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I have to read a line which has the data as follows:
Now, I want to read that line as an array.Code:1.0 1.0 70.0 70.0
I have already defined the name of array as following.
I now there will be four numbers that I am interested.Code:Dim D50() as Double
Then I tried to read as follows:Code:Dim numb as Integer
numb = 4
ReDim D50(numb)
Dim strline as String
strline = " 1.0 1.0 70.0 70.0"
What is the best way to assign D50 to those 4 numbers ?
Well, if you're willing to settle for a Variant rather than a Double array, you can go:
And then if you declare the literals, it should be a variant array of doubles:Code:Dim v As Variant
v = array(1.0,1.0,70.0,70.0)
Code:Dim v As Variant
v = array(1.0#,1.0#,70.0#,70.0#)
I don't want to declare the array but I want to read it.
If you just really must have an array of double rather than a variant array with doubles in it, and you're unwilling to assign each one separately, then you'd need to write a string parser. This would be a bit of a pain, but not really that big of a deal. I'll let someone else do that though, or I might do it later if nobody else does.
I still miss the old "Data" and "Read" statements for anyone who even remembers what they are. They would do exactly what cooljd wants.
Ahhh, I just saw your reply. Well, you still need to declare a variable that it will be read into. As stated above, you could do a string parser. If you wanted to use a dynamic array, you'd need to make two passes. One to count the elements, and then Redim your array, then the second pass to put the data in the array.
Here, I'll knock out a rough string parser:
Something like that. I didn't debug. In fact, I just typed it out here. I'll let others debug it, or it'd be good for you if you did.Code:Dim i as long
Dim t as string
Dim cnt as long
Dim d50() as double
t = trim$(s)
i = instr(t, " ")
do while i <> 0
cnt = cnt + 1
t = trim$(mid$(t, i+1))
i = instr(t, " ")
loop
Redim d50(1 to cnt)
t = trim$(s)
i = instr(t, " ")
cnt = 0
do while i <> 0
cnt = cnt + 1
d50(cnt) = val(left$(t, i-1))
t = trim$(mid$(t, i+1))
i = instr(t, " ")
loop
EDIT: I can tell you one situation that would crash it and that'd be an empty "s" string coming in. Notice I didn't supply the "s" string. That's your job. Also, you may want to make sure that cnt is non-zero at the finish of the first loop through.
EDIT2: Ahhh, there's definitely one bug besides an empty "s" string coming in, but I'm not telling. C programmers tend to call it a "corners" problem, or you may call it a terminating problem. That's all the hint I'll give you.
There are a number or parsing techniques.. Split comes to mind, but it has pitfalls if your line contains double (or more) spaces....Quote:
Originally Posted by cooljd
I 'll write a parsing routine for you to munch on.
I would split the string and then assign the return to a double array.
Code:Private Sub Command1_Click()
Dim strLine As String
Dim D50() As Double
Dim i As Integer
strLine = "1.0 1.0 70.0 70.0"
D50 = StringToDoubleArray(strLine, " ")
For i = 0 To UBound(D50)
Debug.Print D50(i)
Next i
End Sub
Private Function StringToDoubleArray(ByVal InputString, ByVal Delimiter) As Double()
Dim arrString() As String
Dim arrRet() As Double
Dim i As Integer
Dim ub As Integer
arrString = Split(InputString, Delimiter)
ub = UBound(arrString)
ReDim arrRet(ub)
For i = 0 To ub
arrRet(i) = CDbl(arrString(i))
Next i
StringToDoubleArray = arrRet
End Function
Yeah, I don't know why I don't use "Split" more. If you're careful with your delimiters, it's a nice function.
Thank you so much for this example . How about if the spacing is more than 2 or three ? The function may not work properly. What is an easy fix ? If strline looks like following:Quote:
Originally Posted by MarkT
Code:strLine = " 1.0 1.0 70.0 70.0"
It goes like this
Code:Sub doubleParse(astring$, dbl#())
ReDim dbl#(20)
doublecount = 0
arg = Split(astring$, " ")
For i = 0 To UBound(arg)
If Len(arg(i)) Then
dbl#(doublecount) = CDbl(arg(i))
doublecount = doublecount + 1
If doublecount = UBound(dbl) Then
newub = UBound(dbl) + 20
ReDim Preserve dbl#(newub)
End If
End If
Next
ReDim Preserve dbl#(doublecount - 1)
End Sub
Code:astring$ = " 12.0 12.0 34.5 45.6 57.0 "
ReDim dblarray#(0)
doubleParse astring$, dblarray#()
rem array packed tight on return
MsgBox UBound(dblarray#)
Quite the opposite for me, never used it much in the past, but since writing text parsing routines, I put it in all recipes, numeric or string.Quote:
Originally Posted by Elroy
Thank you so much for your answer. Your function works regardless of any spacings in the string. It seems to be the perfect solution among the ones that have been proposed so far.Quote:
Originally Posted by Navion
I hope he gets an "A" Navion. *laughing* :p
You're welcomed. I could have written it as a Function but I always go for readability. Written as a Sub, you never have to go lookup some other piece of code to make sure what the types are.
I get it you pardon my french ? Excusez-moi ;)Quote:
Originally Posted by Elroy
updating the function like this would take care of itQuote:
Originally Posted by cooljd
Code:Private Function StringToDoubleArray(ByVal InputString, ByVal Delimiter) As Double()
Dim arrString() As String
Dim arrRet() As Double
Dim i As Integer
Dim ub As Integer
InputString = Trim(InputString)
Do While InStr(InputString, " ")
InputString = Replace(InputString, " ", " ")
Loop
arrString = Split(InputString, Delimiter)
ub = UBound(arrString)
ReDim arrRet(ub)
For i = 0 To ub
arrRet(i) = CDbl(arrString(i))
Next i
StringToDoubleArray = arrRet
End Function
To add a short Variant (using a Variant <g>) to the pile:
OlafCode:Sub GetDoubles(S As String, D() As Double)
Dim n As Long, V
For Each V In Split(S, " ")
If Len(V) Then ReDim Preserve D(n): D(n) = V: n = n + 1
Next
End Sub