我有代码:
\begin{flalign}
&f_1(y_i; \boldsymbol{\theta}_1) = \lambda e^{-\lambda y_i} \\
&l_1(\boldsymbol{\theta}_1; \textbf{y}, \textbf{x}) = `r num.not.censored`
\log{(\lambda)} - \lambda \sum_{i=1}^{`r num.data`} y_i \\
&f_2(y_i; x_i, \boldsymbol{\theta}_2) = e^{\beta_1 + \beta_2 x_i}
\exp{(-e^{\beta_1 + \beta_2 x_i}y_i)} \\
&l_2(\boldsymbol{\theta}_2; \textbf{y}, \textbf{x}) =
\sum_{i=1}^{`r num.not.censored`} (\beta_1 + \beta_2 x_i)
-\sum_{i=1}^{`r num.data`} (e^{\beta_1 + \beta_2 x_i}y_i) \\
&f_3(y_i; x_i, \boldsymbol{\theta}_3) = \frac{1}{\sqrt{2\pi\sigma^2}}
\exp{\bigg(\frac{-(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2\sigma^2}\bigg)} \\
&l_3(\boldsymbol{\theta}_3; \textbf{y}, \textbf{x}) = - 28
\log(2 \pi \sigma^2) - \sum_{i=1}^{`r num.not.censored`}
\bigg(\frac{(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2 \sigma^2}\bigg)
\space + \\
&\sum_{i=`r num.not.censored + 1`}^{`r num.data`} \log
\bigg(\frac{1}{2} - \frac{1}{2}\text{erf}\Big(\frac{y_i -
(\gamma_1 + \gamma_2 x_i)}{\sigma \sqrt{2}}\Big)\bigg)
\end{flalign}
其呈现形式为:
我怎样才能将第 7 行向右移动,以便它从第 6 行的等号开始?
我还想删除方程 (7) 标签,因为它仍然是方程 (6) 的一部分。
谢谢,
杰克
答案1
您似乎想要的是fleqn
选项,而不是flalign
:
\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{bm}
\DeclareMathOperator{\erf}{erf}
\begin{document}
\begin{gather}
f_1(y_i; \bm{\theta}_1) = \lambda e^{-\lambda y_i} \\
l_1(\bm{\theta}_1; \mathbf{y}, \mathbf{x}) = 56
\log{(\lambda)} - \lambda \sum_{i=1}^{75} y_i \\
f_2(y_i; x_i, \bm{\theta}_2) = e^{\beta_1 + \beta_2 x_i}
\exp{(-e^{\beta_1 + \beta_2 x_i}y_i)} \\
l_2(\bm{\theta}_2; \mathbf{y}, \mathbf{x}) =
\sum_{i=1}^{56} (\beta_1 + \beta_2 x_i)
-\sum_{i=1}^{75} (e^{\beta_1 + \beta_2 x_i}y_i) \\
f_3(y_i; x_i, \bm{\theta}_3) = \frac{1}{\sqrt{2\pi\sigma^2}}
\exp{\biggl(\frac{-(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2\sigma^2}\biggr)} \\
\begin{split}
l_3(\bm{\theta}_3; \mathbf{y}, \mathbf{x})
={}&{-}28\log(2 \pi \sigma^2) - \sum_{i=1}^{56}
\biggl(\frac{(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2 \sigma^2}\biggr)
\\
&+\sum_{i=57}^{75} \log
\biggl(\frac{1}{2} - \frac{1}{2}\erf\Bigl(\frac{y_i -
(\gamma_1 + \gamma_2 x_i)}{\sigma \sqrt{2}}\Bigr)\biggr)
\end{split}
\end{gather}
\end{document}
您可以轻松地将对齐方式更改为等号:
\documentclass{article}
\usepackage[fleqn]{amsmath}
\usepackage{bm}
\DeclareMathOperator{\erf}{erf}
\begin{document}
\begin{align}
f_1(y_i; \bm{\theta}_1) &= \lambda e^{-\lambda y_i} \\
l_1(\bm{\theta}_1; \mathbf{y}, \mathbf{x}) &= 56
\log{(\lambda)} - \lambda \sum_{i=1}^{75} y_i \\
f_2(y_i; x_i, \bm{\theta}_2) &= e^{\beta_1 + \beta_2 x_i}
\exp{(-e^{\beta_1 + \beta_2 x_i}y_i)} \\
l_2(\bm{\theta}_2; \mathbf{y}, \mathbf{x}) &=
\sum_{i=1}^{56} (\beta_1 + \beta_2 x_i)
-\sum_{i=1}^{75} (e^{\beta_1 + \beta_2 x_i}y_i) \\
f_3(y_i; x_i, \bm{\theta}_3) &= \frac{1}{\sqrt{2\pi\sigma^2}}
\exp{\biggl(\frac{-(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2\sigma^2}\biggr)} \\
