Having trouble with TikZ — For Loop and Conditionals

Question

Essentially what I am trying to achieve is a grid of 12 circles with three rows and four columns. All I would like is for the two middle columns to be red while the other two columns are black. Here is my current attempt, which gives me all red circles no matter what I do.

\begin{tikzpicture}

\foreach \x in {0,...,3}{
    \foreach \y in {0,...,2}{
        \ifthenelse{\x in {1,2}}{
            \draw[red,fill=red] (\x,\y) circle (0.4);
        }{
            \draw (\x,\y) circle (0.4);
        }
    }
}

\end{tikzpicture}

Current Result with Above TikZ Code:

Answer 1

You already know matrix dimensions, only twelve nodes. They are easy to manage. Then you can use a matrix and forget foreach and ifthenelse problems ;-)

\documentclass[border=1mm, tikz]{standalone}
\usetikzlibrary{matrix}
\begin{document}
\begin{tikzpicture}
\matrix[matrix of nodes, nodes in empty cells, 
    row sep=2mm, column sep=2mm,
    nodes={circle, draw, minimum width=8mm},
    column 2/.style={nodes={red, fill=red}},
    column 3/.style={nodes={red, fill=red}},
    ]
    { &&&\\&&&\\&&&\\};
\end{tikzpicture}
\end{document}

Answer 2

As said by @greg in the comment, there is no such a thing as a \x in {1,2} test for ifthenelse.

After reading the documentation for the package xifthen, you can:

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usepackage{xifthen}
\begin{document}
\begin{tikzpicture}
\foreach \x in {0,...,3}{
    \foreach \y in {0,...,2}{
        \ifthenelse{\x > 0 \AND \x < 3}{
            \draw[red,fill=red] (\x,\y) circle (0.4);
        }{
            \draw (\x,\y) circle (0.4);
        }
    }
}
\end{tikzpicture}
\end{document}

which results in:

Answer 3

Just for the pleasure of using the modulo function.

\documentclass[tikz,border=5mm]{standalone}
\usepackage{ifthen}

\begin{document}
\begin{tikzpicture}

\foreach \x in {0,...,3}{
    \foreach \y [evaluate =\x as \coloring using {int(Mod(\x,3))}]in {0,...,2}{
     \ifthenelse{\coloring = 0}{
            \draw (\x,\y) circle (0.4);
        }{
            \draw[red,fill=red] (\x,\y) circle (0.4);
        }
    }
}

\end{tikzpicture}
\end{document}

Answer 4

I wouldn't use ifthen. A pgf membership test can be used to implement your condition \x in {1,2},

\documentclass[tikz,border=3mm]{standalone}
\makeatletter
\pgfmathdeclarefunction{memberQ}{2}{%
  \begingroup%
    \edef\pgfutil@tmpb{0}%
    \edef\pgfutil@tmpa{#2}%
    \expandafter\pgfmath@member@i\pgfutil@firstofone#1\pgfmath@token@stop
    \edef\pgfmathresult{\pgfutil@tmpb}%
    \pgfmath@smuggleone\pgfmathresult%
  \endgroup}
\def\pgfmath@member@i#1{%
    \ifx\pgfmath@token@stop#1%
    \else
      \ifnum#1=\pgfutil@tmpa\relax%
      \gdef\pgfutil@tmpb{1}%
      %\typeout{#1=\pgfutil@tmpa}
      \fi%
      \expandafter\pgfmath@member@i
    \fi}
\makeatother
\begin{document}
\begin{tikzpicture}

\foreach \x [evaluate=\x as \isMember using {int(memberQ({1,2},\x))}] in {0,...,3}{
    \foreach \y in {0,...,2}{
        \draw \ifnum\isMember=1
        [red,fill=red]
        \fi (\x,\y) circle[radius=0.4cm];
    }
}
\end{tikzpicture}
\end{document}

This membership test works unless the last entry of the list has a substructure. Similar limitations apply to the dim function, which is built in pgf, but not mentioned in the manual, presumably for that reason. However, as long as you have ordinary lists, both memberQ and dim do work properly. (Proposed alternatives with a much more extensive codes also have drawbacks, sometimes even more severe, and I wish that those who propose them would mention them, too.)

As you can see, an ordinary \ifnum can be built into the path.

\draw \ifnum\itest=1
    [red,fill=red]
    \fi (\x,\y) circle[radius=0.4cm];

You then only have to draw one path. You can also use the conventional ifthenelse instead of the memberQ function.

\documentclass[tikz,border=3mm]{standalone}

\begin{document}
\begin{tikzpicture}

\foreach \x [evaluate=\x as \itest using {int(ifthenelse(\x>0 && \x <3,1,0))}] in {0,...,3}{
    \foreach \y in {0,...,2}{
        \draw \ifnum\itest=1
        [red,fill=red]
        \fi (\x,\y) circle[radius=0.4cm];
    }
}
\end{tikzpicture}
\end{document}

In this case you do not even have to use evaluate, you can simply use the fact that \numexpr truncates the result,

\documentclass[tikz,border=3mm]{standalone}

\begin{document}
\begin{tikzpicture}

\foreach \x  in {0,...,3}{
    \foreach \y in {0,...,2}{
        \draw \ifnum\the\numexpr\x/2\relax=1
        [red,fill=red]
        \fi (\x,\y) circle[radius=0.4cm];
    }
}
\end{tikzpicture}
\end{document}

Answer 5

There is no such test for \ifthenelse. However, you can use a different implementation, see https://tex.stackexchange.com/a/467527/4427

\documentclass{article}
\usepackage{xparse}
\usepackage{tikz} % for the application

\ExplSyntaxOn
\NewExpandableDocumentCommand{\xifthenelse}{mmm}
 {
  \bool_if:nTF { #1 } { #2 } { #3 }
 }

\cs_new_eq:NN \numtest     \int_compare_p:n
\cs_new_eq:NN \oddtest     \int_if_odd_p:n
\cs_new_eq:NN \fptest      \fp_compare_p:n
\cs_new_eq:NN \dimtest     \dim_compare_p:n
\cs_new_eq:NN \deftest     \cs_if_exist_p:N
\cs_new_eq:NN \namedeftest \cs_if_exist_p:c
\cs_new_eq:NN \eqdeftest   \token_if_eq_meaning_p:NN
\cs_new_eq:NN \streqtest   \str_if_eq_p:ee
\cs_new_eq:NN \emptytest   \tl_if_blank_p:n
\prg_new_conditional:Nnn \xxifthen_legacy_conditional:n { p,T,F,TF }
 {
  \use:c { if#1 } \prg_return_true: \else: \prg_return_false: \fi:
 }
\cs_new_eq:NN \boolean \xxifthen_legacy_conditional_p:n
\ExplSyntaxOff

\begin{document}

\begin{tikzpicture}

\foreach \x in {0,...,3}{
    \foreach \y in {0,...,2}{
        \xifthenelse{\numtest{1 <= \x <= 2}}{
            \draw[red,fill=red] (\x,\y) circle (0.4);
        }{
            \draw (\x,\y) circle (0.4);
        }
    }
}

\end{tikzpicture}

\end{document}