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An Array v's A Dictionary
#2
Dictionary's use a different "index" system, keys are used instead of integers and value(s) are associated with each key eg
a dictionionary of Colors the key is a color name and the value could be RGB values

Like with a regular dictionary the keys are the words and the definitions are the values for each key.

TempodiBasic and I did most recent Dictionary stuff here at this forum, try search with our avatar names and Dictionary.

Too much noise in the searches:
here is some excellent code! Smile
Code: (Select All)
Option _Explicit ' Dictionary 3
' b+ remake of TempodiBasic post 2021-04-06
' ref: https://www.qb64.org/forum/index.php?topic=3786.msg131448#msg131448
' 2021-04-07 add some bells and whistles
' 2023-04-11 add Find for faster actions,  add load and save to file

Type Dictionary
    K As String ' keys all caps for faster searches
    V As String
End Type

ReDim MyDict(1 To 1) As Dictionary ' use ubound of array to tell how many values we have

'                                  This code checks all the stuff Dictionary 2 did
' make some new pairs
Print "Show empty MyDict at start of this demo:"
ShowDict MyDict()
Print "Add a KV pair:"
AddModDictionary MyDict(), "mammals", "Cats"
ShowDict MyDict()
Print "Add a KV pair:"
AddModDictionary MyDict(), "trees", "Oak"
ShowDict MyDict()
Print "Add a KV pair:"
AddModDictionary MyDict(), "fish", "Bass"
ShowDict MyDict()
Print "Swap Dogs for Cats in mammals:"
AddModDictionary MyDict(), "mammals", "Dogs"
ShowDict MyDict()
Print "Check current mammals:"
Print "What is current mammal ? answer: "; GetValue$(MyDict(), "mammals")
Print "Remove mammals:"
RemoveKV MyDict(), "mammals"
ShowDict MyDict()
Print "Bring mammals back with Horses AND Dogs,Cats:"
AddAppendDictionary MyDict(), "Mammals", "Horses"
AddAppendDictionary MyDict(), "mammals", "Cats,Dogs"
ShowDict MyDict()
Print "Remove Cats from mammals:"
RemoveValue MyDict(), "mammals", "Cats"
ShowDict MyDict()
Print "Remove Horses from mammals:"
RemoveValue MyDict(), "mammals", "Horses"
ShowDict MyDict()
Print "Remove Unicorns from mammals:"
RemoveValue MyDict(), "mammals", "Unicorns"
ShowDict MyDict()
Print "And finally wipe out mammals again by removing dogs:"
RemoveValue MyDict(), "mammals", "Dogs"
ShowDict MyDict()

' all above seems to work how bout new subs?
Print "Test Save and Load of the Dictionary:"
SaveDictionary MyDict(), "My Dictionary.txt"
_Delay .25
LoadDictionary MyDict(), "My Dictionary.txt"
ShowDict MyDict()
' good

Sub SaveDictionary (Dict() As Dictionary, pathedFileName$)
    Dim As Long i
    Open pathedFileName$ For Output As #1 ' 2 line format key then value list
    For i = 1 To UBound(Dict)
        Print #1, Dict(i).K
        If _Trim$(Dict(i).V) = "" Then Print #1, " " Else Print #1, _Trim$(Dict(i).V)
    Next
    Close #1
End Sub

Sub LoadDictionary (Dict() As Dictionary, pathedFileName$)
    Dim As Long ub ' will track actual amout of items
    ReDim Dict(1 To 1) As Dictionary
    Dict(1).K = "": Dict(1).V = "" ' sometimes var string UDT's have to be zero'd
    If _FileExists(pathedFileName$) Then
        Open pathedFileName$ For Input As #1
        While Not EOF(1)
            ub = ub + 1
            If ub > UBound(Dict) Then ReDim _Preserve Dict(1 To ub + 1000) As Dictionary
            Line Input #1, Dict(ub).K
            Line Input #1, Dict(ub).V
        Wend
        ReDim _Preserve Dict(1 To ub) As Dictionary
    End If
End Sub

