I have a spreadsheet called "MTM Lookup", this is downloaded from a specific site, in column D of this spreadsheet are values, but these values have a formula attached to them. The formula rounds these values to 0. The formulas don't round on a specific cell. They look like this =ROUND(35370.6708773751,0) or =ROUND(48368.0427161566,0). I need the values to come through with all the decimals or rounded to 10 decimals but cannot get this to happen, I can remove the formula and leave the value but it is rounded to zero. Please could anyone assist with some simple vba to either remove the =round(,0) or replace the 0 to 10 ie round(x,10).
I don't see any problem in the formula you provided.
When I put
=ROUND(35370.6708773751,0)
to a cell, I correctly see 35371 in the cell.
There are, however, two things in play here.
Formula
Cell Format
For example, when I enter the following value to the cell:
=ROUND(35370.6708773751,10)
I do see 35370.67088 as a result (after rounding to 10 places, cell format rounds it again to 5 decimal places)
I don't know why entering the value without any formula shows you 0, but this leads me to the same suspscion, i. e. that the problem is in the cell format.
You can check it by right clicking on the cell > Format Cells (in office 2010 at least) or programatically, using the following, for example:
sheets("MTM Lookup").range("A:1").numberformat = "0.0000000000"
Hope that it helps.
Related
I have a data frame in R that I want to analyse. I want to know how many specific numbers are in a data frame column. for example, I want to know the frequency of number 0.9998558 by using
sum(deviation_multiple_regression_3cell_types_all_spots_all_intersection_genes_exclude_50_10dec_rowSums_not_0_for_moran_scaled[,3]== 0.9998558)
However, it seems that the decimal shown is not the actual one (it must be 0.9998558xxxxx) since the result I got from using the above command is 0 (the correct one should be 3468). How can I access that number without knowing the exact decimal numbers so that I get the correct answer? Please see the screenshot below.
The code below gives the number of occurrences in the column.
x <- 0.9998558
length(which(df$a==x))
If you are looking for numbers stating with 0.9998558, I think you can do it in two different ways: working with data as numeric or as character.
Let x be your variable:
Data as character
This way counts exactly what you are looking for
sum(substr(as.character(x),1,9)=="0.9998558")
Data as numeric
This will include all the values with a difference with the reference value lower than 1e-7; this may include values not starting exactly with 0.9998558
sum(abs(x-0.9998558)<1e-7)
You can also "truncate" the numbers in your vector and compare them with the number you want. Here, we write 10^7 because 7 is the number of decimals you want to compare.
sum(trunc(x*10^7)/10^7)==0.9998558)
I have salary data table from 10 years period. Every column has properly set data type (date for "B", number for "C" and "E".
I'm trying to write a formula to calculate average salary for every year. In column "E" I've manually entered all possible years and in column "F" should be an yearly average, according to year from "E".
So, my best try is this formula: =AVERAGEIF(YEAR(B2:B133);"="&E2;C2:C133)
Trying so calculate an average from column C, where year in date from column B equals a year in column E
But all I get is an error Err:504. Figured out, that problem is in YEAR(interval) part, but can't get what exactly...
Can someone point that out?
Thank you!
There are actually many possibilities to solve this.
#JvdV answer;
using an array formula with #JvdV solution;
using an array formula with a combination of AVERAGE() and IF();
using the SUMPRODUCT() function;
and surely many other solutions that I don't know about!
Please beware: I use , instead of ; as formula separator, according to my locale; adapt to your needs.
A side note on "array formulas"
This kind of formulas are applied by mandatory pressing the Ctrl + Shift + Enter key combination to insert them, not only Enter or Tab or mouse-clicking elsewhere on the sheet.
The resulting formula is shown between brackets {}, which are not inserted by the user but are automatically shown by the software to inform that this is actually an array formula.
More on array formulas i.e. on the LibreOffice help system.
Usually you cannot drag and drop array formulas, you have to copy-paste them instead.
Array formula with #JvdV solution
The solution of JvdV could be slighly modified like this, and then inserted as an array formula:
=AVERAGEIFS(C$2:C$133,YEAR($B$2:$B$133),"="&E2)
When you insert this formula with the Ctrl + Shift + Enter key combination, the software puts the formula into brackets, so that you see it like this: {=AVERAGEIFS(C$2:C$133,YEAR($B$2:$B$133),"="&E2)}
You cannot simply drag the formula down, but you can copy-paste it.