\begin{split}
l_3(\bm{\theta}_3; \mathbf{y}, \mathbf{x})
&=-28\log(2 \pi \sigma^2) - \sum_{i=1}^{56}
\biggl(\frac{(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2 \sigma^2}\biggr)
\\
&\mathrel{\hphantom{=}}+\sum_{i=57}^{75} \log
\biggl(\frac{1}{2} - \frac{1}{2}\erf\Bigl(\frac{y_i -
(\gamma_1 + \gamma_2 x_i)}{\sigma \sqrt{2}}\Bigr)\biggr)
\end{split}
\end{align}
\end{document}
需要注意几点。
\boldsymbol
已被弃用,并且\bm
(与bm
包一起)是首选。所有
\textbf
命令都应该是\mathbf
(在斜体上下文中,例如定理陈述,字母将以斜体排版)。而
\text{erf}
应该定义一个合适的运算符\DeclareMathOperator
(理由与前面相同)。
答案2
您将对齐符号放置在&
了不该在的位置,这样就可以用来\notag
抑制数字。
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{flalign}
f_1(y_i; \boldsymbol{\theta}_1) & = \lambda e^{-\lambda y_i} \\
l_1(\boldsymbol{\theta}_1; \textbf{y}, \textbf{x}) & = `r num.not.censored`
\log{(\lambda)} - \lambda \sum_{i=1}^{`r num.data`} y_i \\
f_2(y_i; x_i, \boldsymbol{\theta}_2) & = e^{\beta_1 + \beta_2 x_i} \exp{(-
e^{\beta_1 + \beta_2 x_i}y_i)} \\
l_2(\boldsymbol{\theta}_2; \textbf{y}, \textbf{x}) & = \sum_{i=1}^{`r
num.not.censored`} (\beta_1 + \beta_2 x_i) - \sum_{i=1}^{`r num.data`}
(e^{\beta_1 + \beta_2 x_i}y_i) \\
f_3(y_i; x_i, \boldsymbol{\theta}_3) & = \frac{1}{\sqrt{2\pi\sigma^2}}
\exp{\bigg(\frac{-(y_i - (\gamma_1 + \gamma_2 x_i))^2}{2\sigma^2}\bigg)} \\
l_3(\boldsymbol{\theta}_3; \textbf{y}, \textbf{x}) & = - 28 \log(2 \pi
\sigma^2) - \sum_{i=1}^{`r num.not.censored`} \bigg(\frac{(y_i - (\gamma_1 +
\gamma_2 x_i))^2}{2 \sigma^2}\bigg)\space + \notag\\
&\sum_{i=`r num.not.censored
+ 1`}^{`r num.data`} \log \bigg(\frac{1}{2} - \frac{1}
{2}\text{erf}\Big(\frac{y_i - (\gamma_1 + \gamma_2 x_i)}{\sigma
\sqrt{2}}\Big)\bigg)
\end{flalign}
\end{document}
答案3
标准align
环境(其中所有方程式都与符号对齐=
)至少可以像flalign
环境一样工作。
请注意,我将所有 的实例替换为r num.not.censored
,56
并将 的所有实例替换r num.data
为75
。通过与您发布的屏幕截图进行交叉检查,这似乎是合理的。
\documentclass{article}
\usepackage{amsmath}
\DeclareMathOperator{\erf}{erf}
\begin{document}
\begin{align}
f_1(y_i; \boldsymbol{\theta}_1)
&= \lambda e^{-\lambda y_i} \\
l_1(\boldsymbol{\theta}_1; \mathbf{y}, \mathbf{x})
&= 56\log{(\lambda)} - \lambda \sum_{i=1}^{75} y_i \\
f_2(y_i; x_i, \boldsymbol{\theta}_2)
&= e^{\beta_1 + \beta_2 x_i} \exp{(-e^{\beta_1 + \beta_2 x_i}y_i)} \\
l_2(\boldsymbol{\theta}_2; \mathbf{y}, mathbf{x})
&= \sum_{i=1}^{56} (\beta_1 + \beta_2 x_i)
-\sum_{i=1}^{75} (e^{\beta_1 + \beta_2 x_i}y_i) \\
f_3(y_i; x_i, \boldsymbol{\theta}_3)
&= \frac{1}{\sqrt{2\pi\sigma^2}}
\exp{\biggl(\frac{-(y_i - (\gamma_1 +
\gamma_2 x_i))^2}{2\sigma^2}\biggr)} \\
l_3(\boldsymbol{\theta}_3; \mathbf{y}, \mathbf{x})
&= - 28 \log(2 \pi \sigma^2)
- \sum_{i=1}^{56} \biggl(\frac{(y_i -
(\gamma_1 + \gamma_2 x_i))^2}{2 \sigma^2}\biggr) \\
&\quad +\sum_{i=57}^{75} \log
\biggl(\frac{1}{2} - \frac{1}{2}\erf \Bigl(\frac{y_i -
(\gamma_1 + \gamma_2 x_i)}{\sigma \sqrt{2}}\Bigr) \biggr) \notag
\end{align}
\end{document}