I am looking to find formula which gives me count of -> how many line in multiline of the cell are begining with - (hyphen)
for e.g. if cell contains
how are you keeping up
-I am well and need toy
-"You" are asking wrong question
<you are wrong>
-why should i reply you
sum count of qualified multiline is = 3
can anyone help me out here please
If you first lines never start with an hyphen, or at least do not count towards the total, then try:
Formula in B1:
=(LEN(A1)-LEN(SUBSTITUTE(A1,CHAR(10)&"-","")))/2
If your first line can also start with an hyphen and therefor count towards the total, try:
=(LEN(CHAR(10)&A1)-LEN(SUBSTITUTE(CHAR(10)&A1,CHAR(10)&"-","")))/2
Here is a VBA solution:
Function CountLines(text As String, Optional flag As String = "") As Long
'counts all lines in text which starts with flag
Dim i As Long, count As Long
Dim lines As Variant
lines = Split(text, vbLf)
For i = LBound(lines) To UBound(lines)
If Mid(lines(i), 1, Len(flag)) = flag Then
count = count + 1
End If
Next i
CountLines = count
End Function
If this is in a standard code module, the example text in A1 and in B1 you enter the formula =CountLines(A1,"-"), it will evaluate to 3.
If you want to include the first line in the potential count, then, in Windows Excel 2013+, you can try:
=COUNTA(FILTERXML("<t><s>" & SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,">",">"),"<","<"),"""","""),CHAR(10),"</s><s>") & "</s></t>","//s[starts-with(text(),'-')]"))
Replace illegal xml characters ",<, and >
Create an XML by splitting into nodes based on the LF character
Use xpath //s[starts-with(text(),'-')] to return only those nodes that start with a hyphen.
COUNTA to return the count of those nodes
Hello I have a column with strings (names of products) in it.
Now these are formatted as Name LenghtxWidth, example Green box 20x30. Now I need to change the 20 with the 30 in this example so I get Green box 30x20, any ideas how I can achieve this?
Thanks
Here is both a formula solution, as well as a VBA solution using Regular Expressions:
Formula
=LEFT(A1,FIND(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),A1)-1)&
MID(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),SEARCH("x",TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)))+1,99)&
"x"&
LEFT(TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)),SEARCH("x",TRIM(RIGHT(SUBSTITUTE(A1," ",REPT(" ",99)),99)))-1)
UDF
Option Explicit
Function RevWL(S As String)
Dim RE As Object
Const sPat As String = "(\d+.?\d*)x(\d+.?\d*)"
'If L or W might start with a decimal point, and not a digit,
'Then change sPat to: (\d*.?\d+)x(\d*.?\d+)
Set RE = CreateObject("vbscript.regexp")
With RE
.Global = True
.ignorecase = True
.Pattern = sPat
RevWL = .Replace(S, "$2x$1")
End With
End Function
Here is an example of the kinds of data I tested with:
The Formula works by looking at the last space-separated substring which would be LxW, then reversing the portion after and before the x, then concatenating everything back together.
The regex pattern captures the two numbers (could be integers or decimals, so long as the start with an integer -- although that could be changed if needed), and reversing them.
Here is a more detailed explanation of the regex (and the replacement string) with links to a tutorial:
(\d+.?\d*)x(\d+.?\d*)
(\d+.?\d*)x(\d+.?\d*)
Options: Case insensitive; ^$ don’t match at line breaks
Match the regex below and capture its match into backreference number 1 (\d+.?\d*)
Match a single character that is a “digit” \d+
Between one and unlimited times, as many times as possible, giving back as needed (greedy) +
Match any single character that is NOT a line break character .?
Between zero and one times, as many times as possible, giving back as needed (greedy) ?
Match a single character that is a “digit” \d*
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *
Match the character “x” literally x
Match the regex below and capture its match into backreference number 2 (\d+.?\d*)
Match a single character that is a “digit” \d+
Between one and unlimited times, as many times as possible, giving back as needed (greedy) +
Match any single character that is NOT a line break character .?
Between zero and one times, as many times as possible, giving back as needed (greedy) ?
Match a single character that is a “digit” \d*
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) *
$2x$1
Insert the text that was last matched by capturing group number 2 $2
Insert the character “x” literally x
Insert the text that was last matched by capturing group number 1 $1
Created with RegexBuddy
Here is a VBA solution that will work for you:
Option Explicit
Function Switch(r As Range) As String
Dim measurement As String
Dim firstPart As String
Dim secondPart As String
measurement = Right(r, Len(r) - InStrRev(r, " "))
secondPart = Right(measurement, Len(measurement) - InStr(1, measurement, "x"))
firstPart = Left(measurement, InStr(1, measurement, "x") - 1)
Switch = Left(r, InStrRev(r, " ") - 1) & " " & secondPart & "x" & firstPart
End Function
You can paste this in a regular module in the VBE (Visual Basic Editor) and use it as a regular function/formula. If your value is in cell A1 then type =Switch(A1) in cell B1. Hope it helps!
Ok, so it is really easier to use VBA, but if you want only some formulas you can use some columns to split your text and then concatenate your cells.
