bfill

bfill(obj, [limit])

• If obj is a vector: back fill the null values in obj with the next non-null value.
• If obj is an array vector:
  • For an empty row, fill it with the first non-empty row that follows it.
  • For null values in a column, fill them with the first non-null value that follows within the same column.
• If obj is a matrix or a table: back fill the null values in each column of objwith the next non-null value. according to the above rules.

This operation creates a new object and does not change the input obj. Function bfill! changes the input obj.

obj is a vector/matrix/table/array vector.

limit (optional) is a positive integer indicating the maximum number of consecutive null values to be filled. It cannot be specified when obj is an array vector.

Returns the object with null values filled, with the same type and form as obj.

x=1 2 3 NULL NULL NULL 4 5 6
x.bfill();
// output: [1,2,3,4,4,4,4,5,6]

x=1 2 3 NULL NULL NULL 4 5 6
x.bfill(1);
// output: [1,2,3,,,4,4,5,6]

x.bfill!(2);
x;
// output: [1,2,3,,4,4,4,5,6]

date=[2012.06.12,2012.06.12,2012.06.13,2012.06.14,2012.06.15]
sym=["IBM","MSFT","IBM","MSFT","MSFT"]
price=[,,26.56,,50.76]
qty=[,,4500,5600,6800]
timestamp=[09:34:07,09:35:26,09:36:42,09:36:51,09:36:59]
t=table(date,timestamp,sym,price,qty)
t;

t.bfill()

select date, timestamp, sym, price.bfill() as price, qty.bfill() as qty from t context by sym;

In the following example, the third row of the array vector x is empty, so it is filled with the fourth row [8, 9, 10]; the first element of the third column is null, so it is filled with the next non-null value 10 in that column.

x = array(INT[], 0).append!([1 2 NULL, 4 5, NULL, 8 9 10])
x
// output:[[1,2,NULL],[4,5],[NULL],[8,9,10]]
bfill(x)
// output:[[1,2,10],[4,5],[8,9,10],[8,9,10]]