' replace 2 TempodiBasic Functions with 1 Sub, to handle both new and modified values for keys and dynamic Dict() dbl string array.
' Now just take ubound of dict() and have number of pairs it contains
Sub AddModDictionary (Dict() As Dictionary, K$, V$) ' mod with mod Find
    ReDim ub As Long, i As Long, ky$, f As Long, ip As Long
    ub = UBound(Dict)
    ky$ = UCase$(_Trim$(K$)) 'don't change k$ but make case insensitive?
    If ky$ <> "" Then ' bullet proof sub routine K$ must not be empty!
        If ub = 1 And Dict(1).K = "" Then 'our very first pair!
            Dict(1).K = ky$: Dict(1).V = V$: Exit Sub
        Else
            'For i = 1 To ub ' see if we have that name yet
            '    If ky$ = Dict(i).K Then Dict(i).V = V$: Exit Sub ' yes name is registered so change value
            'Next
            f = Find&(Dict(), ky$, ip)
            If f Then
                Dict(f).V = V$
            Else
                'still here? add var name and value to dictionary
                ReDim _Preserve Dict(1 To ub + 1) As Dictionary ' create one slot at a time such that ubound = number or pairs
                For i = ub To ip Step -1
                    Dict(i + 1) = Dict(i)
                Next
                Dict(ip).K = ky$: Dict(ip).V = V$ ' fill it with key and value
            End If
        End If
    End If
End Sub

Function GetValue$ (Dict() As Dictionary, K$) 'mod
    Dim f As Long, ip As Long
    f = Find&(Dict(), K$, ip)
    If f Then GetValue$ = Dict(f).V
End Function

Sub ShowDict (Dict() As Dictionary)
    Dim i As Long
    Print "Dictionary has "; _Trim$(Str$(UBound(Dict))); " items."
    For i = 1 To UBound(Dict)
        Print i, Dict(i).K, Dict(i).V
    Next
    Print
    Print "zzz... press any to continue"
    Sleep
    Print
End Sub

Sub RemoveKV (Dict() As Dictionary, K$) ' mod
    Dim As Long j, f, ip
    f = Find&(Dict(), K$, ip)
    If f Then
        If f <> UBound(Dict) Then
            For j = f + 1 To UBound(Dict)
                Swap Dict(j - 1), Dict(j)
            Next
        End If
        ReDim _Preserve Dict(1 To UBound(Dict) - 1) As Dictionary
    End If
End Sub

' instead or replacing a value with another we will add the new value delimited by a comma
Sub AddAppendDictionary (Dict() As Dictionary, K$, V$) ' mod
    Dim As Long ub, i, f, ip
    Dim ky$
    ub = UBound(Dict)
    ky$ = UCase$(_Trim$(K$)) 'don't change k$ but make case insensitive?
    If ky$ <> "" Then ' bullet proof sub routine K$ must not be empty!
        If ub = 1 And Dict(1).K = "" Then 'our very first pair!
            Dict(1).K = ky$: Dict(1).V = V$: Exit Sub
        Else
            f = Find&(Dict(), ky$, ip)
            If f Then
                Dict(f).V = Dict(f).V + "," + V$
            Else

                ReDim _Preserve Dict(1 To ub + 1) As Dictionary ' create one slot at a time such that ubound = number or pairs
                For i = ub To ip Step -1
                    Dict(i + 1) = Dict(i)
                Next
                Dict(ip).K = ky$: Dict(ip).V = V$ ' fill it with key and value
            End If
        End If
    End If
End Sub

