I have two columns for the time in and time out of the employee, and I want to determine their lates and undertimes. But I cannot properly do it, I already can do the negative value to zero but I just want to know how to do the positive value to zero.
This is my formula =(G10-C10)*1440 then as per my column values it will show a positive value but I want the positive value become ZERO
Thanks!
You could use IF, which would be the most popular choice:
=IF(G10-C10>0,0,(G10-C10)*1440)
Could be read as, 'if the difference is greater than 0 (positive), then put 0, otherwise put the difference itself'.
Or a little less common, but simpler:
=MIN((G10-C10)*1440,0)
As an alternative to Jerry's answer, after applying your original formula, you can change the format of the cells to something like this:
"0";-0;0
The first 0 is for positive. The quotes around it make sure the zero gets displayed literally.
The second section (each section separated by a semi-colon ;) is for negative and the last one's for zero values.
The zero without quotes here means that one number should get displayed compulsorily.
Here's the output:
Related
I have cells that contain both numbers and special characters such as this:
[1:250:10]
The 'coordinates' shown above can be in the following format.
[(1-9):(1-499):(1-15)] in terms of what numbers can be within each part.
How do I extract these three numbers into three separate cells?
Assuming your data is in Cell A1 the to extract first number use following formula
=MID(A1,2,(FIND(":",A1,1)-2))
for second number use
=SUBSTITUTE(MID(SUBSTITUTE(":" & A1&REPT(" ",6),":",REPT(":",255)),2*255,255),":","")
finally for third number enter
=SUBSTITUTE(TRIM(RIGHT(SUBSTITUTE(A1,":",REPT(" ",LEN(A1))),LEN(A1))),"]","")
Just tossing out some other options.
First number since it only has a length of 1 digit and is on the left side, use the following:
=RIGHT(LEFT(A1,2))
second number will be found by locating the : in the string
=MID(A1,FIND(":",A1)+1,FIND(":",A1,FIND(":",A1)+1)-(FIND(":",A1)+1))
third number will be dealt with in the same way as the second but we will use the second : and the ] as the identifiers as to where to grab from and how much to pull.
=MID(A1,FIND(":",A1,FIND(":",A1)+1)+1,FIND("]",A1)-(FIND(":",A1,FIND(":",A1)+1)+1))
now all those number will actually come through as text. If you want to have them as numbers in the cells, send them through a math operation that will not change their value. Do something like +0, -0, or *1 at the end. Alternatively you could add -- at the start of each formula (yes that is double - incase you were wondering if it was a typo)
Is it possible to count the number of individual entries in a cell?
For example 2+2+4-1 = 4 entries
Using the count formula counts the entries as 1
I want to calculate the number of adjustments made in a particular period.
Each +/- in an individual cell represents 1 adjustment.
Assuming you're referring to a text cell, the trick is to count the symobols you'd like to find. Before we dig into that, if you want to enter this data as text, you can use the ` symbol (Usually to the left of the 1 key on your keyboard) before entering your text to make sure it gets processed as text.
If you want to verify that it is text, you can use the TYPE function and look for a return result of 2 (check the link for other possible return types)
There are no direct functions to count characters in Excel, so the trick is to find the length of the original text and subtract it from the length of a new text where you have removed all of the special characters. You mentioned you were trying to count the entries (i.e. the numbers), but you said your goal was to ultimately count the number of '+/-' operations. Since counting numbers can be tricky with excel formulas (since we'll get hung up on 2 and 3 digit numbers), I am going to approach this problem from the perspective of counting the operations you are looking for. So here is a basic example:
length("2+1") = 3
- length("21") = 2 (we replaced the + with "" [blank])
= 1
So we know there is 1 '+' since we replaced it. The appropriate functions used to accomplish this are LEN and SUBSTITUTE
Since you can only find one symbol at a time using the SUBSTITUTE function, we must take the output of the first formula, and give it to the second formula, and so on and so forth. Ultimately, we can put together as many functions as we need to achieve the desired result.
