topic Re: How to place List values into the column in Discussions

How to place List values into the column

miguello — Sat, 18 Feb 2017 04:54:21 GMT

As a result of some manipulations I have a list of size of number of rows in my table.

I want to create a new column and place these values into this colum.

JSL reference says that this should work:

dt = New Table( "My Data");
dt << New Column( "Last Name", Character, Values( {"Smith", "Jones",
"Anderson"} ) );

IThe list that I want to put into that column is actually list of lists. If I do Eval(names) I get something like {{"John"}, {"Jim" },{"Jane"}}

How to make it one dimensional list to put into the column?

Thanks!

Re: How to place List values into the column

txnelson — Sat, 18 Feb 2017 12:32:12 GMT

This should do it:

Names Default To Here( 1 );
origlist = {{"John"}, {"Jim"}, {"Jane"}};
newlist = {};
For( i = 1, i <= N Items( origlist ), i++,
	Insert Into( newlist, origlist[i] )
);
dt = New Table( "My Data" );
dt << New Column( "Last Name", Character, Values( newlist ) );

Re: How to place List values into the column

maurogerber — Tue, 21 Feb 2017 06:29:23 GMT

disregard

Re: How to place List values into the column

txnelson — Mon, 20 Feb 2017 17:05:37 GMT

Your code does work, however the issue is that the list, "Origlist" is a list of lists, not just a single list, which is the way your have the list defined in your script. So the issue is, how to get the list of lists into the data table.

Re: How to place List values into the column

Phil_Brown — Tue, 21 Feb 2017 14:52:39 GMT

Following on @txnelson post...I actually created a little function to do this, as I found myself in the same position many times. This will "flatten" list of lists 2 levels deep for example.

Flatten2DList = Function( {list_arg},
	{DefaultLocal},
	NewList = {};
	For( i = 1, i <= N Items( list_arg), i++, 
		For( j = 1, j <= N Items( list_arg[i] ), j++, 
			Insert Into( NewList, list_arg[i][j] );
		);
	);
	NewList;
)

Re: How to place List values into the column

miguello — Tue, 21 Feb 2017 21:31:01 GMT

Jim,

Wouldn't it just take the items in the first list (which are already lists) into new list (also in the form of one-tem-lists)?

Anyways, I couldn't find a way to fight it when list already formed, but I did it like this:

In the loop that constracts my list of lists, I have something like this:

for(t=1, t<=NItems(....)....
......
names[t] = someFunctionThatSpitsSingleNameLists[i][j];
);

I do this:

names[t] = Item(2,Char(someFunctionThatSpitsSingleNameLists[rowTool][rowcurrentReplacementDate]), "\!"");

It just evaluates those single name lists like "{"Jim"}" and then takes only Jim - and puts it in names[t] list.