i have a code that copies and rewrites anything thats between "(" and ")", but now i have different type of data which do not end with ")" so, i need it to stop when it reaches the last character in cell. Maybe it is dumb question but i cant seem to find how to fix my problem. I am a student and total newbie in vba (5 days ago i didn't know what vba is...) also sorry for my bad english.
I've tried to search (in here, google, youtube) but i couldnt find anything i need
'zaciatok=start koniec=end dlzka=length
Do While Mid(LookInHere, y, 1) <> ""
If Mid(LookInHere, Z, 1) = "(" Then
zaciatok = Z
End If
If Mid(LookInHere, y, 1) = ")" Then
koniec = y
dlzka = (koniec - 1) - zaciatok
dlzka = Abs(dlzka)
SplitCatcher = Mid(LookInHere, zaciatok + 1, CStr(dlzka))
MsgBox SplitCatcher
End If
y = y + 1
Z = Z + 1
Loop
In your specific implementation, one option is to modify your Do While ... loop to also test against the length of the string. That line would look something like:
Do While Mid(LookInHere, y, 1) <> "" And y < Len(LookInHere)
That modification tells the statement that it should terminate the loop when the iterating variable y goes past the length of the statement.
Another option is to change it from a Do While loop to a For loop. It would read something like:
For i = 1 to Len(LookInHere)
MsgBox Mid(LookInHere, i, 1)
'Input your logic here
Next i
The problem is that each of these versions is relatively inefficient, looping through each letter in a string a performing a calculation. Consider using built-in Excel functions. The Instr returns the position of a character, or a zero if it is not found. As an example, Instr("Abcdef", "b") would return the number 2, and Instr("Abcdef", "k") would return zero. You can replace the entire loop with these two function calls.
Z = Instr(LookInHere, "(")
y = Instr(LookInHere, ")")
If y = 0 Then y = Len(LookInHere)
Final note: if your patterns begin to get more and more complex, consider reviewing and implementing regular expressions.
You can use Right(LookInHere, 1) to get the last character of LookInHere
Related
basically, I need to find a way to check if a String variable contains another String variable(or preferably an item from an array). Here's my case:
I have a list of 15k+ Strings in Excel, and another one with 5k+ Strings.The Second list is composed of 3 character long abbreviations, and the first one might contain from 0 to 15 of those, separated by ", ". My goal is to count each occurrence, of each of those abbreviations.
I've approached this by creating an array containing all the abbreviations. Then I was hoping to loop through the 15k+ long list 5k+ times, to count the occurrences of each of those abbreviations, changing the abbreviation sought every loop. However I'm stuck with comparing the strings...
I've tried to use InStr to see if I get 0 or not. Although it worked for me before when I referred to a String being searched by variable, but it seems that expression sought can't be placed in such a way. I've been trying to use Find or Application.Match instead but had the same issue... I wonder if there is a VBA function I don't know about? Or maybe someone has an idea for a custom function or an entirely different approach? I've hit a brick wall...
Thank you in advance for any advice! I highly appreciate it.
Unfortunately I can't show you the data since it's customer's sensitive data but here's my loop:
With ActiveWorkbook
For p = 1 To size
For j = 6 To lrow
a = .Sheets("Server View").Cells(j, 3).Value
tmp = Apps(p)
test = InStr(1, a, tmp)
If test = 0 Then
Else
b = .Sheets("Server View").Cells(j, 9).Value
If b = PROD Then
cval = .Sheets("temp").Cells(j, 3).Value
cval = cval + 1
.Sheets("temp").Cells(j, 3).Value = cval
Else
cval = .Sheets("temp").Cells(j, 2).Value
cval = cval + 1
.Sheets("temp").Cells(j, 2).Value = cval
End If
End If
Next j
Next p
End With
when I run this codes the output is (" "," "),however it should be ("I","love")!!!, and there is no errors . what should I do to fix it ??
sen="I love dogs"
function Longest_word(sen)
x=" "
maxw=" "
minw=" "
minl=1
maxl=length(sen)
p=0
for i=1:length(sen)
if(sen[i]!=" ")
x=[x[1]...,sen[i]...]
else
p=length(x)
if p<min1
minl=p
minw=x
end
if p>maxl
maxl=p
maxw=x
end
x=" "
end
end
return minw,maxw
end
As #David mentioned, another and may be better solution can be achieved by using split function:
function longest_word(sentence)
sp=split(sentence)
len=map(length,sp)
return (sp[indmin(len)],sp[indmax(len)])
end
The idea of your code is good, but there are a few mistakes.
