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jiasheng
Creator
Creator

Distinct?

 

HI all expert!

I have this set of data,

   

Contact PersonContact NumberCOMPANY
MA11111111ABC
MA11111111DEF
MA22222222GHI
MB22222222JIL
MC33333333MNO
MD33333333PQR
ME33333333STU
ME33333333

VWX

ME3333333YZA

I would like to find out the number of cases where same contact person, multiple company. So in this set of data, it will be 2 (MA & ME both represent multiple company).

I would not like to touch my scripting as much as possible. How can I expression in my KPI to show the number of case.

Regards,

Jia

1 Solution

Accepted Solutions
OmarBenSalem
Partner
Partner

It's :
Count({<contact = {"=Count(Distinct contact&company)>1"}>} DISTINCT contact)

we want to count distinct contact;

So the original expression is count(distinct contact)

Now, we don't want to count all of them but only the contact who has more then a company;

To do so; we create a 'new field' contact&company and count it distinct:

exp: for MD we'll only have MD&PQR : so 1
for ME: we'll have : ME&STE + ME/ VWX +ME&YZA : so 3 > 1 , we count it and so on;

which leads us to 2 contacts;

hope this helps !

View solution in original post

10 Replies
jayaseelan
Creator III
Creator III

Hi,

     Please post you expecting output In .xsl or something else.

    This is your output?

Capture.PNG

Thanks,

jiasheng
Creator
Creator
Author

Hello Jayaseelan,

Thank for ur prompt help, even for my others qns as well.

My expected output for kpi will be

CapturePersonal.PNG

and my expected output in result table will be

   

Contact PersonContact NumberCOMPANY
MA11111111ABC
MA11111111DEF
MA22222222GHI
ME33333333STU
ME33333333VWX
ME33333333YZA
CELAMBARASAN

Hi,

KPI should be Count({<[Contact Person] = {"=Count(Distinct [Contact Numer] &'_' & COMPANY)>1"}>} DISTINCT [Contact Person])

Table

Dimension - [Contact Person], [Contact Number]

Expression - Only({<[Contact Person] = {"=Count(Distinct [Contact Numer] &'_' & COMPANY)>1"}>} COMPANY)

OmarBenSalem
Partner
Partner

It's :
Count({<contact = {"=Count(Distinct contact&company)>1"}>} DISTINCT contact)

we want to count distinct contact;

So the original expression is count(distinct contact)

Now, we don't want to count all of them but only the contact who has more then a company;

To do so; we create a 'new field' contact&company and count it distinct:

exp: for MD we'll only have MD&PQR : so 1
for ME: we'll have : ME&STE + ME/ VWX +ME&YZA : so 3 > 1 , we count it and so on;

which leads us to 2 contacts;

hope this helps !

View solution in original post

jiasheng
Creator
Creator
Author

Hi Celambarasan,

could you explain what  {"=Count(Distinct [Contact Number] &'_' & COMPANY)>1"} does?

and also if I want to change to  same CONTACT NUMBER, multiple company, how do I go around it?

Regards,

Jia

OmarBenSalem
Partner
Partner

Hi Jia,

I explained already explained the process:

Now to count the numeros existing in different companys, here how you proceed:

Count({<numero = {"=Count(Distinct numero&company)>1"}>} DISTINCT numero)

jiasheng
Creator
Creator
Author

Hi Omar,

Thank for the clarification!

Yes yes, guess my post was slightly later, thus missing out your explanation before posting.

So the table wise, correct me if I am wrong. instead of count, we want to show the only the result, thus

Only({<[Contact Person] = {"=Count(Distinct [Contact Person] &'_' & COMPANY)>1"}>} COMPANY)?

jayaseelan
Creator III
Creator III

Hi Jia Sheng Loh,

  Use the following expression you will get you answer.

Count({<[Contact Person] = {"=Count(Distinct COMPANY)>1"}>} DISTINCT [Contact Person])

Capture1.PNG

In the above expression we counting the Distinct Company and Checking with the [Contact person] is > 1 and Count the over all Distinct [Contact person].

Thanks,

jiasheng
Creator
Creator
Author

Hi Jayaseelan,

Seems like the code Celambarasan, Omar and you provided works,

But since Omar came up with Explanation first, I think is fair to choose his as correct answer so that other that tumble upon this question will understand as well.

Ur explanation is correct and helpful too!

Thank you!

Regards,

Jia