Re: Computing n-grams

*To*: mathgroup at smc.vnet.net*Subject*: [mg88942] Re: [mg88913] Computing n-grams*From*: Darren Glosemeyer <darreng at wolfram.com>*Date*: Thu, 22 May 2008 02:34:17 -0400 (EDT)*References*: <200805211849.OAA10371@smc.vnet.net>

Coleman, Mark wrote: > Greetings, > > Imagine one has a list such as {a,b,c,d,e,f,g}. I'm trying to find an > efficient way in Mathematica to compute the n-grams of the list. That is, for > n=2, the n-grams are all the lists of length 2 consisting of consecutive > elements, e.g., > > {a,b},{b,c},{c,d},{d,e},... > > While for n=3, > > {a,b,c},{b,c,d},{c,d,e},..., and so on. > > As I understand it, the built-in Mathematica commands such as Subsets or > Permutations compute all possible list of size n, without regard to the > order of the list elements. > > Thanks, > > Mark > > Partition with an offset of 1 will do the trick. In[1]:= Partition[{a, b, c, d, e, f}, 3, 1] Out[1]= {{a, b, c}, {b, c, d}, {c, d, e}, {d, e, f}} In[2]:= nGram[list_,n_]:=Partition[list,n,1] In[3]:= nGram[{a, b, c, d, e, f},2] Out[3]= {{a, b}, {b, c}, {c, d}, {d, e}, {e, f}} Darren Glosemeyer Wolfram Research

**References**:**Computing n-grams***From:*"Coleman, Mark" <Mark.Coleman@LibertyMutual.com>