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Re: Question about function construction

  • To: mathgroup at smc.vnet.net
  • Subject: [mg119483] Re: Question about function construction
  • From: Albert Retey <awnl at gmx-topmail.de>
  • Date: Mon, 6 Jun 2011 06:23:14 -0400 (EDT)
  • References: <isfnrv$btm$1@smc.vnet.net>

Am 05.06.2011 13:04, schrieb ZiYuan Lin:
> Hi, there.
>    I want to construct a function serving as a hash table. This is how I make it:
>
>    ConstructHashtable[] :=
>     Module[{pairs = {{"key1", "value1"}, {"key2", "value2"}, {"key3",
>          "value3"}, {"key4", "value4"}}}, Clear[Hashtable];
>      Hashtable[key_?StringQ] := (Hashtable[#[[1]]] = #[[2]])&  /@ pairs;]
>
> And then I run the code above together with a test:
>
>    ConstructHashtable[]; Hashtable["key1"]
>
> Yet it returns all values:
>
>    {"value1", "value2", "value3", "value4"}
>
> But if I run it again:
>
>    Hashtable["key1"]
>
> It will get normal:
>
>    "value1"
>
> What is the reason for this? Thank you~

the reason is that the first call to Hashtable will call the code that 
fills the hashtable and that returns the list with the values. Only 
after that for the defined keys the corresponding values are returned. 
Your implementation has another "feature" that I hardly can believe is 
intended: for every key that is not in the hashtable yet, it will 
recreate the entries and again return the list of values. Actually I 
think you want something like this:

ConstructHashtable[] :=
  Module[{pairs = {{"key1", "value1"}, {"key2", "value2"}, {"key3",
       "value3"}, {"key4", "value4"}}}, Clear[Hashtable];
   (Hashtable[#[[1]]] = #[[2]]) & /@ pairs;
   ]

I think you also probably want to add a rule for those keys that are not 
available, and I don't see an advantage to define pairs in the first 
place. This is what I probably would do:

ConstructHashtable[] := (
   Clear[Hashtable];
   Hashtable[___] = Missing["NotAvailable"];
   Scan[(Hashtable[#[[1]]] = #[[2]]) &,
    {{"key1", "value1"}, {"key2", "value2"},
     {"key3", "value3"}, {"key4", "value4"}}
    ]
   )


hth,

albert



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