
我創建了一個表,其中有四列和十五行。然而,有點點,角度欄:0,10,20,30,40,.... 120 沒有顯示。
documentclass[%
preprint,
doublecolumn
%superscriptaddress,
%groupedaddress,
%unsortedaddress,
%runinaddress,
%frontmatterverbose,
%preprint
%preprintnumbers,
%nofootinbib,
%nobibnotes,
%bibnotes,
amsmath,amssymb,
aps,
%pra,
prc,
%rmp,
%prstab,
%prstper,
%floatfix,
]{revtex4-2}
\usepackage{anyfontsize}
\usepackage{float}
\usepackage{mathtools, nccmath}
\usepackage{graphicx}% Include figure files
\usepackage{tabularx}
\usepackage{dcolumn}% Align table columns on decimal point
\usepackage{bm}% bold math
%\usepackage{hyperref}% add hypertext capabilities
\usepackage[mathlines]{lineno}% Enable numbering of text and display math
\usepackage{amsmath}
\linenumbers\relax % Commence numbering lines
%\usepackage{latexsym}
%\usepackage[showframe,%Uncomment any one of the following lines to test
%%scale=0.7, marginratio={1:1, 2:3}, ignoreall,% default settings
%%text={7in,10in},centering,
%%margin=1.5in,
%%total={6.5in,8.75in}, top=1.2in, left=0.9in, includefoot,
%%height=10in,a5paper,hmargin={3cm,0.8in},
%]{geometry}
\begin{document}
\begin{table}
\caption{This table contains Cobalt's count rate:$R(\theta)$, count rate without background:$Y(\theta)$, and normalized count rate:$\frac{Y(\theta)}{Y(0)}$ their uncertainties. To obtain $R(\theta)$, divide "C" from Table.~\ref{Co_Count} by measured time, and its corresponding uncertainty:$\sigma R(\theta)$ is $\sigma C$ over time. For $\Y(\theta)$, subtract background count rate from$R(\theta)$. Its corresponding uncertainty is calculated by using Eq.~(\ref{eq:two}). Normalized count rate is divide $Y(\theta)$ by Y(0). Its uncertainty is from Eq.~(\ref{eq:one})}
\begin{ruledtabular}
\begin{tabular}{lcrr}
\textrm{$\theta$}& \textrm{$R(\theta)\pm \sigma R(\theta)$ }& \textrm{$Y(\theta) \pm \sigma {Y(\theta)}$} & \textrm{$\frac{Y(\theta)}{Y(0)}$ $\pm \sigma$ $\frac{Y(\theta)}{Y(0)}$ }\\
\textrm{($^{\circ}$)}&\textrm{$\frac{Count}{second}$}&\texrm{$\frac{Count}{second}$}& \\
\colrule
0& 2.970$\pm$0.070 &2.967$\pm$0.070&1.000$\pm$0.033
10&2.774$\pm$0.068 & 2.771$\pm$0.068&0.934$\pm$0.032
20&2.800$\pm$0.068 & 2.797$\pm$0.068&0.943$\pm$0.032
30&2.526$\pm$0.065 & 2.523$\pm$0.065&0.850$\pm$0.030
40&2.401$\pm$0.063 & 2.398$\pm$0.063&0.808$\pm$0.029
50&2.399$\pm$0.063 & 2.396$\pm$0.063&0.808$\pm$0.029
60&2.066$\pm$0.059 & 2.063$\pm$0.059&0.695$\pm$0.026
70&2.174$\pm$0.060 & 2.171$\pm$0.060&0.732$\pm$0.027
80&2.161$\pm$0.060 & 2.157$\pm$0.060&0.727$\pm$0.027
90&2.091$\pm$0.059 & 2.088$\pm$0.059&0.704$\pm$0.026
100&2.187$\pm$0.060 & 2.184$\pm$0.060&0.736$\pm$0.027
110&2.272$\pm$0.062 & 2.269$\pm$0.062&0.765$\pm$0.033
120&2.181$\pm$0.060 & 2.177$\pm$0.060&0.734$\pm$0.027
\end{tabular}
\end{ruledtabular}
\end{table}
\end{document)
先感謝您。能解釋一下lcrr是什麼意思嗎?
