Somewhat simple problem:
I need to turn a column A, which contains numbers with up to 1 decimal (20, 142, 2.5, etc.) to a string with a specific format, namely 8 whole digits and 6 decimal digits but without the actual decimal period, like so:
1 = 00000001000000
13 = 00000013000000
125 = 00000125000000
46.5 = 00000046500000
For what it's worth, the input data from column A will never be more than 3 total digits (0.5 to 999) and the decimal will always be either none or .5.
I also need for Excel to leave the zeroes alone instead of auto-formatting as a number and removing the ones at the beginning of the string.
As a makeshift solution, I've been using =CONCATENATE("'",TEXT(A1,"00000000.000000")), then copying the returning value and "pasting as value" where I actually need it.
It works fine, but I was wondering if there was a more direct solution where I don't have to manually intervene.
Thanks in advance!
=TEXT(A1*1000000,"0000000000000") I think that's what you mean.
Related
One of my Excel column of my board have to store numbers of 9 digits.
I'm looking for a solution to keep only the 9 last digits of any bigger number past in this specific column. It's only entire number.
Also if after formatting the number it appear that the number starts with 0 the 0 have to be kept. Is there another solution than adding an '0 at first ?
Here is what I already done : (i is the row number / Range01 is Range("A14:O400"))
If Len(Range01.Cells(i,5).value) = 9 Then
Range01.Cells(i,5).Interior.color = vbGreen
ElseIf Len(Range01.Cells(i,5).value) = 8 Then
Range01.Cells(i,5).value = "'0" & Range01.Cells(i,5).value
ElseIf Len(Range01.Cells(i,5).value) > 9 Then
????
Else
Range01.Cells(i,5).Interior.color = vbRed
End If
Thanks for the help.
The simplest way to get the last nine numbers of an integer is:
=MOD(A1,1000000000)
(For your information, that's one billion, a one with nine zeroes.)
If you're interested in showing a number with leading zeroes, you can alter the cell formatting as follows: (the format simply contains nine zeroes)
If you're interested in keeping the zeroes, you might need to use your number as a string, and precede it with a good number of repeated zeroes, something like:
=REPT("0",9-LEN(F8))&F8
Take the length of your number (which gets automatically converted into a string)
Subtract that from 9 (so you know how many zeroes you need)
Create a string, consisting of that number of zeroes
Add your number behind it, using basic concatenation.
You can simply use the math operator of modulus. If you want the last 9 digit you can write:
n % 10000000000
Where n is the number in the column.
In VBA:
MOD(n,1000000000)
I'm trying to create a CSV file of one of my customer's serial numbers. We print them as barcodes for them to use, and normally I'd use our barcode software to generate the numbers. However, we're using a different method of printing, and it requires a CSV/Excel file of all the numbers to be printed. The barcode is as follows:
MC100VGVA.
The last digit is a check digit created from the rest of the string.
Now, my problem comes with the "VGVA" bit. Column A is the prefix (MC), Column B is the number (100), Column C is the incrementing 4 characters (VGVA), and Column D is the check digit.
I need for the VGVA bit to increment alphanumerically. So, when it gets to VGVZ, I need it to go to VGW0. Then when it gets to VGZZ, it needs to go to VH00 and so on until they reach ZZZZ, in which the next digit would increase Column B to 101, and Column C would become 0000.
I've attempted to use the CHAR formula, as well as CONCATENATE, and MID. But, because I'm not well versed in these formulas, my attempts at editing them to work with 4 digits have been failing me.
I'm not opposed to using VBA if needed, but it's not something I've ever worked with, so you'll have to forgive any ignorance on my part.
Please let me know if you need more information. Thanks!
It looks like you are trying to create a new base, the one based on 27 digits (0 and all letter from 'A' to 'Z'). So I'd advise you to create a conversion from and to 27-digit system.
Let me first explain you what I mean in octal numbering (8 digits, from 0 to 7): in that system we start from (just some examples):
a=0011
b=1237
c=1277
The meaning of those numbers is:
a equals 0*8^3 + 0*8^2 + 1*8^1 + 1*8^0 = 9, so:
a+1 equals 10, and converting this to octal numbering yields:
0012
b equals 1*8^3+2*8^2+3*8^1+7*8^0 = 671, so:
b+1 equals 672, and converting this to octal numbering yields:
1240
c equals 1*8^3 + 2*8^2 + 7*8^1 + 7*8^0 = 703, so:
c+1 equals 704, and converting this to octal numbering yields:
1300
I propose to do exactly the same for your 27-digit system, with following example:
VGZZ equals 22*27^3 + 7*27^2 + 26*27^1 + 26 = 438857
VGZZ+1 equals 438858, and converting this to 27-digit numbering yields:
VH00
You can do this, using a VBA function you need to develop yourself. The converting from the string to the normal number is obvious, and in the other way around, you use =MOD(...,27^3) and other similar functions.
I believe I've found a non-VBA answer to this question, thanks to someone on another forum.
Here's what they suggested and it seems to be working perfectly:
B2
=B1+(C2="0000")
C2
=RIGHT(BASE(DECIMAL(C1,36)+1,36,4),4)
and maybe try this at D1
=MID("0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ-. $/+%",MOD(SUMPRODUCT(SEARCH(MID((A1&B1&C1),ROW($1:$99),1),
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ-. $/+%") )-99,43)+1,1)
I would like to print a series of floats with varying amounts of numbers to the left of the decimal place. I would like these numbers to exactly fill a padding with blank spaces, digits, and a decimal point.
