I prefer a large number be shown in three digit groups. For example 1234567 be shown like 1 234 567.
Is there a way to make this automatic in LaTeX such that wherever I insert 1234567, it is automatically shown as 1 234 567 in the output?
I prefer a large number be shown in three digit groups. For example 1234567 be shown like 1 234 567.
Is there a way to make this automatic in LaTeX such that wherever I insert 1234567, it is automatically shown as 1 234 567 in the output?
The packages numprint and siunitx can do this (here I show numprint)
\documentclass{article}
\usepackage{numprint}
\begin{document}
\numprint{1234567}
\numprint{123456789}
\end{document}
Here's a LuaLaTeX-based solution, which adds thinspaces automatically to the integer parts of numbers that contain between 5 and 12 digits.
Some comments:
Numbers with exactly four digits do not get a thinspace inserted after the first (left-most) digit, in keeping with the comment @egreg posted. If you do want a thinspace separator for numbers with exactly four digits, change the final string.gsub instruction line from
text = string.gsub ( text , "(%d?%d%d)(%d%d%d)", "%1\\,%2" )
to
text = string.gsub ( text , "(%d?%d?%d)(%d%d%d)", "%1\\,%2" )
No thinspaces are inserted in the decimal portions of any long numbers; only the integer portions are processed.
The code keeps track of whether or not we are inside a verbatim or Verbatim environment. If we're inside such an environment, processing is suspended.
% !TEX TS-program = lualatex
\documentclass{article}
\usepackage{luacode}
\begin{luacode}
ok_to_process = true
function ez_read ( line )
if string.find ( line , "\\begin{[vV]erbatim}" ) then
ok_to_process = false
return line
elseif string.find ( line , "\\end{[vV]erbatim}" ) then
ok_to_process = true
return line
else
if ok_to_process then
x = line:gsub ( "([%.]-)(%d+)", function ( back, text )
if back ~= "" then
return back..text
end
text = string.gsub ( text , "(%d?%d?%d)(%d%d%d)(%d%d%d)(%d%d%d)",
"%1\\,%2\\,%3\\,%4" )
text = string.gsub ( text , "(%d?%d?%d)(%d%d%d)(%d%d%d)",
"%1\\,%2\\,%3" )
text = string.gsub ( text , "(%d?%d%d)(%d%d%d)", "%1\\,%2" )
return text
end)
return x
end
end
end
luatexbase.add_to_callback( "process_input_buffer", ez_read, "ez_read")
\end{luacode}
\begin{document}
123456789012, 123456789, 123456, 12345
But: 1234 --- no thinspace after ``1''
123456.123456789, 0.12345, 123456
\end{document}
automaticallyword there. I just tried to improve it grammatically. The answers are good, btw they don't solve my problem. So I haven't accepted them yet. – Minimus Heximus Oct 24 '15 at 08:15123456.789012? Please advise. – Mico Oct 25 '15 at 05:04