Re: Lists: how to?

• To: mathgroup at smc.vnet.net
• Subject: [mg110016] Re: Lists: how to?
• Date: Sat, 29 May 2010 04:44:04 -0400 (EDT)

```Some possible solutions:

In[1]:= L = {{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 2, 6}, {1, 2, 7}, {1, 3,
4},
{1, 3, 5}, {1, 3, 6}, {1, 3, 7}, {1, 4, 5}, {1, 4, 6}, {1, 4, 7},
{1, 5, 6}, {1, 5, 7}, {1, 6, 7}, {2, 3, 4}, {2, 3, 5}, {2, 3, 6},
{2, 3, 7}, {2, 4, 5}, {2, 4, 6}, {2, 4, 7}, {2, 5, 6}, {2, 5, 7},
{2, 6, 7}, {3, 4, 5}, {3, 4, 6}, {3, 4, 7}, {3, 5, 6}, {3, 5, 7},
{3, 6, 7}, {4, 5, 6}, {4, 5, 7}, {4, 6, 7}, {5, 6, 7}};
In[2]:= Cases[L, x_ /; FreeQ[x, 1 | 4 | 5]]
Out[2]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}}
In[3]:= Select[L, Intersection[#1, {1, 4, 5}] === {} & ]
Out[3]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}}
In[4]:= DeleteCases[L, x_ /;  ! FreeQ[x, 1 | 4 | 5]]
Out[4]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}}
In[6]:= Select[L, Complement[#1, {1, 4, 5}] === #1 & ]
Out[6]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}}

2010/5/28 S. B. Gray <stevebg at roadrunner.com>

> Hello:
>
> I have this line
>
> subs3 = Subsets[Range[numbr], {3}] which for numbr=7 gives subs3=
>
> {{1,2,3}, {1,2,4}, {1,2,5}, {1,2,6}, {1,2,7}, {1,3,4}, {1,3,5},
>  {1,3,6}, {1,3,7}, {1,4,5}, {1,4,6}, {1,4,7}, {1,5,6}, {1,5,7},
>  {1,6,7}, {2,3,4}, {2,3,5}, {2,3,6}, {2,3,7}, {2,4,5}, {2,4,6},
>  {2,4,7}, {2,5,6}, {2,5,7}, {2,6,7}, {3,4,5}, {3,4,6}, {3,4,7},
>  {3,5,6}, {3,5,7}, {3,6,7}, {4,5,6}, {4,5,7}, {4,6,7}, {5,6,7}}
>
> Given one of these sublists, say {1,4,5}, I want a nice way to delete
> all members of the subs3 list which have 1,4, or 5 in any one of the
> three positions. The reduced list in this example would be
> {{2,3,6}, {2,3,7}, {2,6,7}, {3,6,7}}.
>
> The way I'm doing it runs DeleteCases nine times. Is there a better way?
>
> Steve Gray
>
>

```

• Prev by Date: Re: Trying to do some stuff with JLINK ???
• Next by Date: Re: Eric weisstein's MathWorld packages
• Previous by thread: Re: Lists: how to?
• Next by thread: Re: Lists: how to?