Question

Mathematica 如何提取指定元素的表？

如何提取指定元素的表？
例如:
输入; s1={{a,b,x,y,c},{c,d,e,x},{x,a,y,d,z},{z,a,b}}
输出; s2={{a,b,x,y,c},{x,a,y,d,z}}
s3={{a,b,x,y,c},{c,d,e,x},{x,a,y,d,z}}
s4={{c,d,e,x},{z,a,b}}
s5={{z,a,b}}
即: s2------含x与y s3------含x 或y s4------不含x与y s5------不含x 或y

Answer 1

s1 = {{a, b, x, y, c}, {c, d, e, x}, {x, a, y, d, z}, {z, a, b}};
s2 = Select[s1, MemberQ[#, x] && MemberQ[#, y] &];
s3 = Select[s1, MemberQ[#, x] || MemberQ[#, y] &];
s4 = Complement[s1, s2];
s5 = Complement[s1, s3];

追问

你好,可否把x与y作为一个表来处理，例如输出s3含x与y:

输入: s1={{a,b,x,y,c},{c,d,e,x},{x,a,y,d,z},{z,a,b}};
s2={ x,y};
输出: s3={{a,b,x,y,c},{x,a,y,d,z}}

追答

In[53]:= s1 = {{a, b, x, y, c}, {c, d, e, x}, {x, a, y, d, z}, {z, a,
b}};
s2 = {x, y};
s3 = Intersection @@ (Cases[s1, {___, #, ___}] & /@ s2);
s4 = Union @@ (Cases[s1, {___, #, ___}] & /@ s2);
{s3, s4}

Out[57]= {{{a, b, x, y, c}, {x, a, y, d, z}}, {{c, d, e, x}, {a, b, x,
y, c}, {x, a, y, d, z}}}