06-26-2023
anandbillava
Fluorite | Level 6
Member since
06-23-2011
- 86 Posts
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Latest posts by anandbillava
Subject Views Posted 1107 06-26-2023 05:23 PM 1133 06-26-2023 02:46 PM 1161 06-26-2023 01:15 PM 426 06-26-2023 01:13 PM 3116 08-06-2019 01:12 PM 3126 08-06-2019 11:23 AM 3155 08-05-2019 04:15 PM 3551 08-30-2018 03:35 PM 2963 08-29-2018 07:32 AM 3001 08-28-2018 08:19 PM -
Activity Feed for anandbillava
- Posted Re: Adding dropdown list to excel files using ods tagsets on SAS Programming. 06-26-2023 05:23 PM
- Posted Re: Adding dropdown list to excel files using ods tagsets on SAS Programming. 06-26-2023 02:46 PM
- Posted Adding dropdown list to excel files using ods tagsets on SAS Programming. 06-26-2023 01:15 PM
- Posted ods tagsets Insert vbscript on SAS Programming. 06-26-2023 01:13 PM
- Posted Re: Hierachy data (Parent Child) on SAS Programming. 08-06-2019 01:12 PM
- Posted Re: Hierachy data (Parent Child) on SAS Programming. 08-06-2019 11:23 AM
- Posted Re: Hierachy data (Parent Child) on SAS Programming. 08-05-2019 04:15 PM
- Posted Re: Hierachy data (Parent Child) on SAS Programming. 08-30-2018 03:35 PM
- Got a Like for Re: Get ancestors for each children. 08-29-2018 01:13 PM
- Posted Re: Get ancestors for each children on SAS Programming. 08-29-2018 07:32 AM
- Posted Re: Get ancestors for each children on SAS Programming. 08-28-2018 08:19 PM
- Posted Get ancestors for each children on SAS Programming. 08-28-2018 03:07 PM
- Posted Re: Hierachy data (Parent Child) on SAS Programming. 08-28-2018 11:30 AM
- Posted Re: Hierachy data (Parent Child) on SAS Programming. 08-27-2018 03:39 PM
- Posted Hierachy data (Parent Child) on SAS Programming. 08-27-2018 01:52 PM
- Posted Re: find the character in a string which is other than in the list on SAS Programming. 06-12-2017 01:01 PM
- Posted Re: find the character in a string which is other than in the list on SAS Programming. 06-12-2017 12:41 PM
- Posted find the character in a string which is other than in the list on SAS Programming. 06-12-2017 12:28 PM
- Posted Re: SYSEVAF function to evaulate a expression stored in character (SKIP syntax) on SAS Programming. 12-28-2015 10:10 AM
- Posted Re: SYSEVAF function to evaulate a expression stored in character (SKIP syntax) on SAS Programming. 12-28-2015 10:09 AM
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My Liked Posts
Subject Likes Posted 1 08-29-2018 07:32 AM
06-26-2023
05:23 PM
Thats right
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06-26-2023
02:46 PM
Thank you for your reply. I am talking about adding data validation list for a perticular cell.
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06-26-2023
01:15 PM
Hi,
Is there anyway to add dropdown list to the excel file using ods tagsets or any other ods engine which creates ods excel file.
Thank you,
Ananda
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06-26-2023
01:13 PM
I am creating a excel file (xml) with ods tagsets and planning to insert some vbcodes as part of the file. Is there a way to do it ?
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08-06-2019
01:12 PM
I will have to create more than 4000 datasets which is am already doing and its taking lot of time to process.
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08-06-2019
11:23 AM
Hi ,
I have lot of records in the dataset not just 3. It will take lot of time to process if i write a macro to subset these. Is there a way i can do it in one step.
Thank you,
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08-05-2019
04:15 PM
Can you help me in solving problem below. I am providing here the appropriate ancestor for each of the record. I need to find this ancestor for each of the record by user. Can you please help.
data a;
user=1; base=4.1; cur=4.2; output; /* 4.1 to 4.2 - Ancestor here is 4.4 */
user=1; base=4.2; cur=4.4; output; /* 4.2 to 4.4 - Ancestor here is 4.4 */
user=1; base=4.6; cur=5.5; output; /* 4.6 to 5.5 - Ancestor here is 6.1 */
user=1; base=5.5; cur=6.1; output; /* 5.5 to 6.1 - Ancestor here is 6.1 */
user=2; base=4.2; cur=4.6; output; /* 4.2 to 4.6 - Ancestor here is 6.1 */
user=2; base=4.6; cur=5.5; output; /* 4.6 to 5.5 - Ancestor here is 6.1 */
user=2; base=5.5; cur=6.1; output; /* 5.0 to 6.1 - Ancestor here is 6.1 */
user=3; base=4.0; cur=4.2; output; /* 4.0 to 4.2 - Ancestor here is 5.5 */
user=3; base=4.2; cur=4.6; output; /* 4.2 to 4.6 - Ancestor here is 5.5 */
user=3; base=4.6; cur=5.5; output; /* 4.6 to 6.1 - Ancestor here is 5.5 */
run;
I am able to achieve this if there is only one user in the dataset using code below. But its not working if there are more than one user.
