\ifthenelse 处理不等式和表达式，使用 \pgfmathparse

Question

我正在尝试编写一个函数，该函数绘制一个矩阵，其中包含一些 1（黑色）和 0（白色），并且 1 和 0 的分布具有一定的模式。我只想绘制 1，但前提是它们在矩阵的边界内，并且需要评估矩阵索引上的一些不等式来测试这一点。这是一个 MWE：

\documentclass{beamer}
\usepackage{tikz}
\usepackage{ifthen}
\usetikzlibrary{calc}
\newcommand{\LStepMatrix}[6] {
  % #1 origin:  top left corner of the matrix
  % #2: N number of cols in the matrix
  % #3: M number of rows in the matrix
  % #4: L size of the step
  % #5: The numeric sequence used for this matrix
  % #6: scale factor for the figure

  % Draw the zeros
  \foreach \i in {1,2,...,#2} {
    \foreach \j in {1,2,...,#3} {
      \node[minimum width=#6cm,minimum height=#6cm,draw,inner sep=0] at ($#1+#6*(\i,-\j)$) {};
    }
  }

  % Draw the ones in the positions of the numeric sequence
  \foreach \i in {1,2,...,#3} { % Iterate over the rows
    \foreach \j in {#5} {
      %% \pgfmathparse{\i*#4-\j > -1? "gtz":"ltz"} \edef\mycol{\pgfmathresult}
      %% \ifthenelse {\equal{\mycol}{gtz}} { % If we are inside the boundaries of the matrix print the bit
      %%   \node[minimum width=#6cm,minimum height=#6cm,fill=black,inner sep=0] at ($#1+#6*(\i*#4-\j+1,-\i)$) {};
      %% } {                                 %Otherwise don't print the bit
      %% }

      \pgfmathparse{\i*#4-\j} \edef\mycol{\pgfmathresult}
      \ifthenelse {2>1} { % If we are inside the boundaries of the matrix print the bit
        \node[minimum width=#6cm,minimum height=#6cm,fill=black,inner sep=0] at ($#1+#6*(\i*#4-\j+1,-\i)$) {};
      } {                                 %Otherwise don't print the bit
      }

    }
  }
}
\begin{document}
\begin{frame}{}
  \begin{tikzpicture}
    \LStepMatrix{(0.5,7)}{64}{8}{7}{1,2,3,4,5,6,7,9,11,13,15,17,19,21}{0.15}
  \end{tikzpicture}
\end{frame}
\end{document}

我正在尝试使用 \pgfmathparse 来评估我想要在矩阵第 i 行中绘制的位的位置。有一个测试 2>1 始终为真，即：它始终绘制位。如果我将该测试更改为 \mycol>0，Latex 不会接受它。我只想在矩阵边界内绘制位。

我找到了一个解决方案，上面有注释。我评估列的表达式，进行比较，并根据测试是在矩阵内部还是外部返回字符串 gtz 或 ltz。然后我在 \ifthenelse 中测试该字符串并相应地绘制或不绘制位。然而，我觉得这是一种次优的方法，需要定义一个中间字符串来确定测试的有效性。我如何评估表达式 \i*#4-\j 并根据此表达式的范围排版不同的代码？

Answer 1

据我所知，

  \foreach \i in {1,2,...,#3} { % Iterate over the rows
    \foreach \j in {#5} {
      \pgfmathparse{\i*#4-\j > -1 ? 1 : 0}
      \ifthenelse {\pgfmathresult>0} 
        {% If we are inside the boundaries of the matrix print the bit
         \node[minimum width=#6cm,minimum height=#6cm,fill=black,inner sep=0] 
           at ($#1+#6*(\i*#4-\j+1,-\i)$) {};
        }                                 
        {}% otherwise do nothing

    }
  }

可以。您尝试的测试的问题是\pgfmathparse{\i*#4-\j}变成\pgfmathresult了一个带有小数部分的数字，不适合 的数字测试\ifthenelse。

Answer 2

我使用简单\ifnum的\numexpr：

\documentclass{beamer}
\usepackage{tikz}
\usetikzlibrary{calc}
\newcommand{\LStepMatrix}[6] {%
    % #1 origin:  top left corner of the matrix
    % #2: N number of cols in the matrix
    % #3: M number of rows in the matrix
    % #4: L size of the step
    % #5: The numeric sequence used for this matrix
    % #6: scale factor for the figure
    % Draw the zeros
    \foreach \i in {1,2,...,#2} {
        \foreach \j in {1,2,...,#3} {
            \node[minimum width=#6cm,minimum height=#6cm,draw,inner sep=0] at 
             ($#1+#6*(\i,-\j)$) {};
        }%
    }%
    % Draw the ones in the positions of the numeric sequence
    \foreach \i in {1,2,...,#3} { % Iterate over the rows
        \foreach \j in {#5} {
             \ifnum \numexpr\i*#4-\j > -1 
               \node[minimum width=#6cm,minimum height=#6cm,fill=black,inner sep=0] at ($#1+#6*(\i*#4-\j+1,-\i)$) {};
             \fi 
             \ifnum \numexpr\i*#4-\j > 1 
                \node[minimum width=#6cm,minimum height=#6cm,fill=black,inner sep=0] at ($#1+#6*(\i*#4-\j+1,-\i)$) {};
             \fi
        }%
    }%
}

\begin{document}
    \begin{frame}{}
        \begin{tikzpicture}
        \LStepMatrix{(0.5,7)}{64}{8}{7}{1,2,3,4,5,6,7,9,11,13,15,17,19,21}{0.15}
        \end{tikzpicture}
    \end{frame}
\end{document}