So we start with + for your example (And assuming your data is in A1)
=LEN(A1)-LEN(SUBSTITUTE(A1,"+",""))
which gives us a result of 2. But we also need to find the - symbol. So we wrap another SUBSTITUTE:
=LEN(A1)-LEN(SUBSTITUTE(SUBSTITUTE(A1,"+",""),"-",""))
You have said you wanted to count the number of +/- in the cell, and this does accomplish that, but if you want to expand it to more mathematical operators, you simply add more SUBSTITUTE functions (here is a complete function where I've added * and /)
=LEN(A1)-LEN(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1,"+",""), "-",""),"*",""),"/",""))
Well, this formula would replace all your numbers with "" and then Count the +/- and adds one, should do it, but is ugly:
=LEN(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(SUBSTITUTE(A1;"0";"");"1";"");"2";"");"3";"");"4";"");"5";"");"6";"");"7";"");"8";"");"9";""))+1
Could probably be done with RegEx, but I don't know how to do that in formulas
This makes 31+12+3-4 to ++-, Counts the LEN (3) and adds 1
First off, thank you to all how have helped get me to this point. I'm so close! On to the scenario, which I apologize in advance is a bit of a work in progress.
I have text in a cell and I need to extract a number. The tricky part is there are various situations to address.
The number may immediately follow a "#" and could vary in length. People on Stack Overflow helped me with coming up with this which works great:
MID(B2,(FIND("#",B2,1)+1),FIND(" ",B2,FIND("#",B2,1)+1)-FIND("#",B2,1))
That was a huge leap forward, but there are also situations where there is no # sign and the cell might have "abc (1205) 645 chan", where I need to extract the 645.
I'm using this, below, in conjunction with an on error statement for when there is no "#"
TRIM(MID(B53,(FIND(" " &{"1","2","3","4","5","6","7","8","9"},B53,1)),FIND(" ",B53,FIND({"1","2","3","4","5","6","7","8","9"},B53,1))-FIND({"1","2","3","4","5","6","7","8","9"},B53)))
So I use the first Mid/Find to avoid the (1205) and find the next " x" where x is a number. The problem is it seems I have trouble when the number I'm searching for has 1 or 3+ numbers in it, but if it has 2 I return the value just fine.
It seems I'm very close but just not there yet.
This formula will return the number that follows either a # or a ) in your string. If that pattern does not exist, it will return a #NUM!` error
=AGGREGATE(14,6,--MID(A1,MIN(FIND({"#",")"},A1&"#)"))+1,{1,2,3,4,5}),1)
Note the array constant as the num_chars argument of the MID function. The maximum number should be at least the largest number of digits (or decimal + digits) plus any spaces between the delimiter and the first digit, that might be expected to be found.
EDIT: If your version of Excel is prior to 2010, and does not have the AGGREGATE function, you may use this array-entered formula instead, so long as the values to be returned will be positive numbers:
=MAX(IFERROR(--MID(A1,MIN(FIND({"#",")"},A1&"#)"))+1,{1,2,3,4,5}),0))
This formula must be entered by holding down ctrl+shift while hitting enter
To make the sale to my customer I need to import numbers from a report into an Excel document. For example the number coming in will be 14.182392. The only reason for my guy not to buy the product is because he only wants to view 14.182 on the Excel sheet. Okay so the other catch is, the number CANNOT be rounded in any shape or form.
So what I need is a way to just show so much of number, WITHOUT ROUNDING.
Is this possible? Any ideas of how I could get around this would be fantastic.
Please try:
=TEXT(INT(A1)+VALUE(LEFT(MOD(A1,1),5)),"00.000")
Firstly =TRUNC is a better answer (much shorter). My version was connected with uncertainty in your requirement (it is odd!) and in the hope it might be easier to adjust if not exactly what you/your boss wanted.
TRUNC literally just truncates the decimals (no rounding!) to a length to suit (ie 3 if to show nn.182 given nn.182392 or say nn.182999).
LEFT may also be a better choice, but that depends upon knowing how large the integer part of your number is. =LEFT(A1,6) would display 14.189 given say 14.189999 in A1. However it would show 1.4189 given 1.4189999 in A1 (ie four decimal places).
The formula above combines text manipulation with number manipulation.:
INT takes just the integer value (here 14.)
MOD takes just the modulus – the residual that is not an integer after division, in this case by 1. So just the .182392 part. LEFT is then applied here in a similar way to as used above, but without needing to concern oneself with the length of the integer part of the source value (ie 14 or 1 etc does not matter).
VALUE then converts the result back into numeric format (string manipulation functions such as LEFT always return text format) so our abbreviated decimal string can then be added to our integer.
Finally, the TEXT part is for formatting but is hard or impossible to justify! About the only use is that it displays the result left-justified in the cell – perhaps a little warning that the number displayed is not the “true” value (eg it won’t SUM) because, as a result of a formula, it won’t be marked with a little green warning triangle.
