I have a number set as shown below;
140080066
90110439
110100255
110090248
10120539
150130043
150130045
40120258
10130133
150130142
50120473
50120521
40130224
Some numbers are 8 order some are 9. I want to add "0" (Zero) to left of the number which has 8 orders. After the process the numbers should be like;
010120539
110100255
Thanks for your answers.
You can use the formula
=IF(LEN(A1)=8,0&A1,A1)
in any column and set format cell to text or set format cell in column as number and put 000000000 in Format code field
I think more context is required.
Since you tagged "libreoffice" you wanna do it with a macro?
If so, you can use the len-operator to test if the value is 8 or 9 digits long.
More info:
http://www.debugpoint.com/2014/11/string-processing-in-libreoffice-calc-macro-with-examples/
Related
OBJECTID E201701 G201701 E201702 G201702 E201703 G201703 E201704 G201704
1 2 5 6 5 -1 NaN 6 5
I want to find if a particular entry (i.e. objectid) has negative value in any of variable (e201701, etc). I tried;
IF(AND(B2>0,C2>0,D2>0,E2>0,F2>0,G2>0,H2>0),1,0)
but it just give me 1 irrespective.
Your formula is asking "Is thisCell > (Greater than) thisCell?" Try using this. Also, change the AND to an OR because AND is going to check all values and if ALL of them are returning true then it will return as such. OR will check EACH individually.
IF(OR(B2<0,C2<0,D2<0,E2<0,F2<0,G2<0,H2<0),1,0)
Which is asking "Is thisCell > (Less than) thisCell?"
If you use excel 2013 or later then you can use below formula. This formula will also give you position number where the negative value is. As your sample data it will return 6 because negative value is in 6th position.
=AGGREGATE(15,6,COLUMN(A1:I1)/(A1:I1<0),1)
For sample same results as #xTwisteDx but shorter and easy to extend (eg to include G201704 just change H to I) :
=1*(MIN(B2:H2)<0)
I have an Excel spreadsheet with over 2000 entries:
Field B1: CustomerID as 000012345
Field B2: CustomerID as 0000432
Field C1: CustomerCountry as DE
Field C2: CustomerCountry as IT
I need to build codes 13 digits long including "CustomerCountry" + "CustomerID" without leading 0 + random number (can be 6 digits, more or less, depends in length of CustomerID).
The results should be like this: D1 Code as DE12345967895 or D2 Code as IT43274837401
How to do it with Excel functions?
UPDATED:
I tried this one. My big problem is to say that random number should be long enough to get 13 characters in all. Sometimes CustomerID is just 3 or 4 digits long, and concatenation of three variables can be just 10 or 9 characters. But codes have to be always 13 characters long.
Use & to concatenate strings.
Use VALUE(CustomerID) to trim the leading zeroes from the ID
Use RAND() to add a random number between 0 and 1 or RANDBETWEEN(x,y) to create one between x and y.
Combine the above and there you are!
If you always want 13 digits you can use LEFT(INT(RAND()*10^13);(13-LEN(CustomerCountry)-LEN(VALUE(CustomerID)))) for the random number to ALWAYS be the right length.
total formula
= CustomerCountry
& VALUE(CustomerID)
& LEFT(INT(RAND()*10^13);(13-LEN(CustomerCountry)-LEN(VALUE(CustomerID))))
=C1 & TEXT(B1,"0") & RIGHT(TEXT(RANDBETWEEN(0,99999999999),"00000000000"),11 - LEN(TEXT(B1,"0")))
that should do it
I don’t understand what is where and OP has accepted answer so have not bothered testing:
=LEFT(RIGHT(C1,2)&VALUE(MID(B1,15,13))&RANDBETWEEN(10^9,10^10),13)
(but I might revert to this if no one else picks the flaws in it first!)
I am trying to convert a single column of numbers in excel to multiple depending on the content.
e.g. Table 1 contains 1 column that contains 1 or more numbers between 1 and 11 separated with a comma. Table 2 should contain 11 columns with a 1 or a 0 depending on the numbers found in Table 1.
