I want to this but i don't know what to do, the only functions it seems to be useful is "DEC.TO.HEX".
This is the problem, i have in one cell this text:
1234
And in the next cell i want the hexadecimal value of each character, the expected result would be:
31323334
Each character must be represented by two hexadecimal characters. I don't have an idea how to solve this in excel avoiding make a coded program.
Regards!
Edit: Hexadecimal conversion
Text value Ascii Value (Dec) Hexadecimal Value
1 49 31
2 50 32
3 51 33
4 52 34
Please try:
=DEC2HEX(CODE(MID(A1,1,1)))&DEC2HEX(CODE(MID(A1,2,1)))&DEC2HEX(CODE(MID(A1,3,1)))&DEC2HEX(CODE(MID(A1,4,1)))
In your version you might need the .s in the function (and perhaps ;s rather than ,s).
DEC2HEX may be of assistance. Use, as follows:
=DEC2HEX(A3)
First split 1234 to 1 2 3 4 by using MID(), then use Code() for each character, and then again concentate. Below is the formula, Y21 is the cell in which 1234 is written
=CONCATENATE(CODE(MID(Y21,1,1)),CODE(MID(Y21,2,1)),CODE(MID(Y21,3,1)),CODE(MID(Y21,4,1)))
1234 >> 49505152
Related
20/11/2022 12:00:52 2 X 15.95 15.95 USD 57 5 689 5 689 1 4111 0 Amazing Lego Team
I need to get the position of No 4111 in the above text string, As an excel beginner any help will be greatly appreciated. Thanks.
All of the Text Strings will have a 4 digit number like 4111 which i have to get the position for.
Have tried using this formula to get four digit number in another column, LOOKUP(10^15,MID(A1,ROW(INDIRECT("1:"&LEN(A1))),5)+0) but I am looking to get position instead.
I have tried using lookup but I could only go so far as a beginner.
Use the formula you provided =LOOKUP(10^15,MID(A1,ROW(INDIRECT("1:"&LEN(A1))),5)+0) to identify the number, and use FIND to get the position of the number.
Cell A1="20/11/2022 12:00:52 2 X 15.95 15.95 USD 57 5 689 5 689 1 4111 0 Amazing Lego Team"
=FIND(LOOKUP(10^15,MID(A1,ROW(INDIRECT("1:"&LEN(A1))),5)+0),A1)
=58
This looks to be in a space (or some other character) delimited text string. If you have a bunch of rows of data like this, use the Text to Columns feature on the Data tab. Use the Delimited option and then click the check box next to space (or other if it's something custom not in the available options) and Excel will split the data into columns for you.
For your string example, if you know that your 4 digit is started 58 char from the beginning of the string, use =MID(A1,58,4) A1 is the cell with your string.
MID function returns a specific number of characters from a text string, starting at the position you specify, based on the number of characters you specify.
=LET(t,TEXTSPLIT(A1,," "),
FIND(FILTER(t,(LEN(t)=4)*(ISNUMBER(--(t)))),A1))
It splits the string at every space and filters the result for being of length of 4 characters and not being an error if converted to number.
20/11/2022 12:00:52 2 X 15.95 15.95 USD 57 5 689 5 689 1 4111 0 Amazing Lego Team
I need to get the position of No 4111 in the above text string, As an excel beginner any help will be greatly appreciated. Thanks.
All of the Text Strings will have a 4 digit number like 4111 which i have to get the position for.
Have tried using this formula to get four digit number in another column, LOOKUP(10^15,MID(A1,ROW(INDIRECT("1:"&LEN(A1))),5)+0) but I am looking to get position instead.
I have tried using lookup but I could only go so far as a beginner.
Use the formula you provided =LOOKUP(10^15,MID(A1,ROW(INDIRECT("1:"&LEN(A1))),5)+0) to identify the number, and use FIND to get the position of the number.
Cell A1="20/11/2022 12:00:52 2 X 15.95 15.95 USD 57 5 689 5 689 1 4111 0 Amazing Lego Team"
=FIND(LOOKUP(10^15,MID(A1,ROW(INDIRECT("1:"&LEN(A1))),5)+0),A1)
=58
This looks to be in a space (or some other character) delimited text string. If you have a bunch of rows of data like this, use the Text to Columns feature on the Data tab. Use the Delimited option and then click the check box next to space (or other if it's something custom not in the available options) and Excel will split the data into columns for you.
For your string example, if you know that your 4 digit is started 58 char from the beginning of the string, use =MID(A1,58,4) A1 is the cell with your string.
MID function returns a specific number of characters from a text string, starting at the position you specify, based on the number of characters you specify.
=LET(t,TEXTSPLIT(A1,," "),
FIND(FILTER(t,(LEN(t)=4)*(ISNUMBER(--(t)))),A1))
It splits the string at every space and filters the result for being of length of 4 characters and not being an error if converted to number.
