Re: matrix problem

• To: mathgroup at smc.vnet.net
• Subject: [mg107134] Re: matrix problem
• From: dr DanW <dmaxwarren at gmail.com>
• Date: Wed, 3 Feb 2010 06:13:18 -0500 (EST)
• References: <hk93ca\$f68\$1@smc.vnet.net>

```Here is the most straightforward way to do this:

m = {{1, 2}, {3, 4}, {5, 6}, {7, 8}};
n = {{1, 8}, {5, 0}, {9, 7}};

m1 = Select[m, Function[x, MemberQ[Union[ n[[All,1]] ], x[[1]] ] ] ]

n1 = Select[n, Function[x, MemberQ[Union[ m[[All,1]] ], x[[1]] ] ] ]

First, avoid using capitals for variable names; N is a function name
in Mathematica.  The rest of this is a lot to swallow in one go, so
lets break it down.

n[[All,1]]

is a list of that is the first element of all rows, also known as the
first column

Union[ n[[All,1]] ]

is a list of all the unique members of that list.  Unnecessary in this
case, but good programming practice.

MemberQ[Union[ m[[All,1]] ], x[[1]] ]

tests to see if the first member of x is a member of that list

Function[x, MemberQ[Union[ m[[All,1]] ], x[[1]] ] ]

makes a function out of it, which takes one parameter, x.  Finally

Select[m, Function[x, MemberQ[Union[ n[[All,1]] ], x[[1]] ] ] ]

selects all elements out of m which match the criterion as laid out in
the Function.

Mathematica eats this kind of thing for lunch.  Study this example,
because once you get used to this kind of thinking, you can really
exploit the power of Mathematica.

Daniel

```

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