You can see what's going wrong by debugging a bit. The easiest way to do this is with #show, which prints out the value of variables. When code doesn't work like you expect, this is the first thing to do -- just ask it what it's doing by printing everything out!
E.g. if you put
if(sen[i]!=" ")
x=[x[1]...,sen[i]...]
#show x
and run the function with
Longest_word("I love dogs")
you will see that it is not doing what you want it to do, which (I believe) is add the ith letter to the string x.
Note that the ith letter accessed like sen[i] is a character not a string.
You can try converting it to a string with
string(sen[i])
but this gives a Unicode string, not an ASCII string, in recent versions of Julia.
In fact, it would be better not to iterate over the string using
for i in 1:length(sen)
but iterate over the characters in the string (which will also work if the string is Unicode):
for c in sen
Then you can initialise the string x as
x = UTF8String("")
and update it with
x = string(x, c)
Try out some of these possibilities and see if they help.
Also, you have maxl and minl defined wrong initially -- they should be the other way round. Also, the names of the variables are not very helpful for understanding what should happen. And the strings should be initialised to empty strings, "", not a string with a space, " ".
#daycaster is correct that there seems to be a min1 that should be minl.
However, in fact there is an easier way to solve the problem, using the split function, which divides a string into words.
Let us know if you still have a problem.
Here is a working version following your idea:
function longest_word(sentence)
x = UTF8String("")
maxw = ""
minw = ""
maxl = 0 # counterintuitive! start the "wrong" way round
minl = length(sentence)
for i in 1:length(sentence) # or: for c in sentence
if sentence[i] != ' ' # or: if c != ' '
x = string(x, sentence[i]) # or: x = string(x, c)
else
p = length(x)
if p < minl
minl = p
minw = x
end
if p > maxl
maxl = p
maxw = x
end
x = ""
end
end
return minw, maxw
end
Note that this function does not work if the longest word is at the end of the string. How could you modify it for this case?
I would like to read a data file with a Fortran program, where each line is a list of integers.
Each line has a variable number of integers, separated by a given character (space, comma...).
Sample input:
1,7,3,2
2,8
12,44,13,11
I have a solution to split lines, which I find rather convoluted:
module split
implicit none
contains
function string_to_integers(str, sep) result(a)
integer, allocatable :: a(:)
integer :: i, j, k, n, m, p, r
character(*) :: str
character :: sep, c
character(:), allocatable :: tmp
!First pass: find number of items (m), and maximum length of an item (r)
n = len_trim(str)
m = 1
j = 0
r = 0
do i = 1, n
if(str(i:i) == sep) then
m = m + 1
r = max(r, j)
j = 0
else
j = j + 1
end if
end do
r = max(r, j)
allocate(a(m))
allocate(character(r) :: tmp)
!Second pass: copy each item into temporary string (tmp),
!read an integer from tmp, and write this integer in the output array (a)
tmp(1:r) = " "
j = 0
k = 0
do i = 1, n
c = str(i:i)
if(c == sep) then
k = k + 1
read(tmp, *) p
a(k) = p
tmp(1:r) = " "
j = 0
else
j = j + 1
tmp(j:j) = c
end if
end do
k = k + 1
read(tmp, *) p
a(k) = p
deallocate(tmp)
end function
end module
My question:
Is there a simpler way to do this in Fortran? I mean, reading a list of values where the number of values to read is unknown. The above code looks awkward, and file I/O does not look easy in Fortran.
Also, the main program has to read lines with unknown and unbounded length. I am able to read lines if I assume they are all the same length (see below), but I don't know how to read unbounded lines. I suppose it would need the stream features of Fortran 2003, but I don't know how to write this.
Here is the current program:
program read_data
use split
implicit none
integer :: q
integer, allocatable :: a(:)
character(80) :: line
open(unit=10, file="input.txt", action="read", status="old", form="formatted")
do
read(10, "(A80)", iostat=q) line
if(q /= 0) exit
if(line(1:1) /= "#") then
a = string_to_integers(line, ",")
print *, ubound(a), a
end if
end do
close(10)
end program
A comment about the question: usually I would do this in Python, for example converting a line would be as simple as a = [int(x) for x in line.split(",")], and reading a file is likewise almost a trivial task. And I would do the "real" computing stuff with a Fortran DLL. However, I'd like to improve my Fortran skills on file I/O.