答案1
\documentclass{article}
\usepackage[tmargin=1cm]{geometry}
\usepackage{tabulary,booktabs}
\begin{document}
\begin{table*}
\centering
\begin{tabular}{cccc}
\toprule
\textrm{$\theta$}& \textrm{$R(\theta)\pm \sigma R(\theta)$ }& \textrm{$Y(\theta) \pm \sigma {Y(\theta)}$} & \textrm{$\frac{Y(\theta)}{Y(0)}$ $\pm \sigma$ $\frac{Y(\theta)}{Y(0)}$ }\\
\textrm{($^{\circ}$)}&\textrm{$\frac{Count}{second}$}&\texrm{$\frac{Count}{second}$}& \\
\midrule
0& 2.970$\pm$0.070 &2.967$\pm$0.070&1.000$\pm$0.033\\
10&2.774$\pm$0.068 & 2.771$\pm$0.068&0.934$\pm$0.032\\
20&2.800$\pm$0.068 & 2.797$\pm$0.068&0.943$\pm$0.032\\
30&2.526$\pm$0.065 & 2.523$\pm$0.065&0.850$\pm$0.030\\
40&2.401$\pm$0.063 & 2.398$\pm$0.063&0.808$\pm$0.029\\
50&2.399$\pm$0.063 & 2.396$\pm$0.063&0.808$\pm$0.029\\
60&2.066$\pm$0.059 & 2.063$\pm$0.059&0.695$\pm$0.026\\
70&2.174$\pm$0.060 & 2.171$\pm$0.060&0.732$\pm$0.027\\
80&2.161$\pm$0.060 & 2.157$\pm$0.060&0.727$\pm$0.027\\
90&2.091$\pm$0.059 & 2.088$\pm$0.059&0.704$\pm$0.026\\
100&2.187$\pm$0.060 & 2.184$\pm$0.060&0.736$\pm$0.027\\
110&2.272$\pm$0.062 & 2.269$\pm$0.062&0.765$\pm$0.033\\
120&2.181$\pm$0.060 & 2.177$\pm$0.060&0.734$\pm$0.027\\
\bottomrule
\end{tabular}
\end{table*}
\end{document}
答案2
- 我會考慮我之前問題的答案,特別是如果它們包含解釋,如何編寫表格,我在哪裡犯了錯誤,以及它們是否解決了我的全部或至少部分(殘留)問題(桌子看起來很難看)
- 在這樣的表格中,我將使用所有可用的工具根據 SI(國際單位制,請參閱例如維基百科)標準,特別是如果它們為我的表提供更短且一致的代碼。這樣的工具是
siunitx
包含使用 SI 單位的優秀文件的軟體包
上表產生的 MWE(最小工作範例)為:
\documentclass[twocolumn]{revtex4-2}
\usepackage{nccmath, mathtools, amssymb}
\usepackage{makecell}
\usepackage{siunitx}
\begin{document}
\begin{table}
\caption{This table contains Cobalt's count rate: $R(\theta)$,
count rate without background: $Y(\theta)$, and normalized count rate:
$Y(\theta)/Y(0)$ their uncertainties. To obtain $R(\theta)$,
divide "C" from Table.~\ref{Co_Count} by measured time,
and its corresponding uncertainty: $\sigma R(\theta)$ is $\sigma C$ over time.
For $Y(\theta)$, subtract background count rate from $R(\theta)$.
Its corresponding uncertainty is calculated by using Eq.~(\ref{eq:two}).
Normalized count rate is divide $Y(\theta)$ by Y(0).
Its uncertainty is from Eq.~(\ref{eq:one})
}
\label{tab:cobalt}
\centering
\sisetup{table-format=1.3(2),
table-figures-uncertainty=6,
separate-uncertainty}
\setcellgapes{3pt}
\makegapedcells
\begin{tabular}{ S[table-format=3.0] @{\quad} *3{S} }
\colrule
{$\theta$}
& {$R(\theta) \pm \sigma R(\theta)$}
& {$Y(\theta) \pm \sigma Y(\theta)$}
& {$\mfrac{Y(\theta)}{Y(0)} \pm \sigma\mfrac{Y(\theta)}{Y(0)}$} \\
{\si{\degree}}
& {$\mfrac{\text{Count}}
{\text{second}}$}
& {$\mfrac{\text{Count}}
{\text{second}}$}
& \\
\colrule
0 & 2.970(70) & 2.967(70) & 1.000(33) \\
10 & 2.774(68) & 2.771(68) & 0.934(32) \\
20 & 2.800(68) & 2.797(68) & 0.943(32) \\
30 & 2.526(65) & 2.523(65) & 0.850(30) \\
40 & 2.401(63) & 2.398(63) & 0.808(29) \\
50 & 2.399(63) & 2.396(63) & 0.808(29) \\
60 & 2.066(59) & 2.063(59) & 0.695(26) \\
70 & 2.174(60) & 2.171(60) & 0.732(27) \\
80 & 2.161(60) & 2.157(60) & 0.727(27) \\
90 & 2.091(59) & 2.088(59) & 0.704(26) \\
100 & 2.187(60) & 2.184(60) & 0.736(27) \\
110 & 2.272(62) & 2.269(62) & 0.765(33) \\
120 & 2.181(60) & 2.177(60) & 0.734(27) \\
