在 alignat* 环境中纠正间距

Question

给出以下代码：

\begin{alignat*}{3}
    h(0) & = g_s(0) - g_s(0 - 1) = 0.25 & -  0.00 & =  \nicefrac{1}{4}\\
    h(1) & = g_s(1) - g_s(1 - 1) = 0.50 & -  0.25 & =  \nicefrac{1}{4}\\
    h(2) & = g_s(2) - g_s(2 - 1) = 0.75 & -  0.50 & =  \nicefrac{1}{4}\\
    h(3) & = g_s(3) - g_s(3 - 1) = 1.00 & -  0.75 & =  \nicefrac{1}{4}\\
    h(4) & = g_s(4) - g_s(4 - 1) = 1.00 & -  1.00 & = 0
\end{alignat*}

输出 但是输出结果中每个数字之间的间距应该与第一个方程相同h(x) = g_s(x) - g_s(x - 1)。（换句话说，减号之间的间距太小。）我该如何获得这个结果？

Answer 1

两种方式取决于你如何看待它{}-强制二进制 - 但你可以将形式视为只有第二组和第三组的右侧，所以如果你用标记空的 lhs，&&那么-在右侧的开头就会得到通常的空间。

\documentclass{article}

\usepackage{amsmath,nicefrac}

\begin{document}

aaa
\begin{alignat*}{3}
    h(0) & = g_s(0) - g_s(0 - 1) = 0.25 & -  0.00 & =  \nicefrac{1}{4}\\
    h(1) & = g_s(1) - g_s(1 - 1) = 0.50 & -  0.25 & =  \nicefrac{1}{4}\\
    h(2) & = g_s(2) - g_s(2 - 1) = 0.75 & -  0.50 & =  \nicefrac{1}{4}\\
    h(3) & = g_s(3) - g_s(3 - 1) = 1.00 & -  0.75 & =  \nicefrac{1}{4}\\
    h(4) & = g_s(4) - g_s(4 - 1) = 1.00 & -  1.00 & = 0
\end{alignat*}

bbb
\begin{alignat*}{3}
    h(0) & = g_s(0) - g_s(0 - 1) = 0.25 & {}-  0.00 & =  \nicefrac{1}{4}\\
    h(1) & = g_s(1) - g_s(1 - 1) = 0.50 & {}-  0.25 & =  \nicefrac{1}{4}\\
    h(2) & = g_s(2) - g_s(2 - 1) = 0.75 & {}-  0.50 & =  \nicefrac{1}{4}\\
    h(3) & = g_s(3) - g_s(3 - 1) = 1.00 & {}-  0.75 & =  \nicefrac{1}{4}\\
    h(4) & = g_s(4) - g_s(4 - 1) = 1.00 & {}-  1.00 & = 0
\end{alignat*}

ccc
\begin{alignat*}{3}
    h(0) & = g_s(0) - g_s(0 - 1) = 0.25 && -  0.00 && =  \nicefrac{1}{4}\\
    h(1) & = g_s(1) - g_s(1 - 1) = 0.50 && -  0.25 && =  \nicefrac{1}{4}\\
    h(2) & = g_s(2) - g_s(2 - 1) = 0.75 && -  0.50 && =  \nicefrac{1}{4}\\
    h(3) & = g_s(3) - g_s(3 - 1) = 1.00 && -  0.75 && =  \nicefrac{1}{4}\\
    h(4) & = g_s(4) - g_s(4 - 1) = 1.00 && -  1.00 && = 0
\end{alignat*}


\end{document}

Answer 2

使用{}=解决间距问题：

\documentclass{article}

\usepackage{amsmath}
\usepackage{nicefrac}

\begin{document}

\begin{alignat*}{3}
h(0) & = g_s(0) - g_s(0 - 1) = 0.25 & {}- 0.00 & = \nicefrac{1}{4} \\
h(1) & = g_s(1) - g_s(1 - 1) = 0.50 & {}- 0.25 & = \nicefrac{1}{4} \\
h(2) & = g_s(2) - g_s(2 - 1) = 0.75 & {}- 0.50 & = \nicefrac{1}{4} \\
h(3) & = g_s(3) - g_s(3 - 1) = 1.00 & {}- 0.75 & = \nicefrac{1}{4} \\
h(4) & = g_s(4) - g_s(4 - 1) = 1.00 & {}- 1.00 & = 0
\end{alignat*}

\end{document}

（我没有nicefrac安装该软件包。我建议使用\tfrac而不是\nicefrac。大多数数学家更喜欢水平分数线。）

编辑

以下是另一个从内容结构上来说可能更有意义的版本：

\documentclass{article}

\usepackage{amsmath}
\usepackage{nicefrac}

\begin{document}

\begin{alignat*}{3}
h(0) & = g_s(0) - g_s(0 - 1) && = 0.25 - 0.00 && = \nicefrac{1}{4} \\
h(1) & = g_s(1) - g_s(1 - 1) && = 0.50 - 0.25 && = \nicefrac{1}{4} \\
h(2) & = g_s(2) - g_s(2 - 1) && = 0.75 - 0.50 && = \nicefrac{1}{4} \\
h(3) & = g_s(3) - g_s(3 - 1) && = 1.00 - 0.75 && = \nicefrac{1}{4} \\
h(4) & = g_s(4) - g_s(4 - 1) && = 1.00 - 1.00 && = 0
\end{alignat*}

\end{document}

大卫·卡莱尔显示了另一个将分隔符放置在他的回答回答这个问题。所有版本都有优点和缺点，这取决于数字的格式和其他内容，哪一个最好。

Answer 3

一种tabstackengine方法。在这里，\TABbinary强制将前导负片视为二进制，允许\setstackalingap{}消除对齐组之间的水平间隙，并\setsatckgap{L}{}允许设置基线跳跃。

\documentclass{article}
\usepackage{amsmath,nicefrac,tabstackengine}
\TABstackMath
\begin{document}
\[
\TABbinary
\setstackaligngap{0pt}
\setstackgap{L}{1.2\baselineskip}
\alignCenterstack{
h(0) & = g_s(0) - g_s(0 - 1) = 0.25 & - 0.00 & = \nicefrac{1}{4} \\
h(1) & = g_s(1) - g_s(1 - 1) = 0.50 & - 0.25 & = \nicefrac{1}{4} \\
h(2) & = g_s(2) - g_s(2 - 1) = 0.75 & - 0.50 & = \nicefrac{1}{4} \\
h(3) & = g_s(3) - g_s(3 - 1) = 1.00 & - 0.75 & = \nicefrac{1}{4} \\
h(4) & = g_s(4) - g_s(4 - 1) = 1.00 & - 1.00 & = 0
}
\]
\end{document}