Am 18.04.2015 um 18:55 schrieb Otared Kavian
: Hi everyone,
In the example below I define a macro which chooses at random a name from a list of names. But I wonder whether this can be done in a more clever way without using a numerical macro created with math.random in Lua. The shortcoming of the macro below is that before hand I must know the nomber of elements in the list of names (for instance 5 in the example below), while it may happen that I need to create as many as random names that there are elements in the list, but sometimes I don’t know what is this number.
Thanks for any insight and help. Best regards: OK %%%% begin random-names.tex \setuprandomize[2015] % set a seed
\starttext
\startluacode Name = {'F', 'G', 'u', 'v', 'W'} \stopluacode
\define[3]\RandomName{% \setevalue{Named#1}{\ctxlua{tex.print(math.random(#2,#3))}}} \define\RandomFunctionName{\ctxlua{tex.print(Name[\NamedFunctionNumber])}}
\dorecurse{10}{\RandomName{FunctionNumber}{1}{5}% Give an example of a function $\RandomFunctionName : {\Bbb R} \longrightarrow {\Bbb R}$ which has a derivative only at the origin, and such that $\RandomFunctionName(0) = 1$.\par \hairline\par}
\stoptext %%%% begin random-names.tex
You can access the size of your Name table with #Name but have to replace # with \letterhash when you use it in a TeX command because # is already taken for the TeX arguments. \starttext \startluacode Name = {'F', 'G', 'u', 'v', 'W'} \stopluacode \define\RandomFunctionName {\startlua local listsize = \letterhash Name ; local randomvalue = math.random(1,listsize) ; context(Name[randomvalue]) \stoplua} \dorecurse{10}{Give an example of a function $\RandomFunctionName : {\Bbb R} \longrightarrow {\Bbb R}$ which has a derivative only at the origin, and such that $\RandomFunctionName(0) = 1$.\par \hairline\par} \stoptext Wolfgang