I don't claim it is the shortest possible, but it is much shorter than yours. And once you have it, you can reuse it. I don't completely agree with these claims how Fotran is bad at string processing, I do tokenization, recursive descent parsing and similar stuff just fine in Fortran, although it is easier in some other languages with richer libraries. Sometimes you can use the libraries written in other languages (especially C and C++) in Fortran too.
If you always use the comma you can remove the replacing by comma and thus shorten it even more.
function string_to_integers(str, sep) result(a)
integer, allocatable :: a(:)
character(*) :: str
character :: sep
integer :: i, n_sep
n_sep = 0
do i = 1, len_trim(str)
if (str(i:i)==sep) then
n_sep = n_sep + 1
str(i:i) = ','
end if
end do
allocate(a(n_sep+1))
read(str,*) a
end function
Potential for shortening: view the str as a character array using equivalence or transfer and use count() inside of allocate to get the size of a.
The code assumes that there is just one separator between each number and there is no separator before the first one. If multiple separators are allowed between two numbers, you have to check whether the preceding character is a separator or not
do i = 2, len_trim(str)
if (str(i:i)==sep .and. str(i-1:i-1)/=sep) then
n_sep = n_sep + 1
str(i:i) = ','
end if
end do
My answer is probably too simplistic for your goals but I have spent a lot of time recently reading in strange text files of numbers. My biggest problem is finding where they start (not hard in your case) then my best friend is the list-directed read.
read(unit=10,fmt=*) a
will read in all of the data into vector 'a', done deal. With this method you will not know which line any piece of data came from. If you want to allocate it then you can read the file once and figure out some algorithm to make the array larger than it needs to be, like maybe count the number of lines and you know a max data amount per line (say 21).
status = 0
do while ( status == 0)
line_counter = line_counter + 1
read(unit=10,, iostat=status, fmt=*)
end do
allocate(a(counter*21))
If you want to then eliminate zero values you can remove them or pre-seed the 'a' vector with a negative number if you don't expect any then remove all of those.
Another approach stemming from the other suggestion is to first count the commas then do a read where the loop is controlled by
do j = 1, line_counter ! You determined this on your first read
read(unit=11,fmt=*) a(j,:) ! a is now a 2 dimensional array (line_counter, maxNumberPerLine)
! You have a separate vector numberOfCommas(j) from before
end do
And now you can do whatever you want with these two arrays because you know all the data, which line it came from, and how many data were on each line.
I am trying to make a certain string with Digits change form, so that it looks like below.
Modified to show with Digits instead of X.
So let´s set, i get a number, like this.
234.123123123
I need to change how it appears, for 3 digits before a Dot, it would need to become.
0.234123123123
And let´s say it´s 2.23123123123, that would become, 0.0023123123123
Now i will try to explain the Formula, bare with my English.
The formula needs to change the position of the Dot ".".
You can see that it changes place, But you can also see that i am adding 0s.
That is important, the 0s needs to be added IF the dot get´s to the far left (the beginning of the string).
So how much must the Dot move?
The Dot must Always move 3 steps to the left.
And if it hit´s the wall (the start of the string) you need to add 0s, and then one 0s before the dot.
So if the number is, 2.222222
I will first have to move the dot to the left, let´s show step by step.
Step 1:
.2222222
Step 2:
.02222222
Step 3:
0.02222222
It must Always be a 0 Before the Dot, it it ever hit the Start.
It sounds more complicated then it is, it´s just that i don't know how to explain it.
EDIT:
My attempt:
TextPos = InStr(1, TextBox4.Value, ".") - 2
If (TextPos = 4) Then
sLeft = Left(TextBox4.Value, Len(TextBox4.Value) - Len(TextBox4.Value) + 2)
sRight = Right(TextBox4.Value, Len(TextBox4.Value) - 2)
sFinal = (sLeft & "." & Replace(sRight, ".", ""))
TextBox4.Value = Replace(sFinal, "-", "")
End If
If (TextPos = 3) Then
TextBox4.Value = "0." + Replace(TextBox4.Value, ".", "")
End If
If (TextPos = 2) Then
TextBox4.Value = "0.0" + Replace(TextBox4.Value, ".", "")
End If
If (TextPos = 1) Then
TextBox4.Value = "0.00" + Replace(TextBox4.Value, ".", "")
End If
TextBox4.Value = Replace(TextBox4.Value, "-", "")
If i /1000 it, may work, but it seems like it will alter the number, leaving it to fail.