\colrule
\end{tabular}
\end{table}
\end{document}
筆記:
- 在 MWE 的 MWE 前導碼中,僅考慮與該表相關的套件
- 您的問題可以透過正確終止的表格行來解決。這一點在你之前問題的回答中已經強調過。
c
,l
,r
是列類型意義中心,左邊和正確的列內容的對齊。由他們指定表格。在你的情況下lcrr
意味著你的表有四列,第一列有左邊對齊單元格的內容,在第二個單元格的內容是居中在最後兩列中有正確的對齊的內容。如欲了解更多詳情,請參閱wiki LaTeX 表格。S
正如您所看到的,在建議的解決方案中,使用了列的類型而不是它們。它們能夠將單元格中的數字對齊到小數點,並簡單地寫入報告測量值的公差/不確定性。
答案3
一些觀察與評論:
你必須提供
\\
(“雙反斜線”)指令來指示應該出現換行符的位置。輸入填充中的換行符是不夠的。你問:“你能解釋一下什麼
lcrr
意思嗎?”我假設你指的是聲明\begin{tabular}{lcrr}
LaTeX 核心設定了多種列類型以供在
tabular
和array
環境中使用。其中l
,c
, 和r
, 分別代表我左對齊,C進入,並且r分別右對齊。就您的表格而言,我看不出使用
r
任何列的充分理由。我將用於c
數據列。請刪除諸如
\textrm
“包裝器”之類的無用程式碼。除了增加程式碼混亂之外,它們絕對沒有做任何事情。由於環境的全部內容
tabular
都處於數學模式,因此我不會使用tabular
環境。相反,使用array
環境。就目前情況而言,只有那些貪圖懲罰的人才會仔細閱讀你的表格。請始終記住,如果人們注意到您努力以有吸引力的方式組織您希望呈現的訊息,他們就更有可能關注您。
將此標準應用於整個
table
環境,我想說,以更有吸引力的方式組織圖例和表格材料將是一個好主意。當然,不要只是以長而曲折的標題的形式將圖例「傾倒」在讀者的頭上。相反,使標題簡短明快,例如,\caption{Count rates}
並像普通的運行文本一樣組織圖例:使用段落分隔符,並使用完整的句子。array
不要在( 或) 環境中顯示大量沒有結構的 13 行數字tabular
,而是在每第四行或第五行後面加上一些空白來提供一些視覺興趣。或者,使用S
列類型(由套件提供siunitx
)將第一列中的數字與其(隱式)小數標記對齊。
\documentclass[twocolumn, % the option is "twocolumn", NOT "doublecolumn"
amsmath,amssymb,aps,prc]{revtex4-2}
%% Simplified the preamble to include just the bare minimum needed:
\usepackage{booktabs} % for \toprule, \midrule, \bottomrule
\usepackage{siunitx} % for 'S' column type
\usepackage{ragged2e} % for '\justifying' command
\begin{document}
\begin{table}
\caption{Count rates}
\justifying
This table contains Cobalt's count rate, $R(\theta)$, the count rate without
background, $Y(\theta)$, and the normalized count rate, $Y(\theta)/Y(0)$,
along with their uncertainties.
To obtain $R(\theta)$, divide ``C'' from Table~\ref{Co_Count} by measured
time; its corresponding uncertainty, $\sigma R(\theta)$, is $\sigma C$ over
time. For $Y(\theta)$, subtract background count rate from $R(\theta)$; its
corresponding uncertainty is calculated by using Eq.~(\ref{eq:two}). The
normalized count rate is $Y(\theta)$ divided by~$Y(0)$; its uncertainty is
from Eq.~(\ref{eq:one}).
\medskip % insert a bit of vertical whitespace
\centering
$\begin{array}{@{} S[table-format=3.0] ccc @{}}
\toprule
{\theta} &
R(\theta)\pm \sigma R(\theta) &
Y(\theta) \pm \sigma Y(\theta) &
\frac{Y(\theta)}{Y(0)} \pm \sigma \frac{Y(\theta)}{Y(0)} \\[1ex]
{[^{\circ}]}
& \bigl[\frac{\text{Count}}{\text{second}}\bigr]
& \bigl[\frac{\text{Count}}{\text{second}}\bigr]
& \\
\midrule
0 & 2.970\pm0.070 & 2.967\pm0.070 & 1.000\pm0.033\\
10 & 2.774\pm0.068 & 2.771\pm0.068 & 0.934\pm0.032\\
20 & 2.800\pm0.068 & 2.797\pm0.068 & 0.943\pm0.032\\
30 & 2.526\pm0.065 & 2.523\pm0.065 & 0.850\pm0.030\\
40 & 2.401\pm0.063 & 2.398\pm0.063 & 0.808\pm0.029\\
\addlinespace
50 & 2.399\pm0.063 & 2.396\pm0.063 & 0.808\pm0.029\\
60 & 2.066\pm0.059 & 2.063\pm0.059 & 0.695\pm0.026\\
70 & 2.174\pm0.060 & 2.171\pm0.060 & 0.732\pm0.027\\
80 & 2.161\pm0.060 & 2.157\pm0.060 & 0.727\pm0.027\\
\addlinespace
90 & 2.091\pm0.059 & 2.088\pm0.059 & 0.704\pm0.026\\
100 & 2.187\pm0.060 & 2.184\pm0.060 & 0.736\pm0.027\\
110 & 2.272\pm0.062 & 2.269\pm0.062 & 0.765\pm0.033\\
120 & 2.181\pm0.060 & 2.177\pm0.060 & 0.734\pm0.027\\
\bottomrule
\end{array}$
\end{table}
\end{document}