I have a report which is downloaded from a Warehouse Management System.
On this report there is a time column which unfortunately puts the time into a string of numbers that can be anywhere from 5-8 digits long.
I.e.
22434900 = 22:43:49:00 with 22 being the hour, 43 the minutes, 49 the seconds.
2480000 = 02:48:00:00 with 2 being the hour, 48 the minutes etc.
54300 = 00:05:43:00
The 00 on the end (milliseconds) doe not change in each number so is quite irrelevant.
Is there an easy way to format the text in these cells so it shows as a time as oppose to a number?
Thanks in advance.
I know I'm late, but here's an alternate solution:
=TIMEVALUE(TEXT(A1/100,"00\:00\:00.00"))
Again, as mentioned in Jerry's answer, you'll need to use cell formatting of hh:mm:ss.00
You can use TIME with some math functions:
=TIME(INT(A1/1000000),MOD(INT(A1/10000),100),MOD(A1/100,100))
TIME takes 3 parameters: hours, minutes and seconds.
To get the hours, I'm dividing by 1000000, then INT rounds it down to the closest integer.
To get the minutes, I'm first dividing by 10000, but there is still the hours in that result. So I use MOD (which gives the remainder when a number is divided by another number). In the first example, the division gives 2243, and the remainder when dividing this by 100 is 43, which is the number of minutes I'm looking for.
To get the seconds, I divide the number by 100 and similar to the minutes, I use MOD to remove the minutes and hours parts. I am not using INT here in case there are milliseconds, which will be kept using this formula.
Also note that I am using the formatting hh:mm:ss.00, because excel complains if I try using hh:mm:ss:00.
For your Warehouse Management System query you may want to try something like this:
• Taking the 6 digit numeric time and making this into a string that the time function can handle
• Using the digits function to avoid issues with varying lengths of data (i.e. 64512 vs 1113012)
• Use the function Time over this string to return the value (this way we can add hours or minutes as the example below)
Here is an example to experiment with this and :
select MyTimeField
, time(int(MyTimeField/10000) || ':' ||
substring(digits(MyTimeField),3,2) || ':' ||
substring(digits(MyTimeField),5,2))
from MyTable where MyCompany = 1 and MyInvoiceDate = Current_date
Related
I have some time data that, without VBA, I need to round to the nearest 10 milliseconds. For example:
input: 01:02:03.017 output: 01:02:03.020
input: 03:12:44.123 output: 03:12:44.120
Current approach is to convert to an integer number of milliseconds; round that to the nearest 10; finally convert back to time:
=ROUND(A1*86400000,10)/86400000
I must be making a really stupid error, just don't see it.
EDIT:
The formula is returning the same value as the input?!?
Change the 10 to 0
By multiplying the value you want the round to the nearest integer, then divide again. By using 10 you are rounding to the 10th decimal place after creating a integer time.
=ROUND(A1*86400000,0)/86400000
Assuming your time data is in column A:
=TEXT(A1,"hh:mm:ss.00")+0
Change the number format with TEXT, then add the ending 0 with +0
Below excel formula is working fine but in some cases its not give me proper value .
Input:
19:20:42
24:58:36
26:11:18
After using this formula:
=IF(TIMEVALUE(K7)>TIMEVALUE("09:00:00"),TRUE,FALSE)
I got the below output:
FALSE
TRUE
TRUE
What I Observe if the time value is > or = 24:00:00 it will not give me the proper answer.
How do I fix this?
As an alternative to Captain's excellent answer, you could also use:
=IF(K7>(9/24),TRUE,FALSE)
DateTime values are internally stored as a number of days since 1/1/1900,
therefore 1 = 1 day = 24 hours. So 9/24 = 0.375 = 9 hours :-)
You can easily verify this by clearing the format of your DateTime cells.
Edit: note that such Boolean formula can be expressed in a shorter way without losing legibility:
=K7>(9/24)
When you go over 24 hours, Excel counts it as the next day... and then the TIMEVALUE is the time the next day (i.e. 00:58:36 and 02:11:18 in your examples) and can, therefore, be before 0900.
You could do DATEVALUE(K7)+TIMEVALUE(K7) to ensure that you count the day part too...
i need to know how can i exclude 0 from rows and get the MIN Value.
But also i need to exlude the F1 Cell.
Ex:
A B C D E F
1 0 18 20 0 150 = 18
but if i do this In excel with =MIN(A1,B1,C1,D1,E1) return 0.
Any help is appreciated.
Try this formula
=SMALL((A1,C1,E1),INDEX(FREQUENCY((A1,C1,E1),0),1)+1)
Both SMALL and FREQUENCY functions accept "unions" as arguments, i.e. single cell references separated by commas and enclosed in brackets like (A1,C1,E1).