Original: 25.1521584671082
/1000 : 2.51521584671082E-02
As you can see, it doesn't work as expected.
It should become 0.0251521584671082E-02 (And of course i don't want " E- " to be there.
So if it's possible to do this, but ONLY moving and not actually, dividing (i think excel is limited to how many numbers it can calculate with) then it will work.
Worth Noting, the numbers shown are actually less (shorter) then the real number for some reason, if i write it down to a text file, it becomes this:
2.51521584671082E-02251521584671082E-02
So as you can see, i probably hit a Wall in the calculation, so it changes to Characters (I know math uses those, but no idea what they mean, and they are useless for me anyway).
And i think it's because of them that it fails, not sure though.
EDIT:
Okay, by limiting to 15 decimals it will return the correct place, but i would like not to do this round about, should Excel really be limited to only 15 Decimals, or is it "Double,Float etc" that does this?
EDIT 2:
Tony's example provides this:
Original: 177.582010543847177582010543847
Tony's: 0.1775820110177582011
Round(15decimals): 0.1775820105438470177582010543847
Not really sure, but isn't it Incorrect, or perhaps i did something wrong?
If possible, i want ALL decimals and number to stay in place, as everything is already correct, it's just that they are in the wrong place.
/1000 solves this, but not in the best way, as it will recalculate instead of just moving.
Normally i wouldn't care, the results are the same, but here it does matter as you can see.
I can however think of a solution, where i till look for the position of the Dot, then cut out the last decimals, divide the Result by 1000, then later add the last decimals, though that is more of a hack , not sure if it will actually work.
I conclude that this is Answered, it does what i want, and the limitations is in the Calculation itself, not the Solution.
All Solutions work pretty much in the same way, but i chose Tony's as he was first to comment the solution in my question.
Many Thanks everyone!
It appears to me through all your edits that you want to divide the number by one thousand. To do this, you can simply use
TextBox4.Value = TextBox4.Value / 1000
You code sometimes fails because it does not handle every situation. You calculate TextPos so:
TextPos = InStr(1, TextBox4.Value, ".") - 2
You then handle TextPos being 4, 3, 2 or 1. But if, for example, TextBox4.Value = "2.22222" then TextPos = 0 and you have no code for that situation.
At 11:45 GMT, I suggested in a comment that dividing by 1000 would give you the result you seek. At 14:08 GMT, tunderblaster posted this as an answer. It is now 15:23 but you have not responded to either of us.
Having tried your code, I now know these answer do not match the result it gives when it works because you remove the sign. The following would give you almost the same as a working version of your code:
TextBox4.Value = Format(Abs(Val(TextBox4.Value)) / 1000, "#,##0.#########")
The only deficiency with this statement is that it does not preserve all the trailing decimal digits. Is this really important?
While I was preparing and posting this answer, you edited your question to state that dividing by 1000 would sometimes give the result in scientific format. My solution avoids that problem.
If you want to avoid floating point error when dividing by 10, you may use the Decimal type
Private Sub CommandButton1_Click()
Dim d As Variant
d = CDec(TextBox1.Text)
TextBox1.Text = d / 1000
End Sub
Is this what you are trying?
