RE: Assigning to a sublist
- To: mathgroup at smc.vnet.net
- Subject: [mg35171] RE: [mg35161] Assigning to a sublist
- From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
- Date: Fri, 28 Jun 2002 02:30:59 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
> -----Original Message----- > From: Bob Harris [mailto:nitlion at mindspring.com] To: mathgroup at smc.vnet.net > Sent: Thursday, June 27, 2002 6:24 AM > To: mathgroup at smc.vnet.net > Subject: [mg35171] [mg35161] Assigning to a sublist > > > 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 > > Bob, observe 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]] = Append[zz[[3]][[2]], {1, 1, 1, 1, 1}] Out[2]= {{2, 1}, {1, 1, 1, 1, 1}} In[3]:= zz Out[3]= {{2, {{1, 1}}}, {10, {{3, 1}}}, {5, {{2, 1}, {1, 1, 1, 1, 1}}}} So here at the lhs of Set zz[[3]][[2]] is not the same as zz[[3, 2]] In[4]:= Hold[zz[[3]][[2]]] // FullForm Out[4]//FullForm= Hold[Part[Part[zz, 3], 2]] In[5]:= Hold[zz[[3, 2]]] // FullForm Out[5]//FullForm= Hold[Part[zz, 3, 2]] Set looks at the first element of Part at the (unevaluated) lhs, this must be symbol, and zz[[3]] is none, as the error message tells. Set does not look down further, clearly on reasons of performance. So just use simple Part. -- Hartmut