So the formula uses FREQUENCY and INDEX to find the number of zeroes in a range and if you add 1 to that you get the k value such that the kth smallest is always the minimum value excluding zero.
I'm assuming you don't have negative numbers.....
Enter the following into the result cell and then press Ctrl & Shift while pushing ENTER:
=MIN(If(A1:E1>0,A1:E1))
Not entirely sure what you want here, but if you want to discount blank cells in the range and pass over zeros then this would do it; if a little contrived:
=MIN(IF(A1:E1=0,MAX(A1:E1),A1:E1))
With Ctrl+Shift+Enter as an array.
What I'm doing here is replacing zeros with the maximum value in the list.
if all your value are positive, you can do -max(-n)
Solutions listed did not exactly work for me. The closest was Chief Wiggum - I wanted to add a comment on his answer but lack the reputation to do so. So I post as separate answer:
=MIN(IF(A1:E1>0;A1:E1))
Then instead of pressing ENTER, press CTRL+SHIFT+ENTER and watch Excel add { and } to respectively the beginning and the end of the formula (to activate the formula on array).
The comma "," and "If" statement as proposed by Chief Wiggum did not work on Excel Home and Student 2013. Need a semicolon ";" as well as full cap "IF" did the trick. Small syntax difference but took me 1.5 hour to figure out why I was getting an error and #VALUE.
Throwing my hat in the ring:
1) First we execute the NOT function on a set of integers,
evaluating non-zeros to 0 and zeros to 1
2) Then we search for the MAX in our original set of integers
3) Then we multiply each number in the set generated in step 1 by the MAX found in step 2, setting ones as 0 and zeros as MAX
4) Then we add the set generated in step 3 to our original set
5) Lastly we look for the MIN in the set generated in step 4
{=MIN((NOT(A1:A5000)* MAX(A1:A5000))+ A1:A5000)}
If you know the rough range of numbers, you can replace the MAX(RANGE) with a constant. This speeds things up slightly, still not enough to compete with the faster functions.
Also did a quick test run on data set of 5000 integers with formula being executed 5000 times.
{=SMALL(A1:A5000,COUNTIF(A1:A5000,0)+1)}
1.700859 Seconds Elapsed |
5,301,902 Ticks Elapsed
{=SMALL(A1:A5000,INDEX(FREQUENCY(A1:A5000,0),1)+1)}
1.935807 Seconds Elapsed |
6,034,279 Ticks Elapsed
{=MIN((NOT(A1:A5000)* MAX(A1:A5000))+ A1:A5000)}
3.127774 Seconds Elapsed |
9,749,865 Ticks Elapsed
{=MIN(If(A1:A5000>0,A1:A5000))}
3.287850 Seconds Elapsed |
10,248,852 Ticks Elapsed
{"=MIN(((A1:A5000=0)* MAX(A1:A5000))+ A1:A5000)"}
3.328824 Seconds Elapsed |
10,376,576 Ticks Elapsed
{=MIN(IF(A1:A5000=0,MAX(A1:A5000),A1:A5000))}
3.394730 Seconds Elapsed |
10,582,017 Ticks Elapsed
min() fuction exlude BOOLEAN and STRING values.
if you replace your zeroes with "" (empty string) - min() function will do its job as you like!
In Microsoft 365 you can use the new function MINIFS
=MINIFS(A1:E1;A1:E1;">0")
gives 1
=MINIFS(A1:E1;A1:E1;">1")
gives 18
*replace ; with , if using english version
All you have to do is to delete the "0" in the cells that contain just that and try again. That should work.
I have an Excel 2010 workbook which contains a cell with the value of, say, 9876:54:32 (manually entered) representing 9876 hours, 54 minutes and 32 seconds of, say, phone talk time.
Then I have a cell with the value of, say, 1000 (manually entered) representing 1000 calls.
I want to divide the values to get the average talk time of 592.615 minutes per call.
I'm doing a regular =A1/B1 and it gives me an error.
* EDITED *
Thanks Brain Webster for correcting my math. I mean 9.876 hours. But the point is that Excel gives me an error, not my manual math. Playing around with it I discovered that Excel is fine with me with values up to 9999:59:59. Once I try with 10000:00:00 and up, it doesn't recognize it as a time value.
I love these seemingly easy riddles, so here is my solution as a formula and as a VBA attempt:
my original:
= (LINKS(A38;FINDEN(":";A38)-1)/24)+ZEITWERT("0"&RECHTS(A38;LÄNGE(A38)-FINDEN(":";A38)+1))
translated:
= (LEFT(A38,FIND(":",A38)-1)/24)+TIMEVALUE("0"&RIGHT(A38,LEN(A38)-FIND(":",A38)+1))
This will get you the right value to a given 10k text of a time duration. You would only have to setup the format of the cell to [h]:mm:ss. Then those values will look the same, but one would be a string and the other a number - and that is a major difference ;)
In vba it looks much more easier, and once defined, you can use it as a worksheetfunction.
