12-06-2007 01:49 PM
12-06-2007 01:59 PM
kharris126,
On the Array pallet, look at the Reshape Array.vi. Use index array to extract your resistance column, then wire this to reshape and specify 51 for first dimension (rows) and 10 for second dimension (columns). You will need to resize the vi to add the second input.
-cb
12-06-2007 02:00 PM - edited 12-06-2007 02:01 PM
12-06-2007 02:04 PM
One problem is that you don't have an original array going into the replace array subset function you have circled.
With doing any insert into array function, you will have to be careful with the array functions and may need to do transpositions here or there to make it work. For example, you can't put a 1-D array of 10 rows and put it into a 10 column by 51 row array without doing a transpostion to make it a 10 column by 1 row array. Any work you do will require that the one of the dimensions will match in size. If you put a 2 row by 2 column array into a 10 row by 10 column array, which way does it insert, to the right or down? What goes into the empty elements that get created in either the remainin column, or remaining rows?
12-06-2007 02:05 PM - edited 12-06-2007 02:06 PM
Notice how I left the input of the Insert into array empty?
Let me clean up the picture and repost!
12-06-2007 02:11 PM - edited 12-06-2007 02:12 PM
12-06-2007 02:18 PM - edited 12-06-2007 02:19 PM
10Degee is correct. Reshape array is the proper function. Here's a quick draft, see if it suits your purpose. 🙂
(It will even pad the last row if the number is not divisible by 10).
12-06-2007 02:28 PM
12-06-2007 02:35 PM - edited 12-06-2007 02:37 PM
12-06-2007 02:37 PM