忽略命令中的空格

Question

我正在使用setspace，\spacing。我有许多函数可以创建数组，当我使用\begin{spacing} {1.5}这些特殊函数时，我会得到空格，w \documentclass{article}

\usepackage{cancel,amsmath}
\usepackage{setspace}
\newcommand{\stackBelowCancel}[2]{\begin{array}[t]{@{}c@{}}\cancel{#1}\\#2\end{array}} % stacking BELOW canceling number(NOT FRACTION)


\begin{document}
    \begin{flushleft}
        \begin{spacing}{1.5}
        1.  $\dfrac{1}{3\cancel{0}}\cdot\dfrac{1\cancel{0}}{\stackBelowCancel{27}{3}}=\dfrac{1}{9}$ (Have spacing)
        \end{spacing}
        2. $\dfrac{1}{3\cancel{0}}\cdot\dfrac{1\cancel{0}}{\stackBelowCancel{27}{3}}=\dfrac{1}{9}$ (Not in spacing)
    \end{flushleft}

\end{document}

我想实现第二个行（2.）在间距环境中。如何在命令中忽略第 1 行中的间距？

Answer 1

array这是伴随的普遍问题setspace，可以通过重置\arraystretch来抵消\baselinestretch。

\documentclass{article}
\usepackage{cancel,amsmath}
\usepackage{setspace}
\usepackage{xfp}

\makeatletter
\renewcommand{\setstretch}[1]{%
  \def\baselinestretch{#1}\@currsize
  \edef\arraystretch{\fpeval{1/(#1)}}%
}
\makeatother

\newcommand{\stackBelowCancel}[2]{%
  \begin{array}[t]{@{}c@{}}\cancel{#1}\\#2\end{array}%
} % stacking BELOW canceling number(NOT FRACTION)

\begin{document}

\begin{spacing}{1.5}
1.  $\dfrac{1}{3\cancel{0}}\cdot\dfrac{1\cancel{0}}{\stackBelowCancel{27}{3}}=
\dfrac{1}{9}$ (Have spacing)
\end{spacing}

\begin{doublespacing}
2. $\dfrac{1}{3\cancel{0}}\cdot\dfrac{1\cancel{0}}{\stackBelowCancel{27}{3}}=
\dfrac{1}{9}$ (Have spacing)
\end{doublespacing}

3. $\dfrac{1}{3\cancel{0}}\cdot\dfrac{1\cancel{0}}{\stackBelowCancel{27}{3}}=
\dfrac{1}{9}$ (Not in spacing)

\end{document}

顺便说一句，这些简化是错误的。

Answer 2

忘记array...只是\stackunder。请注意，默认是文本模式堆叠。如果您想要数学模式，\stackMath序言中的声明将执行此操作。

\documentclass{article}
\usepackage{cancel,amsmath,stackengine}
\usepackage{setspace}
\newcommand{\stackBelowCancel}[2]{\stackunder[4pt]{\cancel{#1}}{#2}} % stacking BELOW canceling number(NOT FRACTION)


\begin{document}
    \begin{flushleft}
        \begin{spacing}{1.5}
        1.  $\dfrac{1}{3\cancel{0}}\cdot\dfrac{1\cancel{0}}{\stackBelowCancel{27}{3}}=\dfrac{1}{9}$ (Have spacing)
        \end{spacing}
        2. $\dfrac{1}{3\cancel{0}}\cdot\dfrac{1\cancel{0}}{\stackBelowCancel{27}{3}}=\dfrac{1}{9}$ (Not in spacing)
    \end{flushleft}

\end{document}

Answer 3

\underset也许可以代替array?来使用。

\documentclass{article}
\usepackage{cancel,amsmath}
\usepackage{setspace}

% always textstyle
\newcommand{\stackBelowCancel}[2]{\underset{\textstyle#2}{\cancel{#1}}}

% adapt to outer style
%\newcommand{\stackBelowCancel}[2]{\mathpalette\dostackBelowCancel{{#1}{#2}}}
%\newcommand{\dostackBelowCancel}[2]{\dostackBelowCancelIndeed{#1}#2}
%\newcommand{\dostackBelowCancelIndeed}[3]{\underset{#1#3}{\cancel{#2}}}

\begin{document}
    \begin{flushleft}
        \begin{spacing}{1.5}
        1.  $\dfrac{1}{3\cancel{0}}\cdot\dfrac{1\cancel{0}}{\stackBelowCancel{27}{3}}=\dfrac{1}{9}$ (Have spacing)
        \end{spacing}
        2. $\dfrac{1}{3\cancel{0}}\cdot\dfrac{1\cancel{0}}{\stackBelowCancel{27}{3}}=\dfrac{1}{9}$ (Not in spacing)
    \end{flushleft}

\end{document}