Here is a little example:
Of course B1-4 are optional. It is here only to have something more readable, but you can do use only one formula
=CONCATENATE(LEFT(A1, SEARCH(" ",A1,1)-1)," ",RIGHT(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),LEN(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)))-SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),1)),"x",LEFT(RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)), SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH(" ",A1,1)),1)-1))
If you have several spaces in your names, you can use this formula that will search the last space in the text
=CONCATENATE(LEFT(A1, SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))-1)," ",RIGHT(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),LEN(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))))-SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),1)),"x",LEFT(RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))), SEARCH("x",RIGHT(A1,LEN(A1)-SEARCH("^^",SUBSTITUTE(A1," ","^^",LEN(A1)-LEN(SUBSTITUTE(A1," ",""))))),1)-1))
I am interested in removing leading alphabetical (alpha) characters from cells which appear in a column. I only wish to remove the leading alpha characters (including UPPER and LOWER case): if alpha characters appear after a number they should be kept. Some cells in the column might not have leading alpha characters.
Here is an example of what I have:
36173
PIL51014
4UNV22001
ZEB54010
BICMPAFG11BK
BICMPF11
Notice how there are not always the same number of leading alpha characters. I cannot simply use a Left or Right function in Excel, because the number of characters I wish to keep and remove varies.
A correct output for what I am looking for would look like:
36173
51014
4UNV22001
54010
11BK
11
Notice how the second to last row preserved the characters "BK", and the 3rd row preserved "UNV". I cannot simply remove all alpha characters.
I am a beginner with visual basic and was not able to figure out how to use excel functions to address my issue. How would I do this?
Here is an Excel formula that will "strip off the leading alpha characters" Actually, it looks for the first numeric character, and returns everything after that:
=MID(A1,MIN(FIND({0;1;2;3;4;5;6;7;8;9},A1&"0123456789")),99)
The 99 at the end needs to be some value longer than the longest string you might be processing. 99 usually works.
Here's a formula based solution complete with test results:
=MID(A1,MIN(SEARCH({0,1,2,3,4,5,6,7,8,9},A1&"0123456789"),255),100)
Change the 100 at the end if any string may be longer than 100 characters. Also the 255 is not needed, but it won't hurt.
This short UDF should strip off leading alphabetic characters.
Function noLeadAlpha(str As String)
If Not IsNumeric(str) Then
Do While Asc(str) < 48 Or Asc(str) > 57
str = Mid(str, 2)
If Not CBool(Len(str)) Then Exit Do
Loop
End If
noLeadAlpha = str
End Function
Koodos Jeeped, you beat me to it.
But here is an alternative anyway:
Function RemoveAlpha(aString As String) As String
For i = 1 To Len(aString)
Select Case Mid(aString, i, 1)
Case "0" To "9"
RemoveAlpha = Right(aString, Len(aString) - i + 1): Exit For
End Select
Next i
End Function
How to find a word with space (eg: i want to find a word "X1 X2 X3"). I am using the following code
put the text of field "f4" into xx
find xx in field "f1"
In LiveCode, the space character delimits words -- a single word doesn't contain spaces. If there's only one instance of the string of characters you're searching for, you could use the offset function:
put "X1 X2 X3" into theString
put the text of fld "f4" into temp
put offset(theString,temp) into theNum
The variable theNum will contain the number of characters before the first character of the string, or will contain 0 if the string is not found in the field's text.
I'm writing a macro that parses a string in a cell from my excel sheet and should return the three coordinates in that string. I can get the script to parse the string fine and create a "coordinateHolder" array to hold the three coordinates. My issue is that when I update cells to show the coordinates excel shows does not show the entire coordinate.
For example, if the coordinates string is originally "1234.1324123, 12345.23521, 2384.1234253", my code will update my x, y, and z coordinate cells as "1234.132", "12345", "2384.1234"
Image of what I mean:
(This one shows scientific notation in the cell and a shortened double in the formula builder bar)
My Code:
Dim i, j As Integer
Dim coordinates As String
Dim coordHolder As Variant
i = 2
j = 1
Range("I2:K2").Value = Range("E2:G2").Value
Do While Cells(i, j) <> ""
coordinates = Cells(i, j)
coordinates = Replace(coordinates, ",", "")
coordHolder = Split(coordinates, " ")
For a = 0 To UBound(coordHolder)
Cells(i, 7 + a) = coordHolder(a)
Next a
i = i + 1
Loop
Excel has a limit of 15 digits for numbers. Any number with more digits will lose precision in the lower magnitudes to enable the number display. Your data has values that go beyond the limit and will be truncated.
513402938412.123 shows just 15 digits. The remaining decimal places have been removed. The significance of 4 or more decimals pales in comparison with the magnitude of the value, therefore Excel considers the digits after the third deicmal as dispensable.
If you want to retain all digits in the value, you need to convert it to text and make sure it remains text and is not converted to a number again. To do that, you can precede any number with the apostrophe sign.
If a cell contains the value 5134029388412.12341234 it will be truncated. A cell containing the value '5134029388412.12341234 will be treated as text and remain intact.