Lists: how to?
in [Mathematica]
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From: Thomas Dowling on 29 May 2010 04:46 Hello, One way, perhaps: fn[{x_, y_, z_}] := Select [subs3, FreeQ[#1, x] && FreeQ[#1, y] && FreeQ[#1, z] &] In[35]= fn[subs3[[10]]] Out[35]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}} Tom Dowling On Fri, May 28, 2010 at 12:21 PM, S. B. Gray <stevebg(a)roadrunner.com> wrote: > 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 > >
From: Leonid Shifrin on 29 May 2010 04:46 Hi Steve, Your test list: In[1] := test = {{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}}; Here is one way using Complement: In[2]:= Select[test, Complement[#, Complement[#, {1, 4, 5}]] === {} &] Out[2]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}} Somewhat similar one using MemberQ: In[3]:= DeleteCases[test, x_ /; MemberQ[x, Alternatives @@ {1, 4, 5}]] Out[3]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}} Here is another one using patterns: In[4]:= With[{ptrn = Alternatives @@ {1, 4, 5}}, DeleteCases[test, {ptrn, _, _} | {_, ptrn, _} | {_, _, ptrn}]] Out[4]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}} The above pattern can be constructed programmatically: In[5]:= With[{ptrn = Alternatives @@ {1, 4, 5}}, DeleteCases[test, Alternatives @@ NestList[RotateRight, {ptrn, _, _}, 2]]] Out[5]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}} Regards, Leonid On Fri, May 28, 2010 at 3:21 PM, S. B. Gray <stevebg(a)roadrunner.com> wrote: > 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 > >
From: Bill Rowe on 29 May 2010 04:45 On 5/28/10 at 7:21 AM, stevebg(a)ROADRUNNER.COM (S. B. Gray) wrote: >I have this line >subs3 = Subsets[Range[numbr], {3}] which for numbr=7 gives subs3= <output snipped> >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? Here is one way In[12]:= DeleteCases[subs3, _?(Intersection[{1, 4, 5}, #] != {} &)] Out[12]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}}
From: Peter Breitfeld on 29 May 2010 04:46 "S. B. Gray" wrote: > 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 > This may work, but I don't know if it's fast. First create a pattern for the items to be deleted: testform=Permutations[{1,4,5,_,_},{3}] Then delete all elements matching testform: DeleteCases[subs3,Alternatives @@ testform] you get your result. //Peter -- _________________________________________________________________ Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de
From: J. Clarke on 30 May 2010 23:45
On 5/28/2010 7:21 AM, S. B. Gray wrote: > 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? I'm just learning Mathematica myself so have no idea if this is a good or bad way to do it, but: Select[subs3, Intersection[{1, 4, 5}, #] == {} &] will I think do what you're asking. |