Abinitio Interview Questions

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    Why go for sort within groups?

    We have sort and sort within groups components. we can achieve the sort within group functionality by placing two keys in sort group. Then why we have to go for sort within groups?

    Mahesh

    • May 26th, 2021

    One of the other examples could be :
    graph A has done PBKS on field A
    and when the output of graph A is being read by graph B it needs data to be sorted on fieldA and fieldB so major key is fieldA and minir key is fieldB.This saves time and increases performance.

    sundar

    • May 5th, 2015

    Hey Munmun,

    Do we need sorted data to sort with in group or component will taking care?

    Thanks

  •  

    Output for sort and dedup sort with null key

    I have file containing 5 unique rows and I am passing them through SORT component using null key and and passing output of SORT to Dedup sort. What will happen, what will be the output.

    Amaan Sajid Ansari

    • Nov 14th, 2018

    In Case of sort output would be same as input (unsorted) in case of dedup will depend on no of partition and the dedup method used. Please refer to the other similar dedup question i have explain it in that

    san

    • Aug 7th, 2018

    1. In case of Null ket{} in sort component nothing will happen. 2. In case of Dedup sorted - keep - fist then you will get first record, Keep- Last then you will get last one, Keep- Unique only - in ...

  •  

    Abinitio EME

    In my sandbox i am having 10 graphs, i checked-in those graphs into EME. Again i checked-out the graph and i do the modifications, i found out the modifications was wrong. what i have to do if i want to get the original graph..?

    Amruta

    • Jan 9th, 2021

    Do revert changes...
    Another option is to checked out previous version of object from eme

    Jatin Chawla

    • Sep 21st, 2020

    Air object rollback to rollback to previous version

  •  

    How to achieve the below scenario in abinitio

    Input file has below contents:
    Ball Run
    1 1
    2 1
    3 1
    4 1
    5 1
    6 1
    1 0
    2 1
    3 1
    4 1
    5 1
    6 1
    1 1
    2 1
    3 1
    4 1
    5 1
    6 0

    Required Output :
    1 6
    2 5
    3 5

    UNKNOWN User

    • Oct 23rd, 2024

    Use rollup with key change function
    (in1.ball==6 && in2.ball==1)
    This will cover if the 6th ball is a extra ball

    Shubham Kandpal

    • Jun 22nd, 2024

    Use key change and rollup , this code is 100% working

    Code
    1. let decimal(x01) over=0;
    2.  
    3. out::key_change(in1,in2)=
    4. begin
    5. out::in2.ball%6 ==1;
    6. end;
    7.  
    8. out::rollup(in)=
    9. begin
    10. over=over+1;
    11. out.ball :: over;
    12. out.run :: sum(in.run);
    13. end;

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