Array formula with a combination of AVERAGE() and IF():
For your example, put this formula in cell F2 (for the year 2010):
=AVERAGE(IF(YEAR($B$2:$B$133)=E2,$C$2:$C$133))
When you insert this formula with the Ctrl + Shift + Enter key combination, the software puts the formula into brackets, so that you see it like this {=AVERAGE(IF(YEAR($B$2:$B$133)=E2,$C$2:$C$133))}
You cannot simply drag the formula down, but you can copy-paste it.
SUMPRODUCT() formula:
My loved one...
Plenty of resources on the web to explain this formula.
In your situation, this would give:
=SUMPRODUCT($C$2:$C$133,--(YEAR($B$2:$B$133)=E2))/SUMPRODUCT(--(YEAR($B$2:$B$133)=E2))
This one you can drag down to your needs.
Unfortunately AVERAGEIF() expects a range reference instead of a calculated array. Therefor it will error out. That's the theory at least for Excel, and I expect this to be the same for LibreCalc.
One way around it is using the AVERAGEIFS() function and check against first and last days of the year, for example:
=AVERAGEIFS(C$2:C$133;B$2:B$133;">="&DATE(E2;1;1);B$2:B$133;"<="&DATE(E2;12;31))
Drag the formula down.
I don't want the display format like this: 2.150209e+06
the format I want is 2150209
because when I export data, format like 2.150209e+06 caused me a lot of trouble.
I did some search found this function could help me
formatC(numeric_summary$mean, digits=1,format="f").
I am wondering can I set options to change this forever? I don't want to apply this function to every variable of my data because I have this problem very often.
One more question is, can I change the class of all integer variables to numeric automatically? For integer format, when I sum the whole column usually cause trouble, says "integer overflow - use sum(as.numeric(.))".
I don't need integer format, all I need is numeric format. Can I set options to change integer class to numeric please?
I don't know how you are exporting your data, but when I use write.csv with a data frame containing numeric data, I don't get scientific notation, I get the full number written out, including all decimal precision. Actually, I also get the full number written out even with factor data. Have a look here:
df <- data.frame(c1=c(2150209.123, 10001111),
c2=c('2150209.123', '10001111'))
write.csv(df, file="C:\\Users\\tbiegeleisen\\temp.txt")
Output file:
"","c1","c2"
"1",2150209.123,"2150209.123"
"2",10001111,"10001111"
Update:
It is possible that you are just dealing with a data rendering issue. What you see in the R console or in your spreadsheet does not necessarily reflect the precision of the underlying data. For instance, if you are using Excel, you highlight a numeric cell, press CTRL + 1 and then change the format. You should be able to see full/true precision of the underlying data. Similarly, the number you see printed in the R console might use scientific notation only for ease of reading (SN was invented partially for this very reason).
Thank you all.
For the example above, I tried this:
df <- data.frame(c1=c(21503413542209.123, 10001111),
c2=c('2150209.123', '100011413413111'))
c1 in df is scientific notation, c2 is not.
then I run write.csv(df, file="C:\Users\tbiegeleisen\temp.txt").
It does out put all digits.
Can I disable scientific notation in R please? Because, it still cause me trouble, although it exported all digits to txt.
Sometimes I want to visually compare two big numbers. For example, if I run
df <- data.frame(c1=c(21503413542209.123, 21503413542210.123),
c2=c('2150209.123', '100011413413111'))
df will be
c1 c2
2.150341e+13 2150209.123
2.150341e+13 100011413413111
The two values for c1 are actually different, but I cannot differentiate them in R, unless I exported them to txt. The numbers here are fake numbers, but the same problem I encounter very day.
I am faced with rewriting an Excel project in R. I see a table in which a cell {= TABLE (F2, C2)} is shown. I understand how to create a Table like this (What-If Analysis, Data Table...).
As I have to understand this to rewrite in R, how can I find the original formula which stands behind that cell?
EXAMPLE: I have created a Data Table as shown here and the sheet looks like this:
In my case, I don't know how the sheet was created, and I want to know the initial formula. Now this is shown as {=TABLE(,C4)}.
(In the example I know the answer, it is in the cell (D10), but where is reference for this cell in Data Table?)
