Time Difference

[Alteryx_Geek]

I am looking to find the delta between below timestamps (string format)

Example 1: 

                   2022-07-01 12:19:15.700
                   2022-07-01 12:19:15.800

                   The answer should be 0.1 sec as the output.

Example 2: 

                   2022-07-01 12:19:15.700
                   2022-07-01 12:19:20.700

                   The answer should be 5.0 sec as the output.

[Andreina]

Hello @Alteryx_Geek 

I think one of the solutions of  the  following post might help you:

https://community.alteryx.com/t5/Alteryx-Designer-Discussions/How-to-get-datetimediff-with-milli-sec...

Hope it helps!!!

Regards,

[Alteryx_Geek]

@Andreina - Thanks for sharing the post, I tried this method but it doesn't work when there is vertical Timestamp column like below,

2022-07-05 12:17:48.003
2022-07-05 12:17:52.066
2022-07-05 12:19:02.919
2022-07-05 12:19:18.586
2022-07-05 12:19:34.857
2022-07-05 12:20:04.996
2022-07-01 12:21:30.717

[Alteryx_Geek]

@binuacs / @OllieClarke - Any idea ?

[binuacs]

@Alteryx_Geek Can you provide the expected output for the above DateTime values?

[OllieClarke]

@Alteryx_Geek 

how about this:

IF 
Tonumber(Substring([Time],19))-ToNumber(substring([row-1:Time],19))>=0.5 
THEN
datetimediff(LEFT([time],19),LEFT([row-1:time],19),'seconds')+1
ELSE
datetimediff(LEFT([time],19),LEFT([row-1:time],19),'seconds')
ENDIF

So we add 1 second if the millisecond difference is more than 0.5

Hope that helps,

Ollie

[Andreina]

Hello @Alteryx_Geek , 

Maybe with a combination of some MultiRow and Formula transformations  you can get what you need. Here you have an example that might help you:

Try to calculate Time Differences betweeen rows and then make the transformations in order to get days, minutes, secs and millisecs. 

Check the workflow that I attached. 

Hope it helps. 

Andreina

[OllieClarke]

Hey @Alteryx_Geek 

Apologies, I'd misread your actual question. This I think gives what you want:

ROUND(
datetimediff(LEFT([time],19),LEFT([row-1:time],19),'seconds')
+
tonumber(substring([time],19))-tonumber(substring([row-1:time],19))
,0.1)

It outputs the number of seconds between a row and the row above to 0.1 of a second.

Hope that helps,

Ollie

[Alteryx_Geek]

@OllieClarke - Apologize for late reply. Thanks for sharing Ollie. looks like your output is coming along but when you have answer in msec, its rounding off the number.

As highlighted below, the output should have been "0.11" instead of 1

[Alteryx_Geek]

@binuacs - I just share the expected output in above snap.