Re: Assigning to a sublist

*To*: mathgroup at smc.vnet.net*Subject*: [mg35166] Re: [mg35161] Assigning to a sublist*From*: Andrzej Kozlowski <andrzej at lineone.net>*Date*: Fri, 28 Jun 2002 02:30:51 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Use: zz[[3,2]] = Append[zz[[3]][[2]], {1, 1, 1, 1, 1}] instead of zz[[3]][[2]] = Append[zz[[3]][[2]], {1, 1, 1, 1, 1}] The latter does not work, because, as the error message tell's you the in assignments of the form Part[something,...]= ..., the "something" must be a symbol. The former assignment above has the FullForm Part[zz,3,2] so it satisfies this condition while the latter has the form Part[Part[zz, 3], 2] so it doesn't. Andrzej Kozlowski Toyama International University JAPAN http://platon.c.u-tokyo.ac.jp/andrzej/ On Thursday, June 27, 2002, at 01:23 PM, Bob Harris wrote: > Howdy, > > I'm trying to assign a new value to an entry in a sublist (of another > list), > and I can't understand why it won't work. > > For example, I do the following: > > In[1]:= zz = {{2, {{1, 1}}}, {10, {{3, 1}}}, {5, {{2, 1}}}} > Out[1]= {{2, {{1, 1}}}, {10, {{3, 1}}}, {5, {{2, 1}}}} > > In[2]:= zz[[3]][[2]] > Out[2]= {{2, 1}} > > In[3]:= Append[zz[[3]][[2]], {1, 1, 1, 1, 1}] > Out[3]= {{2, 1}, {1, 1, 1, 1, 1}} > > In[4]:= zz[[3]][[2]] = Append[zz[[3]][[2]], {1, 1, 1, 1, 1}] > Set::"setps : zz[[3]] in assignment of part is not a symbol." > Out[4]= {{2, 1}, {1, 1, 1, 1, 1}} > > For some reason it doesn't like the assignment. What confuses me is > that is > zz[[3]][[2]] were just a variable, it would work. Further, if it were > just > an entry at the *top* level of a list, it would work, as this example > shows: > > In[5]:= yy = zz[[3]] > Out[5]= {5, {{2, 1}}} > > In[6]:= yy[[2]] > Out[6]= {{2, 1}} > > In[7]:= yy[[2]] = Append[yy[[2]], {1, 1, 1, 1, 1}] > Out[7]= {{2, 1}, {1, 1, 1, 1, 1}} > > So it seems like the issue is just that deeply nested things don't > behave > like things that are not as deeply nested. Am I right about that? How > can > I modify an entry in a sublist? > > Thanks, > Bob H > > > >