I have a row which pulls back a Decimal if (row["Time"] and I need an statement which divides the output by 60.
Still quite a novice do apologies.
The answer for this question is:
(row["TIME"]) / 60
Related
I apologize for how I worded the title, but I am having a hard time really distilling my issue into a snappy soundbite.
My problem is this: Using the data below, for each User ID I need to find each instance that a date differs from a previous one by 90 or more days. BUT then, once 90 day difference is found, it needs to "reset" and look for dates 90 or more days from that.
Basically, I want a formula that can reproduce the Result column in the below table.
For example, note how for User ID 1 that the first 8/30 date yields a positive result, but neither the second 8/30 nor the 9/21 do, since they are both less than 90 days distant from 8/30. But then 12/26 does yield a positive result, since it is 118 days past the last positive result.
I really appreciate any help you can offer.
Thanks!!
Eq#1: =DATEDIF(B3,C3,"D")
Eq#2: =IF(E3>90,"Do a little dance","Frowny Face")
I want to round up the minutes of my worktime in ten-steps (e.g 12 -> 20) in MS Excel.
I know the formula for doing so in Excel is (in German): =AUFRUNDEN(NUMBER;-1)
But problematically Excel sometimes does 10 -> 10 and othertimes 10 -> 20.
Thanks for sharing file. I was able to give an answer because of it :)
Is not an error. That annoying 10 is really 10,0000000000001. Yes, with 13 decimals, but you are using ROUNDUP so it gets rounded up to 20.
If I change format in that column to show a lot of decimals, you will get this:
So is not a whole number, and that tiny part (13 decimals!!!) makes Excel to round up to 20 instead of 10.
Easy way would be making sure the value stored in the cell is an integer, with a normal round. In column G your formula is:
=IF(F2;(F2-E2)*1440;D2*60)
Replace it with:
=ROUND(IF(F2;(F2-E2)*1440;D2*60);0)
And now it works as you want:
Hope you can adapt this to your needs.
I need to analyse empirical research data:
Given the following data:
The explanation: 5 people answered Question 1 with "entirely agree", which is equivalent to a score of 7. Zero people answered Question 1 with "entirely disagree", which would be a score of 0.
I want now to calculate the number of responses strictly less than the Median and the number of answers equal to the Median.
The result should look like this (the Median is given thanks to Research data analysis in Excel: Median of x times column value)
I tried with the functions =INDIRECT and =INDEX; without success.
Would be great, if someone could help me. Thank you.
Use SUMIF:
Below:
=SUMIF($B$6:$H$6,"<" & I2,B2:H2)
Equal
=SUMIF($B$6:$H$6,"=" & I2,B2:H2)
A working, but not a better solution is to sum up the row conditionally:
the highlighted cell has this formula:
=IF(B6<I2,B2,0)+IF(C6<I2,C2,0)+IF(D6<I2,D2,0)+IF(E6<I2,E2,0)+IF(F6<I2,F2,0)+IF(G6<I2,G2,0)+IF(H6<I2,H2,0)
I hope it helps.
I keep a running avg of kids grades in Excel over a 2 week period. The way i have the code now
AVERAGE(OFFSET('1'!E4,COUNTA('1'!D:D)-29,):OFFSET('1'!E4,COUNTA('1'!D:D),))
it returns an error if i don't have 2 weeks of data. I found a way around this by doing this
=IFERROR(AVERAGE(OFFSET('1'!E4,COUNTA('1'!D:D)-29,):OFFSET('1'!E4,COUNTA('1'!D:D),)),IFERROR(AVERAGE(OFFSET('1'!E4,COUNTA('1'!D:D)-28,):OFFSET('1'!E4,COUNTA('1'!D:D),)),IFERROR(AVERAGE(OFFSET('1'!E4,COUNTA('1'!D:D)-27,....
Im Sure there is a better way to do this any help would be appreciated.
As pointed out in the comments - it's difficult to give a definitive answer without some more knowledge - it's not easy to tell, for instance, where the numeric data starts (E4 or E5)?
Firstly you can simplify your original formula which appears to AVERAGE the last 30 rows of data (not sure how 30 rows equates to 2 weeks of data) - you can do that with just:
=AVERAGE(OFFSET('1'!E4,COUNTA('1'!D:D)-29,0,30))
Now I assume the error comes about when COUNTA('1'!D:D) is < 29 so you can simply add an If function which AVERAGES all the data if that function returns a number < 29, i.e.
=IF(COUNTA('1'!D:D)<29,AVERAGE('1'!E4:E33),AVERAGE(OFFSET('1'!E4,COUNTA('1'!D:D)-29,0,30)))
That formula may need some small adjustments to cater for the specifics of your layout but the general approach is valid
In Excel 2007, I want the differences for the following string date/times:
A B C
1 Date1 Date2 Difference of A and B
2 2009.11.28 01:25:46:0287 2009.11.28 01:25:46:0287 ?
3 2009.11.28 01:25:46:0443 2009.11.28 01:25:46:0443 ?
I want the differences by converting strings to date/time and then the results as differences of two converted date/times.
A rather long-winded way to calculate zero (for the examples):
=SUBSTITUTE(LEFT(A2,10),".","/")+MID(A2,12,8)+RIGHT(A2,4)/86400000-(SUBSTITUTE(LEFT(B2,10),".","/")+MID(B2,12,8)+RIGHT(B2,4)/86400000)
By special request and very slightly shorter:
=SUBSTITUTE(LEFT(A2,10),".","/")+REPLACE(RIGHT(A2,13),9,1,".")-(SUBSTITUTE(LEFT(B2,10),".","/")+REPLACE(RIGHT(B2,13),9,1,"."))
I couldn't come up with a really nice way of doing this... hopefully someone else will. Having said that, the following may give you what you need.
To convert the main date part, use the following formula (this assumes the string date is in A1):
=DATE(MID(A1,1,4),MID(A1,6,2),MID(A1,9,2)) +
TIME(MID(A1,12,2),MID(A1,15,2),MID(A1,18,2))
To convert the fractional second part, use:
=VALUE(MID(A1,21,4))/10000
The date / time part can easily be subtracted, as can the fractional second part.
The place I ran into trouble was recombining those parts into a whole that Excel will actually display in a sensible way. I finally took the difference between the two dates and multiplied by 86400 (= 24 * 60 * 60 - number of seconds in a typical day) then added the difference in the fractional second part.
Hope this helps.
Regards,
Richard
P.S. There are quite a lot of things I don't like about this solution, the biggest of which is the fragility of the formulas - if the date format changes, the formulas will need to be adjusted.