If I have a list of rules such as this,
rules = {a -> tt, b -> ff, c -> tt, d -> ff} ,
then what is the correct code to replace the rule c->tt in the list with a new rule, like c->ff?
Perhaps
In[1]:= rules = {a -> tt, b -> ff, c -> tt, d -> ff};
rules = rules /. (c -> tt) -> (c -> ff)
Out[2]= {a -> tt, b -> ff, c -> ff, d -> ff}
Thanks for the reply. I guessed that method too, but am not sure whether it is completely justifiable.
The use of parentheses makes (c → tt) the first item to be evaluated, so why does the code not give, for instance, {a → tt, b → ff, tt → ff, d → ff} instead?
Yes, the expression has the following tree form, but what in the Wolfram language determines the sequence of operations in this case?
Hi,
maybe “The Standard Evaluation Procedure” reference is helpful for you. Wolfram usually starts from the head of the function to read attributes. Afterwards the default evaluation process starts from the innermost subexpression(leaves). But here you dont have an evaluation but a rule .You could write it as
rules = ReplaceAll[rules, (c -> tt) -> (c -> ff)]
and it just replaces the part “c->tt” by “c->ff” and gives the final result. If you would try
rules = ReplaceAll[rules, c -> ff]
just the “c” would be replaced by “ff” and you would obtain
{a -> tt, b -> ff, ff -> tt, d -> ff}
The question regarding your TreeForm might be answered by the example
ReplaceAll[Sin[Pi],Pi->a]
which gives just 0 as an output, since here the Trace[] is, that first Sin[Pi] is evaluated and thus the Pi vanished and cant be replaced anymore. If you want to have a different sequence of evaluations than innermost subexpression you might use Attributes like HoldAll , to change the behaviour to “Non?Standard Evaluation”,
Thank you very much! So that ought to be the right way of doing it. You showed me where to look in the reference.
You can also take the associations based approach:
Normal@<|rules, c -> ff|>
{a -> tt, b -> ff, c -> ff, d -> ff}
Thanks. This one is surely correct.