data _tmp1(rename=(base=base1 cur=cur1));
set a;
run;
**Repeat the training module for each of the delta version between base and current module;
data _tmp2;
retain base cur;
set _tmp1; by u;
cur = cur1;
base = base1;
output;
drop base1 cur1 i;
run;
**Get all the base version for the current modue which has training then need to check the
corresponding base version full access;
data _lineage;
set _tmp2; by u;
array hist(6);
count=1;
hist(count)=base;
do i=1 to last while (count<dim(hist)-1);
set _tmp2(rename=(base=base1 cur=cur1)) nobs=last point=i;
if cur=base1 then do;
count+1;
hist(count)=base1;
base=base1;
cur=cur1;
i=1;
end;
end;
count+1;
hist(count)=cur;
ancest = cur;
run;
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08-30-2018
03:35 PM
Thank you.. I am able to solve this issue with another approach. Thanks for all your help
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08-29-2018
07:32 AM
1 Like
Here it is
data test; input child parent; datalines; 23 11 99 88 77 66 72 68 88 77 66 1 68 55 69 55 70 55 55 56 03 99
;
data lineage(keep=hist1-hist10 flag); set test;
array hist(10); count=1; hist(count)=child; do i=1 to last while (count<dim(hist)-1);
set test(rename=(child=child1 parent=parent1)) nobs=last point=i;
if parent=child1 then do; count+1; hist(count)=child1; child=child1; parent=parent1; i=1; end; end; count+1; hist(count)=parent; flag = parent; run;
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08-28-2018
08:19 PM
Thank you. I am able to do the same thing through another means using point and last in the data step
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08-28-2018
03:07 PM
data HIERARCHY;
input Parent Child;
datalines;
. 1
1 2
1 3
1 8
1 9
2 4
3 5
5 6
6 7
8 10
10 11
10 13;
run;
for the above dataset i want to find the ancestor lineage for each of the records. Can you please help.
Output would look as below.
datalines;
parent child ancestor_lineage
1 2 1
1 3 1
1 8 1
1 9 1
2 4 2,1
3 5 3.1
5 6 5,3,1
6 7 6,5,3,1
8 10 8
10 11 10, 8
10 13; 10,8
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08-28-2018
11:30 AM
Thank you very much. Its kind of working. But I need to output this based on value a and c. In this case the variable value for a is 1 and c is 5.5.
data a; a=1; b=5.1; c=5.5; output; a=1; b=5.2; c=5.5; output; a=1; b=5.0; c=5.2; output; a=1; b=4.7; c=5.0; output; a=1; b=4.6; c=5.2; output; a=1; b=.; c=4.6; output;
a=2; b=5.4; c=6.1; output; a=2; b=5.3; c=6.1; output; a=2; b=5.0; c=5.3; output; a=2; b=4.7; c=5.0; output; a=2; b=4.5; c=4.7; output; run;
For the second set of record where a=2 and lets say value c is 6.1 the children output must be as below
5.4
4.5
Only these two do not have children.
I am sorry for not mentioning this case earlier.
run;
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08-27-2018
03:39 PM
I this case the output will be
new variable
5.1
4.7
4.6
All this do not have any child nodes or children for the ancestor or parent 5.5.
Basically when user asks variable c with value 5.5 then output should be look like above.
Steps are
1. The program reads c variable where value is 5.5 and then looks for the children which is variable b and the values are 5.1 and 5.2
2. For each of b value which is 5.1 and 5.2 the program looks for children and their children are for 5.1 no children so it will appear in the report and 5.2 has children 5.0... and it goes on for all the childrens...
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08-27-2018
01:52 PM
Hi There,
I have a data like this
data a; a=1; b=5.1; c=5.5; output; a=1; b=5.2; c=5.5; output; a=1; b=5.0; c=5.2; output;
a=1; b=4.7; c=5.0; output; a=1; b=4.6; c=5.2; output;
a=1; b=.; c=4.6; output;
run;
I am trying to create a output in which for a given parent nodie (variable c) it should get all the childs/nodes (variable b) which should not have any child/nodes. for e.g in the above example for parent node with value 5.5 it should output 5.1, 4.7 and 4.6 which do not have any child.
Here is how the output is picked.
5.5 has 5.1 and 5.2 child, then 5.2 has child 5.0, 5.0 has child 4.7 and 5.2 has child 4.6. 4.6, 4.7 and 5.1 do not have any childs.
Please let me know if you need more expalnation.
Thanks in advance for the help.
Anand
.
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06-12-2017
01:01 PM
Thank you all. I am able to do this using findc function.
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