Sub Sample()
Dim sNum As String, SFinal As String
Dim sPref As String, sSuff As String
Dim nlen As Long
sNum = "234.123123123123123123131231231231223123123"
SFinal = sNum
If InStr(1, sNum, ".") Then
sPref = Trim(Split(sNum, ".")(0))
sSuff = Trim(Split(sNum, ".")(1))
nlen = Len(sPref)
Select Case nlen
Case 1: SFinal = ".00" & Val(sPref) & sSuff
Case 2: SFinal = ".0" & Val(sPref) & sSuff
Case 3: SFinal = "." & Val(sPref) & sSuff
End Select
End If
Debug.Print SFinal
End Sub
Try this. It works by initialy padding the string with leading 0's, shifting the DP, then removing any unnecassary remaining leading 0's
Function ShiftDecimalInString(str As String, Places As Long) As String
Dim i As Long
If Places > 0 Then
' Move DP to left
' Pad with leading 0's
str = Replace$(Space$(Places), " ", "0") & str
' Position of .
i = InStr(str, ".")
str = Replace$(str, ".", "")
If i > 0 Then
' Shift . to left
str = Left$(str, i - Places - 1) & "." & Mid$(str, i - Places)
' strip unnecassary leading 0's
Do While Left$(str, 2) = "00"
str = Mid$(str, 2)
Loop
If str Like "0[!.]*" Then
str = Mid$(str, 2)
End If
ShiftDecimalInString = str
Else
' No . in str
ShiftDecimalInString = str
End If
ElseIf Places < 0 Then
' ToDo: Handle moving DP to right
Else
' shift DP 0 places
ShiftDecimalInString = str
End If
End Function
This "answer" responses to the issue that you want as much precision in the result as possible.
When I was at school this was called spurious accuracy. I have forgotten most of my mathematics in this area but I will try to give you an understanding of the issue.
I have a value, 2.46, which is accurate to 2 decimal places. That is the true value is somewhere in the range 2.455 to 2.465.
If I feed these values into a square root function I will get:
sqrt(2.455) = 1.566843961599240
sqrt(2.46) = 1.568438714135810
sqrt(2.465) = 1.570031846810760
From this I see the square root of the true value is somewhere between 1.567 and 1.570; any extra decimal digits are spurious.
If I understand correctly, you wish to use these numbers to make adjustments. Are you capable of making adjustments to 15 decimal digits?
If you remember your calculus you can look up "propogation of error" for a better explanation.
Here's the question I have to answer for my assignment:
Count the number of words in the string "tx_val" that have 3,4,5 or 6 chatacters. Show these four counts on a single line seperated by commas in the span block id="ans12".
Here's what I've come up with, the output is incorrect and I'm not sure why. I'll post below. Thought I'd give you all a update of where I was at with it.
threematch = 0
fourmatch = 0
fivematch = 0
sixmatch = 0
totalmatch = ""
cntArr = Array()
cntArr = Split(tx_val," ")
i=0
For i=0 To Ubound(cntArr) Step 1
If len(cstr(cntArr(i))) = 3 Then
threecount = threecount + 1
ElseIf len(cstr(cntArr(i))) = 4 Then
fourcount = fourcount + 1
ElseIf len(cstr(cntArr(i))) = 5 Then
fivecount = fivecount + 1
ElseIf len(cstr(cntArr(i))) = 6 Then
sixcount = sixcount + 1
End If
i=i+1
Next
totalmatch = (threecount & ", " & fourcount & ", " & fivecount & ", " & sixcount & ".")
document.getElementById("ans12").innerHTML = totalmatch
First, and this is what is causing the wrong behaviour, you are explicitly incrementing your counter i, even though the For-Next loop already does that for you. The result is that for each pass through the loop, i actually gets incremented by 2.
Remove the i=i+1 line and your script will work as intended.
Second, your variable names are inconsistent, being initialised as e.g. threematch and later used as threecount. You should always declare your variables explicitly (Dim statements) and write Option Explicit at the top of your code to catch such obvious mistakes at compile time. By pure chance this mistake does not actually cause any errors in your particular case.
If you are comfortable using regular expressions in JavaScript, why not use them in VBScript? They both use the same ECMA-262 so patterns are identical between the two languages. VBScript's RegExp object can do the same thing as your example.
Set re = New RegExp
re.IgnoreCase = True ' equivalent to /i modifier
re.Global = True ' equivalent to /g modifier
re.Pattern = "\b\w{3}\b" ' regex pattern without delimiters or modifiers
Set colMatches = re.Execute(someStringOfWords)
intCount = colMatches.Count
To learn more about Regular Expressions in VBScript, stop by the MSDN and read Microsoft Beefs Up VBScript with Regular Expressions.
Break the problem up:
1) Extract all words (break on whitespace) into a list
2) Iterate over the list, checking which words have the specified number of characters, increment a counter each time a matching word length is seen.
3) Write out the total counts