Re: Q: efficient in-place list element replacement?
- To: mathgroup at smc.vnet.net
- Subject: [mg21140] Re: Q: efficient in-place list element replacement?
- From: Bojan Bistrovic <bojanb at physics.odu.edu>
- Date: Fri, 17 Dec 1999 01:21:41 -0500 (EST)
- Organization: Old Dominion Universityaruba
- References: <83208o$gik@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
This might shed more light on your problem if not solve it. After typing
in your example, try the following:
In[6]:= b=Unevaluated[a]
Out[6]= {4,5,a,b,b,7,67,89}
In[7]:= c=a
Out[7]= {4,5,a,b,b,7,67,89}
The output of "a", "b" and "c" will be the same, but they aren't stored
in memory in the same way. This is how they are stored internaly:
In[8]:= Definition[a]
Out[8]= {4,5,Sequence["a","b","c"],7,67,89}
In[9]:= Definition[b]
Out[9]= a
In[10]:= Definition[c]
Out[10]= {4,5,a,b,b,7,67,89}
Therefore, doing something like a[[7]]=Pi doesn't work because Set has
attribute HoldFirst which prevents the left hand side of the equation
from being evaluated. You might think that something like
Unprotect[Set];
ClearAttributes[Set, HoldFirst]
would help, but it won't (I've tried); it will start producing weird
error messages. Functions like Part, Length or FullForm evaluate their
arguments first so you don't see the Sequence fuction. On the other
hand, commands like Unevaluated[a] or Hold[a] will prevent the List
{4,5,...} from being assigned to the symbol "a" so the result will be
just Unevaluated[a] or Hold[a]
Last, but not least, there's a "trick" that will solve your problem
(altho it probably involves copying). Try following:
In[11]:= a=a
Out[11]= {4,5,a,b,b,7,67,89}
In[12]:= Definition[a]
Out[12]= {4,5,a,b,b,7,67,89}
What happened is this: Set evaluates the right-hand side of the
aquation, but not the left-hand; so the right-hand "a" was evaluated,
replaced by {4,5,a,b,b,7,67,89} and then assigned to the unevaluated
symbol "a" on the left-hand side.