Sub RemoveValue (Dict() As Dictionary, K$, RemoveV$) ' mod
    ReDim As Long ub, j, f, ip
    ReDim ky$, b$
    ub = UBound(Dict)
    ky$ = UCase$(_Trim$(K$)) 'don't change k$ but make case insensitive?
    If ky$ <> "" Then ' bullet proof sub routine K$ must not be empty!
        If ub = 1 And Dict(1).K = "" Then 'our very first pair!
            Exit Sub
        Else
            f = Find&(Dict(), ky$, ip)
            If f Then
                If InStr(Dict(f).V, ",") > 0 Then
                    ReDim t$(1 To 1)
                    Split Dict(f).V, ",", t$()
                    For j = 1 To UBound(t$)
                        If t$(j) <> RemoveV$ Then
                            If b$ = "" Then
                                b$ = t$(j)
                            Else
                                b$ = b$ + "," + t$(j)
                            End If
                        End If
                    Next
                    Dict(f).V = b$
                ElseIf Dict(f).V = RemoveV$ Then
                    Dict(f).V = ""
                End If
            End If
        End If
    End If
End Sub

' 2023-04-11 mod for Dictionary Type and inserting new words
Function Find& (SortedArr() As Dictionary, x$, insertPlace&)
    Dim As Long low, hi, test
    Dim xcap$
    xcap$ = UCase$(x$)
    low = LBound(SortedArr): hi = UBound(SortedArr)
    While low <= hi
        test = Int((low + hi) / 2)
        If SortedArr(hi).K < xcap$ Then insertPlace& = hi + 1 Else insertPlace& = low
        If SortedArr(test).K = xcap$ Then
            Find& = test: Exit Function
        Else
            If SortedArr(test).K <= xcap$ Then low = test + 1 Else hi = test - 1
        End If
    Wend
End Function

' note: I buggered this twice now, FOR base 1 array REDIM MyArray (1 to 1) AS ... the (1 to 1) is not same as (1) which was the Blunder!!!
'notes: REDIM the array(0) to be loaded before calling Split '<<<< IMPORTANT dynamic array and empty, can use any lbound though
'This SUB will take a given N delimited string, and delimiter$ and create an array of N+1 strings using the LBOUND of the given dynamic array to load.
'notes: the loadMeArray() needs to be dynamic string array and will not change the LBOUND of the array it is given.  rev 2019-08-27
Sub Split (SplitMeString As String, delim As String, loadMeArray() As String) ' from Handy library
    Dim curpos As Long, arrpos As Long, LD As Long, dpos As Long 'fix use the Lbound the array already has
    curpos = 1: arrpos = LBound(loadMeArray): LD = Len(delim)
    dpos = InStr(curpos, SplitMeString, delim)
    Do Until dpos = 0
        loadMeArray(arrpos) = Mid$(SplitMeString, curpos, dpos - curpos)
        arrpos = arrpos + 1
        If arrpos > UBound(loadMeArray) Then ReDim _Preserve loadMeArray(LBound(loadMeArray) To UBound(loadMeArray) + 1000) As String
        curpos = dpos + LD
        dpos = InStr(curpos, SplitMeString, delim)
    Loop
    loadMeArray(arrpos) = Mid$(SplitMeString, curpos)
    ReDim _Preserve loadMeArray(LBound(loadMeArray) To arrpos) As String 'get the ubound correct
End Sub
b = b + ...
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Messages In This Thread
An Array v's A Dictionary - by Dimster - 10-07-2024, 07:18 PM
RE: An Array v's A Dictionary - by bplus - 10-07-2024, 08:03 PM
RE: An Array v's A Dictionary - by TerryRitchie - 10-07-2024, 08:23 PM
RE: An Array v's A Dictionary - by DSMan195276 - 10-08-2024, 02:27 AM
RE: An Array v's A Dictionary - by Dimster - 10-08-2024, 10:26 AM
RE: An Array v's A Dictionary - by DSMan195276 - 10-08-2024, 12:52 PM
RE: An Array v's A Dictionary - by Dimster - 10-08-2024, 02:22 PM
RE: An Array v's A Dictionary - by Dimster - 10-08-2024, 02:46 PM
RE: An Array v's A Dictionary - by Kernelpanic - 10-08-2024, 04:31 PM
RE: An Array v's A Dictionary - by bplus - 10-08-2024, 05:10 PM
RE: An Array v's A Dictionary - by TempodiBasic - 10-08-2024, 06:59 PM



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