The displayed values can use the TRUNC function like this,
=TRUNC(A1, 3)
But you must use A1 in any calculations to retain the precision of the raw value.
Easiest way I know:
=LEFT(A1; x)
where x = the amount of characters You want. Mind that the dot counts as a character as well.
I have something like this:
A1: ABC-DE442
B1: 0069-1234-12
I'm trying to make a formula that will look at A1, get the number 442, and then check if B1 is in the format of 4 numbers, a dash, 4 numbers, a dash, and then finally 2 numbers. I want to make sure that this could work with any number besides 442 though. So if I had like 123 instead of 442, it would need to be 1 number, dash, 2 numbers, dash, and three numbers.
It would also need to give me an error if I had something like this:
A1: ABC-DE442
B1: 004-2345-34
because it only has three numbers in the first section before the dash.
Hopefully I have been clear enough in what I'm trying to do. Any help would be greatly appreciated.
So in the end, this is the giant formula I had to create:
=IF(LEN(LEFT(J4,FIND("-",J4)-1))=VALUE(MID(I4,9,1)),IF(LEN(LEFT(MID(J4,FIND("-",J4)+1,999),FIND("-",MID(J4,FIND("-",J4)+1,999))-1))=VALUE(MID(I4,10,1)),IF(LEN(MID(MID(J4,FIND("-",J4)+1,999),FIND("-",MID(J4,FIND("-",J4)+1,999))+1,999))=VALUE(MID(I4,11,1)),"True","Doesn't match code part 3"),"Doesn't match code part 2"),"Doesn't match code part 1")
J15: 0069=LEFT(J5,FIND("-",J5)-1)
J16: 4=VALUE(MID(I5,7,1))
J17: 4=LEN(J15)
J18: 0469-56=MID(J5,FIND("-",J5)+1,999)
J19: 0469=LEFT(J18,FIND("-",J18)-1)
J20: 4=VALUE(MID(I5,8,1))
J21: 4=LEN(J19)
J22: 56=MID(J18,FIND("-",J18)+1,999)
J23: 2=VALUE(MID(I5,9,1))
J24: 2=LEN(J22)
Hopefully all of these formulas will make it easier to understand how I combined everything together.
I essentially put every formula into one giant one using the information used in the answer I said was right. For the first part, I took the LEN(D1) but had it all written out, and then set it equal to the =VALUE(MID(A1,7,1)).
Then, for the true part of that if statement, I had to first get A1 without D1, then get it to just be the middle section using the LEFT formula, and then finally taking the LEN of that. Checked to see if it was equal to the =VALUE(MID(I4,10,1)), which was the next number in B1.
If that was true, I went to the final IF statement which again had to go through the same process of getting the last section of A1 on it's own, taking the length of that, and seeing if it was equal to =VALUE(MID(I4,11,1)).
Finally, I just created different print statements for the last true, and every false section of the IF statements.
Moral of the story, I was able to make a formula that works with every single number combination because it turns out B1 could only be in the form of AB - CDE###, so I could always find that number in B1 and compare it.
This should be possible with Excel formulae but it will get a bit complicated. The functions you're probably going to want to make use of are FIND, LEFT, LEN, MID, and VALUE.
For example,
C1: =FIND("-",B1)
will return the position of the first dash in the string to be tested, which for your example is 5. Then
D1: =LEFT(B1,C1-1)
would return the characters before that dash, i.e. '0069'. Meanwhile you need to get the lengths of the three runs of digits out of your first string (the one in A1). Will that string always be the same length? If so you can use e.g.
=VALUE(MID(A1,7,1))
which returns the 7th character of the string in A1 and converts it to a number (4). Now you can test whether that value is equal to the length of the first run of characters, i.e. LEN(D1).
But are those characters all numeric? Well, you can try converting them to a number with VALUE(D1) which will return #VALUE if the argument can't be made into a number.
OK, that checked the first run of characters, what about the rest of them? Well,
=MID(B1,FIND("-",B1)+1,999)
will give you the remainder of the original string after the first dash, which you can analyse in the same way, and so on.
It's easiest to develop these formulae by using lots of cells to hold the intermediate values as you work them out, but once you're confident these are working OK you can consolidate them into fewer cells. Whatever you come up with, make sure you test it with lots of different inputs to check that your formulae respond to them in the way that you expect. (In particular, check what happens if you have one or more spaces after a dash, as VALUE will ignore leading spaces - you might want to use a SUBSTITUTE function on your initial text to remove any spaces.)