Question relates to Excel (Office365):
I am seeking a solution that will extract a number with a length of 4 digits from a string.
A couple of examples of the type of strings I am referring to are:
"16016KT 9999 SCT030"
"PROB30 0500 FG BKN001"
"MOD TURB BLW 5000FT TILL302300"
"INTER 6000 SHRA SCT015"
In each of the above strings there are a combination of letters and numbers of varying lengths and no set pattern.
The sequence of characters that I am interested in are the 4 digit numbers (in BOLD). Not, the 5000 in 5000ft.
The sequence of 4 digits is unique to all the strings I will be evaluating.
Thanks!
You may use:
=IFERROR(TEXT(FILTERXML("<t><s>"&SUBSTITUTE(A1," ","</s><s>")&"</s></t>","//s[.*0=0][string-length()=4]"),"0000"),"Non found")
On more recent versions of Excel, you may try:
=RegexpFind(A1, "\b[0-9]{4}\b", 0)
See here for how to activate regex support in Excel.
another solution:
=IFERROR(TEXT(UNIQUE(SEQUENCE(9999)/(FIND(" " & TEXT(SEQUENCE(9999),"0000") &" ",A2)>0),,1),"0000"),"")
Another option
In B1, formula copied down :
=IFERROR(TEXT(0+MID(A1,SEARCH(" ???? ",A1)+1,4),"0000"),"not found")
I have a string that contain results of the the SUM that have dynamic length that look like:
12345678
I have to display it with space separators:
12 345 678 //in looger
How can I do this with Google Script?
Thank you in advance!
Just formatted it with .00 at the end to use RegEx ('.' is a reference point). Then just sliced last three characters. Not elegant, but it works.
sum.toFixed(2).replace(/\d(?=(\d{3})+\.)/g, '$& ').slice(0,-3)
Vim removes zeros from in front of some digits when decrementing:
If I take a text file with the following:
a02
a03
a04
a05
a06
a07
a08
a09
a10
a11
And use ctrl+V to highlight the second and third columns, and then hit ctrl+X to decrement, I am left with:
a01
a02
a03
a04
a05
a06
a7
a8
a9
a10
I am running Vim version 7.4.1689 and I loaded it without my .vimrc via
$ vim -u NONE
This is happening because Vim will automatically recognize and convert octal values.
From the help (:h variables):
Conversion from a Number to a String is by making the ASCII representation of
the Number.
Examples:
Number 123 --> String "123"
Number 0 --> String "0"
Number -1 --> String "-1"
Conversion from a String to a Number is done by converting the first digits to
a number. Hexadecimal "0xf9", Octal "017", and Binary "0b10" numbers are
recognized. If the String doesn't start with digits, the result is zero.
Examples:
String "456" --> Number 456
String "6bar" --> Number 6
String "foo" --> Number 0
String "0xf1" --> Number 241
String "0100" --> Number 64
String "0b101" --> Number 5
String "-8" --> Number -8
String "+8" --> Number 0
Your values 02 through 07 are being recognized as valid octal values and preserved as such, decremented to octal 01 through 06.
When you reach 08 it is not a valid octal value. It is treated as the string 08, converted to decimal value 8, and decremented to 7. This happens again with 09, which ends up being 8.
The 10 and 11 values are decremented as decimal as well. Because 10 was decimal, not octal, you don't get a leading 0 in the resulting 9 value.
I'm not aware of a way to do what you want with the decrement command.
EDIT: After finding this answer, I tested this expression and it does what you are trying to do in this specific case:
:%s/\v[0-9]+/\=printf("%02d", substitute(submatch(0), '^0\+', '', 0)-1)/
I'm not sure whether this solves your general use case, because it's quite different from the original operation using a selection. But for the file you provided, it achieves the result you were after.
Dissecting this a bit to explain it:
First we start by calling the global sub command %s and passing the \v flag to turn on "very magic" mode. This may or may not change the behavior depending on your settings, but this is a public example, so it is included here to ensure that mode is active.
:%s/\v
Then, we find all the contiguous sequences of digits. This will find 02, 03, and so on from your example.
[0-9]+
Then in the replacement portion we have this command, which does the real work:
\=printf("%02d", substitute(submatch(0), '^0\+', '', 0)-1)
The substitute() function determines what the new value is. submatch(0) means to use the entire match. Using a pattern of ^0\+ and a replacement of (empty string) says to strip the leading zero from any number which has one. The 0 at the end isn't too important; it just says there are no flags to the substitute() function.
The result of the substitute command is a number. Say 02 has been stripped down to be 2. Using the - 1 at the end, we subtract 1 from that result (decrement).
Finally, this result is passed to the printf function. Using a format string %02d says to print the values as decimal, in 2-digit wide format, padding with leading zeroes.
If you want Vim to treat all numbers as decimals, you may want to add the following line to your .vimrc:
set nrformats=