Hello everyone,   I am using VA 7.2 and I have a question about mapping a column to two different columns in an Info Window. I know that a column can only be linked to one other column, so I tried duplicating the column and link the duplicate to the second column. However, I want to display all the rows where either the first column equals the chosen value or the second column equals the chosen value. Let me illustrate the problem with a simple example. I have 5 players that play a simple game (coinflip), and I want to see all the games a player has played by clicking on the player. Let the List Table containing the players be given by:   Player      Wins A              3 B              1 C              0 D              2 E              1   Now, I want to click on the row for player A and see all the games played by player A. So I want to have an Info Window, containing a List Table that looks like this:   Player 1     Player2     Winner    A                B                1  A                D                2  E                A                1  A                C                1  B                A                2   Hence, I want to map the column Player from the first List Table to Player 1 and Player 2 in the second List Table. However, by duplicating the column Player, and linking one to Player 1 and one to Player 2, the Info Window will only show the rows where both Player 1 and Player 2 are equal to A, which are none. I know a solution could be to duplicate the entire dataset with games and create 2 rows for every game and write both players once as player 1. However, I prefer not to duplicate the data set due to its size, and the 2 columns do have a different interpretation in my data set.    Does anybody have an other solution how to achieve this?   Thanks in advance.   Mick ... View more

Hello everyone,   I am using SAS Visual Analytics 7.2 and I have a question about the filter interaction between a list table and a list. My list table filters my list that contains one of the columns/variables from the list table. Now, when I use this list to filter a second list table that also contains the column/variable from the list, the second list table is not only filtered by the chosen value from the list, but the entire selected row in the first list table is passed through as a filter. I will try to explain my problem with a simple example. Let the first list table be given by:   Name          Age           Sex Pete            21               M Rose           23               F Hank           45               M Jenny          67               F  James         23               M   Now, this list table is linked to a list containing the variable Sex, and when I select the first row of the list table (Pete), I want to obtain the value for the variable Sex of that row, which is M, and use this as a filter for the second list table that for example looks like this:   Name          Age           Sex Holly           33               F Pete            21               M Rose           23               F Hank           45               M Jenny          67               F  James         23               M   I want to use the selection in the first list table to filter on the variable Sex in the second list table. So, if I select the row for Pete in the first list table, I want to have all the rows with Sex equal to M in the second list table. My idea was to place a list containing the variable Sex between the two list tables. Then by selecting Pete in the first list table, which is linked to the list, I have that in the list for Sex only M is selected. Then, I would think that by linking this list to my second list table, it would filter the second list table on Sex = M, however, it filters on all the information on the row for Pete, hence it only shows the exact same row (Pete) in the second list table, instead of all rows with Sex = M. Now, my question: Is it possible to select a row in a list table, and use only the value of one particular column/variable of that row which can be used to filter another list table (containing this same column/variable).    I hope my explanation is clear enough and you guys know a solution.   Thanks in advance.   Mick ... View more