I am using the following formula at present:
=IF(ISNUMBER(SEARCH("1",A2)),1,0)
The next column contains the following:
=IF(ISNUMBER(SEARCH("2",A2)),1,0)
All the way to 11
=IF(ISNUMBER(SEARCH("11",A2)),1,0)
The problem with this however is that the code for finding references to 1 also find the references to 11. Is it possible to write a formula that can tell the difference so that if I have the following in Table 1:
2, 5, 11
It doesn't put a 1 in column 1 of Table 2?
Thanks.
Use, for list with just comma:
=IF(ISNUMBER(SEARCH(",1,", ","&A2&",")),1,0)
If list is separated with , (comma+space):
=IF(ISNUMBER(SEARCH(", 1,", ", "&A2&",")),1,0)
A version of LS_dev's answer that will cope with 0...n spaces before or after each comma is:
=IF(ISNUMBER(SEARCH(", 1 ,",", "&TRIM(SUBSTITUTE(A2,","," , "))&" ,")),1,0)
The SUBSTITUTE makes sure there's always at least one space before and after each comma and the TRIM replaces multiple spaces with one space, so the result of the TRIM function will have exactly one space before and after each comma.
How about using the SUBSTITUTE function to change all "11" to Roman numeral "XI" prior to doing your search:
=IF(ISNUMBER(SEARCH("1",SUBSTITUTE(A2, "11", "XI"))),1,0)
If you want to eliminate "11" case, but this is all based on hardcoded values, there should be a smarter solution.
=IF(ISNUMBER(SEARCH(AND("1",NOT("11")),A2)),1,0)
I have column which consists of text including 13 digit number. How Can I keep 10 digit number and delete all other text?
Please help me I am new to excel..
thanks in advance
Edited:
Cell Format
[6/11/2013 3:26:37 PM] 1234503776599, ksdfl 038ddf63Ksdf)
[6/12/2013 3:26:37 PM] 0234503664599, ksdfadssdfl 038ddf6dfsd3Ksdf)
[6/13/2013 3:26:37 PM] 7234503666099, 45sdsdfadssdfl 03845ddf6dfsd3Ksdf)
Here, In second column I want to keep 13 digit and delete all text after that 13 digit.
Is the number always at the beginning of the cell? If so you can use =LEFT(cell_ref, 10) to extract the first 10 characters, which in this case are numbers and will be treated as such by the spreadsheet.
I'm guessing you will need something like this
=MID(A1,32,FIND(",",A1,32) -32)
MID will get text from the middle of a string
FIND will get the location of the first comma
Thinking that the 13 digit number is in the middle of the string, with 9 spaces to the left and a comma to the right we can use the following formula in a separate cell:
=MID(A1,LEN(LEFT(A1,FIND("]",A1)+1))+9,13)
The LEN function determines the length of the part up to and including ] (big bracket). Then the number 9 is added to include nine spaces. Then the number 13 is the length of the 13 digit number. All these are used as parameters for the MID function.
After you get the result just drag down using the autofill handle to get the mid 13 digit number for all the rows.
=LOOKUP(10^11,MID(A1,ROW(INDIRECT("1:"&LEN(A1)-9)),11)+0)
This works for me.
To extract 6 continuous numeric digits, use the below code:
=LOOKUP(10^6,MID(A1,ROW(INDIRECT("1:"&LEN(A1)-5)),6)+0)
Replace 6 in the above code with the number of digits that are required to be extracted.
I have price column in which the prices are displayed in 3 decimals and 4 decimals places, for example 123.456 or 123.4357.
So irrespective of the number of decimal places i want the value to be with only two decimals.
So i am selecting the column and in the VBA i am using Selection.NumberFormat = "0.00"
Which is resulting me the rounded value i.e when i format 123.456 and 123.4357 to 0.00 i am getting 123.46 and 123.44 but i want it to be 123.46 and 123.43.
So just wondering is there a way we can just trim the values instead of rounding.
Please give me some sample examples.
Thank you in advance.
Excel has a built-in function, trunc, that should do the trick for you.
This is what I placed in a1:b2.
123.456 =trunc(A1,2)
123.4357 =trunc(A2,2)
This will display
123.456 123.45
123.4357 123.43
You could treat the result as a string and extract the number of characters you need as in:
dim a as double
dim s as string
a = 123.4357
s = MID(a,1,FIND(".",a))&MID(a,FIND(".",a)+1,2)
I don't believe there is a built in mask that will truncate without rounding, you could instead use another column containing =TRUNC(A1, 2).