I'm using Excel 2007 but have no reason to believe things differ in other versions.
#Stanislav was right to reject my comment suggestion that TABLE was a name; it is an EXCEL function. But it is a very strange function :-}
There isn't any help on the TABLE function in the local help, it isn't listed in "List of worksheet functions (alphabetical)".
You can't manually enter or edit the TABLE function; error "That function is not valid".
Copy/Pasting cells containing the TABLE function pastes their values, not their formulae, even when you specify Paste Special > Formulas
You can't insert rows/columns immediately above/left of cells containing the TABLE function; error "Cannot change part of a data table".
Pace #pnuts using Formulas > Formula Auditing cells containing the TABLE function shows no precedents and no cells show them as dependents. Although in a VBA sheet auditing tool which I use the Range.DirectDependents Property finds the "formula range" dependent on the "margin" cells containing the formulas, but not those containg the values (see below for explanation of those terms).
I haven't been able to find anything I regard as decent documentation of TABLE(). I have found lots of illustrations of how to produce and use that function, but nothing clearly specifying the arguments and result. The best I've found is https://support.office.com/en-us/article/Calculate-multiple-results-by-using-a-data-table-e95e2487-6ca6-4413-ad12-77542a5ea50b. I'd be pleased if anyone can point me to better documentation.
I deduce the bahaviour as described here:
TABLE(Rowinp,Colinp) is an array formula in a contiguous array of cells. I'll refer to that contiguous array as the "formula range" of the data table.
The cells immediately above/left of the formula range are also part of the data table, even though they do not contain a TABLE() function and can be edited; I'll refer to those cells as the "margins" of the data table.
Rowinp and Colinp must be blank or references to single cells.
Rowinp and Colinp must be different (or error "Input cell reference is not valid"), they must not both be blank.
The values in the formula range are calculated by taking formula(s) from the margin(s) and substituting references to Rowinp and/or Colinp with values from the margin(s).
There are three mutually exclusive possibilities, corresponding to Rowinp blank or not.
TABLE(Rowinp, ) Colinp blank. The formula is that in the left margin of the same row with instances of Rowinp replaced by values from the upper margin of the same column.
TABLE( , Colinp) Rowinp blank. The formula is that in the top margin of the same column with instances of Colinp replaced by values from the the left margin of the same row.
TABLE(Rowinp, Colinp) Neither blank. The formula is that in the cell at the intersection of the left and top margins with instances of Rowinp replaced by values from the upper margin of the same column and instances of Colinp replaced by values from the the left margin of the same row.
I think that should let you work out what the effective formula is in each cell of the formula range.
But I wouldn't be surprised to learn that any of the above is wrong :-0
I welcome pointers to anything more authoritative.
I think in your example the F2 and C2 are effectively only the addresses of parameters for a function (TABLE) where that may be located anywhere, with the associated formula in the table's top left cell.
So I suggest go to C2, FORMULAS > Formula Auditing and click Trace Dependents, repeat for F2 and see where the arrows converge.
I am making an editor for a field with numbers. I tried a text field, but since it's a Number datatype coming in, it didn't go smoothly -- despite recasting strings as numbers etc.. it kept giving me NaN as the value. So I decided it would be best to go with a numeric stepper.
When I initially loaded it up it would drop all my decimals and only display my numbers as integers. I changed the stepIncrement to 0.1 and now it does show the decimals (a weird requirement imo).. but when I step up it occasionally gives me a value like '17.700000000000003' when I would expect 17.7. All of the numbers in my data have a single decimal place. I know I can write a dataformatter, but it seems like it shouldn't be necessary in this situation.
Is there another way I could deal with this?
You've stumbled upon the compromise of trying to represent decimal numbers in floating point binary formats like IEEE 754. Not all decimal numbers can be exactly represented. You can read up on this issue in great detail here:
http://en.wikipedia.org/wiki/Floating_point#Representable_numbers.2C_conversion_and_rounding
You can use Number.toFixed(fractionDigits:uint) to display to an arbitrary number of decimal places.
You can use the valueFormatFunction which takes the numeric value and formats it to a string. You will need to set explicit widths on your numeric steppers to make they fit though.
in your MXML
<s:NumericStepper valueFormatFunction="stepperFormatter"/>
in your script
protected function stepperFormatter(newValue:Number):String
{
return Math.ceil(newValue).toString()
}