbol{S}_p(\omega)\boldsymbol{\Phi}^I$}
};
% second column
\node(13)[node2,anchor=north] at ($(7.south) + (0,-1)$)
{\shortstack{$\tilde{\boldsymbol{S}}_u(\omega) = \tilde{\boldsymbol{H}}(\omega)\tilde{\boldsymbol{S}}_p(\omega)\tilde{\boldsymbol{H}}(\omega)^H$}};
\node(14)[node2,anchor=north] at ($(13.south) + (0,-1)$)
{\shortstack{
$\boldsymbol{S}_u(\omega) = \boldsymbol{\Phi}^I\tilde{\boldsymbol{S}}_u(\omega)(\boldsymbol{\Phi}^I)^T$}
};
% arrows
\draw[-stealth] (1.south) -- (2.north);
\draw[-stealth] (2.south) -- (3.north);
\draw[-stealth] (3.south) |- ($(7.west) + (0,0.5)$);
\draw[-stealth] (4.south) -- (5.north);
\draw[-stealth] (5.south) -- (6.north);
\draw[-stealth] (6.south) -- (7.north);
\draw[-stealth] (8.south) --++ (0,-1);
\draw[-stealth] (9.south) --++ (0,-1);
\draw[-stealth] (10.south) -- (11.north);
\draw[-stealth] (11.south) -- (12.north);
\draw[-stealth] (12.south) |- (13.east);
\draw[-stealth] (7.south) -- (13.north) node[pos=0.5,right] {\shortstack{$\tilde{\boldsymbol{H}}(\omega)$}};
\draw[-stealth] (13.south) -- (14.north);
% labels
\foreach \i [count=\ii from 1] in {1,...,14}{%
\node(lab-\i)[circ,anchor=east] at ($(\ii.west) + (-0.5,0)$) {\i};
}
% filling
\begin{pgfonlayer}{background}
\node[fit=(1)(3)(lab-1), fill=green!15!] {};
\node[fit=(4)(6)(lab-4), fill=blue!15!] {};
\node[fit=(12)(lab-8), fill=orange!15!] {};
\end{pgfonlayer}
\end{tikzpicture}
\end{document}