Public Function GetDurationValue(ByVal strInput As String) As Double
Dim arrResult As Variant
arrResult = Split(strInput, ":") 'saves you parsing
GetDurationValue = (arrResult(0) / 24) + _
TimeSerial(0, arrResult(1), arrResult(2))
End Function
A38 = "10971:12:14"
=GetDurationValue(A38)
=457.13349537037
You can use LEFT and RIGHT function to retreive parts of the time value and then sum and multiply these values by 60 [minutes] (resp. 3600 [hours]).
Something like this for the hours, minutes, seconds (A1 is the cell with time value):
B1=VALUE(LEFT(A1;FIND(":";A1)))*3600
B2=VALUE(LEFT(A1;FIND(":";A1; FIND(":";A1))))*60
B3=VALUE(LEFT(A1;FIND(":";A1; FIND(":";A1; FIND(":";A1)))))
Now you can sum that:
C1=SUM(B1;B2;B3)
Then divede by calls count (A2 is the cell with the calls count):
D1=C1/A2
Finally format it like time:
E1=TEXT(D1/(24*3600);"d \day\s hh:mm:ss")
BTW: I tried that in Excel 2013 and when I enter 111:22:33 into a cell it automatically converts to a time. So then I can divide it like you try...
It appears that hours > 10000 are not recognised as such by Excel. Therefore we need to introduce an IF() to see whether this is the case and determined the alternative formula for the case where hours >10000
=IF(ISERROR(FIND(":",A2)),A2/B2, <SCRIPT IN CASE OF >10000>)
<SCRIPT IN CASE OF >10000> will now be:
VALUE(LEFT(A2,FIND(":",A2)))/24+VALUE(LEFT(A2,FIND(":",A2, FIND(":",A2))))/(24*60)+VALUE(LEFT(A2,FIND(":",A2, FIND(":",A2,FIND(":",A2)))))*(24*60*60)
combine and enjoy!
Assuming you don't exceed 100,000 hours in A1, and you always display hours, minutes and seconds then this formula should suffice
=IFERROR(A1/B1,(LEFT(A1)*10000/24+RIGHT(A1,10))/B1)
format result cell as [h]:mm:ss to get the result as a time value
With 10971:12:14 in A1 and 1000 in B1 that should give a result of 10:58:16 [or format result cell as [m]:ss to get minutes and seconds like 658:16]
This version will work with any number of hours and with or without seconds
=IFERROR(A1/B1,(LEFT(A1,FIND(":",A1)-1)/24+RIGHT(A1&IF(COUNTIF(A1,":*:"),"",":00"),5)/60)/B1)
I have a device which outputs the time in the format hh:mm:ss.000, e.g., 00:04:58.727 and I need to convert these to milliseconds.
I can't change the way the device outputs the times so I have to do it in Excel, but I don't know VB so am looking for a cell-by-cell solution.
Let's say that your time value is in cell A1 then in A2 you can put:
=A1*1000*60*60*24
or simply:
=A1*86400000
What I am doing is taking the decimal value of the time and multiply it by 1000 (milliseconds) and 60 (seconds) and 60 (minutes) and 24 (hours).
You will then need to format cell A2 as General for it to be in milliseconds format.
If your time is a text value then use:
=TIMEVALUE(A1)*86400000
UPDATE
Per #dandfra's comment this solution may not work in the Italian version of Excel.
Using some text manipulation we can separate each unit of time and then sum them together with their millisecond coefficients.
To show the formulas in the cells use CTRL + `
Use
=LEFT(B2, 2)*3600000 + MID(B2,4,2) * 60000 + MID(B2,7,2)*1000 + RIGHT(B2,3)
you can do it like this:
cell[B1]: 0:04:58.727
cell[B2]: =FIND(".";B1)
cell[B3]: =LEFT(B1;B2-7)
cell[B4]: =MID(B1;11-8;2)
cell[B5]: =RIGHT(B1;6)
cell[B6]: =B3*3600000+B4*60000+B5
maybe you have to multiply B5 also with 1000.
=FIND(".";B1) is only necessary because you might have inputs like '0:04:58.727' or '10:04:58.727' with different length.
Rather than doing string manipulation, you can use the HOUR, MINUTE, SECOND functions to break apart the time. You can then multiply by 60*60*1000, 60*1000, and 1000 respectively to get milliseconds.
Here it is as a single formula:
=(RIGHT(D2,3))+(MID(D2,7,2)*1000)+(MID(D2,4,2)*60000)+(LEFT(D2,2)*3600000)
try this:
=(RIGHT(E9;3))+(MID(E9;7;2)*1000)+(MID(E9;5;2)*3600000)+(LEFT(E9;2)*216000000)
Maybe you need to change semi-colon by coma...