Lists: how to?
in [Mathematica]
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From: S. B. Gray on 28 May 2010 07:21 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: Bob Hanlon on 29 May 2010 04:43 subs3[numbr_Integer] := Subsets[Range[numbr], {3}] data = subs3[7]; Select[data, FreeQ[#, 1 | 4 | 5] &] {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}} Cases[data, _?(FreeQ[#, 1 | 4 | 5] &)] {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}} Bob Hanlon ---- "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: Adriano Pascoletti on 29 May 2010 04:44 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}} Adriano Pascoletti 2010/5/28 S. B. Gray <stevebg(a)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 > >
From: Raffy on 29 May 2010 04:44 On May 28, 4:21 am, "S. B. Gray" <stev...(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 If you know the set ahead of time, you can just construct the desired result: Subsets[Complement[Range[7], {1,4,5}], {3}]
From: David Skulsky on 29 May 2010 04:45
On May 28, 4:21 am, "S. B. Gray" <stev...(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 Here's one way: In[29]:= Select[subs3, (Length[Intersection[#, {1, 4, 5}]] == 0) &] Out[29]= {{2, 3, 6}, {2, 3, 7}, {2, 6, 7}, {3, 6, 7}} |