Paraphrasing the data and code I have now
floats = [321.1234561, 21.1234561, 1.1234561, 0.123456, 0.02345, 0.0034, 0.0004567]
for number in floats:
print('{:>8.6f}'.format(number))
This outputs
321.123456
21.123456
1.123456
0.123456
0.02345
0.0034
0.000457
I am looking for a way to print the following in a for loop assuming I don't know the amount of digits that will be to the left of the decimal place and the number of digits to the left never exceeds the padding which is 8 for this example.
321.1234
21.12345
1.123456
0.123456
0.02345
0.0034
0.000457
Similar questions have been asked about printing floating points with a certain width but the width they were talking about appeared to be the precision rather than the total number of character used to print the number.
Edit:
I have added a number to the end of the list for the following reason. The use of the specifier 'g' with 7 significant figures was recommended by attdona. This prevents the padding from being exceeded for numbers greater than or equal to 1 but not for numbers less than 1 with precision greater than 6. Using {:>8.7g} instead gives
321.1234
21.12345
1.123456
0.123456
0.02345
0.0034
0.0004567
Where the only one that exceeds the padding is the newly added one.
Use the General format type specifier g:
'{:>8.7g}'.format(number)
reference: https://docs.python.org/3/library/string.html#format-specification-mini-language
Update: For small numbers this format fails to align correctly. In this case you may adopt a mixed approach, but keep in mind that very small numbers will round to zero
for number in floats:
fstr = '{:>8.7g}'.format(number)
if len(fstr) > 8:
fstr = '{:>8.6f}'.format(number)
print(fstr)
for i in floats:
print('{:>8}'.format(f'{i:{8}.{8-len(str(int(i)))-1}f}'.rstrip('0')))
321.1235
21.12346
1.123456
0.123456
0.02345
0.0034
I have a sheet with dates as MMDDYYY with no leading 0's if month number is single digit. For example, 1012018 or 12312018. Each record has a date, and each date is either 7 or 8 characters in length.
Here is the code I am using to convert the numbers to dates:
if Text.Length([ContractDate]) = 7
then
Text.Range([ContractDate],0,1)&"/"&Text.Range([ContractDate],1,2)&"/"&Text.Range([ContractDate],4,4)
else
Text.Range([ContractDate],0,2)&"/"&Text.Range([ContractDate],2,2)&"/"&Text.Range([ContractDate],4,4)
The code works fine for the "else" condition but I am getting error "Expression.Error: The 'count' argument is out of range. Details: 4" for all records where Text.Length() = 7. I verified this by adding a second column to get Length of ContractDate.
What am I missing?
EDIT: Problem Solved - I'm an idiot. I was getting an error because in the "then" condition, I am extracting a substring of (4,4) from a value that only has Len=7. I can't get 4 characters out of a 7 character string when starting at index of 4.
I know you found the issue with your code, but worth pointing out some things that might be good to know.
Text.Range with no character count will pull in all characters past the start point (so Text.Range([ContractDate], 4) would work for both).
Text.Middle operates like Text.Range but will not cause an error if you select a range that expands past the size of the string. This can be useful if for some reason you were dealing with variable size strings where you need a specific number of characters up to a limit past a certain position.
You could also use Text.PadStart([ContractDate], 8, "0") to pad the 7 length strings with a 0 at the start, and avoid the need for a conditional check all together.
I am putting a string into excel. The string is often only numeric digits but can have alpha characters or hypens etc.
When I don't set the number format or set it like this
(Where xlSheet(0) is Excel.Worksheet)
xlSheet(0).Columns("N:N").EntireColumn.Columns.NumberFormat = "#"
It outputs in scientific notation.
When I use this code:
xlSheet(0).Columns("N:N").EntireColumn.Columns.NumberFormat = "0"
It rounds up the number to the nearest 100,000 so that the last five digits are 0's when they shouldn't be.
Should be: 1539648751235678942
But is: 1539648751235600000
The cells that have a hyphen or a letter aren't affected and work fine.
Any help would be greatly appreciated.
EDIT:
I add the data like this:
I loop through and put in xlSheet(0).Cells(i, 14) = rs!value_number
Where rs is my ADODB.Recordset
EDIT2: Herbert Sitz got it by adding an apostrophe before the text! Thanks everyone.
I think problem is that the number you're trying to enter can't be accommodated exactly by Excel. Excel has limitations on what numbers it display/represent because of the way numbers are stored internally. In Excel's case numbers are limited to 15 digit precision (see http://office.microsoft.com/en-us/excel-help/excel-specifications-and-limits-HP010073849.aspx ), which is not enough to represent your number.
You can enter the number as a string ("152..42") and all digits will be displayed, but you won't be able to perform exact mathematical operations with it.
For numbers, Excel can only handle 15 significant digits.
If you want to store a number that is more than 15 digits long without losing data, you have to store the data as text.
Doing what you've been doing will resolve the issue:
You can do either of the following to add your numbers as text:
xlSheet(0).Cells(i, 14).Numberformat = "#"
xlSheet(0).Cells(i, 14) = rs!value_number
Or
xlSheet(0).Cells(i, 14) = "'" & rs!value_number