Formula to Extract Numbers from an Alphanumeric String

gonesalsa

New Member
Joined
May 5, 2014
Messages
7
Hello
I'm wondering, if there is formula to extract numbers from an alphanumeric string. The numbers will be 5 digits, but the placement within the string will vary. I was using MID, but I have to keep changing the formula based on the placement of the numbers in the string and it's not very efficent. Thanks!

Examples are:
22047 - Cjljj
S022509 Chhheefe 5/12/14
Self Pay # 022597 jjjkjkj
jljlj cheque #022378

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</tbody>



<colgroup><col style="width: 170pt; mso-width-source: userset; mso-width-alt: 8265;" width="226"> <tbody>
</tbody>





<tbody>
</tbody><colgroup><col></colgroup>



<tbody>
</tbody><colgroup><col></colgroup>
 

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What would the answers be for the examples above?

Are you omitting the 0's(Zeroes)
 
Upvote 0
See which of these gives your desired result


Excel 2010
AB
122047 - Cjljj22047
2S022509 Chhheefe 5/12/1422509
3Self Pay # 022597 jjjkjkj22597
4jljlj cheque #02237822378
Sheet12
Cell Formulas
RangeFormula
B1{=MID(A1,MATCH(TRUE,IF(LEFT(MID(A1,ROW(INDIRECT("1:" &LEN(A1)-4)),5),1)<>"0",IF(LEFT(MID(A1,ROW(INDIRECT("1:" &LEN(A1)-4)),5),1)<>" ",ISNUMBER(1*MID(A1,ROW(INDIRECT("1:" &LEN(A1)-4)),5)))),0),5)}
Press CTRL+SHIFT+ENTER to enter array formulas.


OR


Excel 2010
AB
122047 - Cjljj22047
2S022509 Chhheefe 5/12/1402250
3Self Pay # 022597 jjjkjkj02259
4jljlj cheque #02237802237
Sheet13
Cell Formulas
RangeFormula
B1{=MID(A1,MATCH(TRUE,IF(LEFT(MID(A1,ROW(INDIRECT("1:" &LEN(A1)-4)),5),1)<>" ",ISNUMBER(1*MID(A1,ROW(INDIRECT("1:" &LEN(A1)-4)),5))),0),5)}
Press CTRL+SHIFT+ENTER to enter array formulas.
 
Upvote 0
Thank you so much! The first formula works.

Can you explain this formula? I want to learn how to solve these problems myself.
 
Upvote 0
Thank you so much! The first formula works.

Can you explain this formula? I want to learn how to solve these problems myself.

Let me give this a try although I might leave out some parts you deem as important :)

Feel free to ask for further clarification, I will be sure to provide the ones i have answers to

EXPLANATION

The Logic behind the formula is to pick all consecutive 5 characters(could be numbers) that can be extracted from the string 1.e from the 1st character, pick 5 characters, from the second character pick 5 and on and on which in your case would be "22047","2047 ","047 - ","47 - C", "7 - Cj"
That’s what this section does MID(A1,ROW(INDIRECT("1:" &LEN(A1)-4)),5)

What you want to check now is to ensure that the strings9all the set of 5 characters) it has selected only contains numbers, hence i used ISNUMBER which tells me true for numbers and false for non-numbers, the new formula at this stage gives us ISNUMBER(1*MID(A1,ROW(INDIRECT("1:" &LEN(A1)-4)),5)) and it will return TRUE for 22047, TRUE for "2047", FALSE for "047 - "

What I then do is to use the MATCH Function to get the first instance of TRUE 'cos we would be having something like
{TRUE,TRUE,FALSE,FALSE........)

The match in this case gives us 1 as the first instance of TRUE meaning the first string in the set of strings meets our criteria which is 5 digits and they are all numbers
So you wrap that in the MID to say, give me five characters starting from where the match starts from an give me 5 characters
That’s the basic thing(I hope I have explained well)

But you will see I use like two IF statements in there
1. One IF is to ensure that we don’t have a string of 5 numbers starting with 0, cos you seem to want the numbers but not starting with the zero
so IF(LEFT(MID(A1,ROW(INDIRECT("1:" &LEN(A1)-4)),5),1)<>"0"

2. second IF to be sure the first or last character/number in the string is not a SPACE cos "2047 " will give you true when i use ISNUMBER, so to avoid that I used an IF i.e. IF(LEFT(MID(A1,ROW(INDIRECT("1:" &LEN(A1)-4)),5),1)<>" " but this only takes care of the first character bot being a space 'cos " 2047" will give TRUE but you don’t want this

So the summary of the whole formula is
1. PICK 5 characters starting from position 1,2,3.......length(string)-4 so you don’t pick characters that don’t exist in the string, that’s why i stop at len(string)-4
2. Check that the first character is not a 0 and that the first character is also not a SPACE, once these 2 conditions are met then
3. USE MATCH to give me the first set of 4-characters that satisfy 2 above and are all numbers
4. Once you get the starting position of the SET that satisfies the condition, pick 5 characters.
 
Upvote 0
Thank you so much! The first formula works.
Maybe it works... depending... can you ever have any "short" numbers mixed in with text before the 5-digit number you want? I ask because the formula will return a wrong value for something like this...

123E4 - lf Pay # 022597 jjjkjkj
 
Upvote 0
Maybe it works... depending... can you ever have any "short" numbers mixed in with text before the 5-digit number you want? I ask because the formula will return a wrong value for something like this...

123E4 - lf Pay # 022597 jjjkjkj

Nice point, Mr Rothstein. Didn't see that one coming :) I hope he doesn't have data like that.

Am sure you can give us a formula that would get us the desired result
 
Upvote 0
Here is a variation of Momentman's formula that also handles the example from Rick Rothstein. Since Excel evaluates 123E4 as an exponential, my array formula first changes all E's to spaces.
Code:
=MID(A1,MATCH(TRUE,MID(LEFT(SUBSTITUTE(A1,"E"," "),LEN(A1)),ROW(INDIRECT("1:" &LEN(A1)-4)),5)*1>=10000,0),5)
 
Upvote 0
Here is a variation of Momentman's formula that also handles the example from Rick Rothstein. Since Excel evaluates 123E4 as an exponential, my array formula first changes all E's to spaces.
Code:
=MID(A1,MATCH(TRUE,MID(LEFT(SUBSTITUTE(A1,"E"," "),LEN(A1)),ROW(INDIRECT("1:" &LEN(A1)-4)),5)*1>=10000,0),5)

It should also work without the left in there

Code:
=MID(A1,MATCH(TRUE,MID(SUBSTITUTE(A1,"E"," "),ROW(INDIRECT("1:" &LEN(A1)-4)),5)*1>=10000,0),5)

Don't forget CTRL+SHIFT+ENTER
 
Last edited:
Upvote 0

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