cut

Syntax

cut(X, size|cutPositions)

Arguments

X is a scalar/vector/matrix/table.

size is a positive integer that must be no greater than the size of X.

cutPositions is a vector with increasing elements, which is used to specify the starting position of each vector in the result.

Details

This function divides X based on the specified size or cutPositions, and returns a tuple:

  • When X is a scalar, size can only be specified as 1.

  • When X is a vector:

    • if size is specified, it divides X into a list of scalars (size=1) or vectors (size>1) of length size.

    • if cutPositions is specified, it divides X into a list of vectors at the specified positions.

  • When X is a matrix (table):

    • if size is specified, it divides X into several matrices (tables) with size columns (rows).

    • if cutPositions is specified, it divides X into several matrices (tables) at the specified positions.

Refer to function flatten for the reverse operation.

Examples

$ a=1..10;

$ cut(a,2);
([1,2],[3,4],[5,6],[7,8],[9,10])

$ cut(a,3);
([1,2,3],[4,5,6],[7,8,9],[10])

$ cut(a,9);
([1,2,3,4,5,6,7,8,9],[10])

$ b = cut(a,2);
$ b;
([1,2],[3,4],[5,6],[7,8],[9,10])

$ flatten b;
(1,2,3,4,5,6,7,8,9,10)

$ cut(a, 0 2 7);
([1,2],[3,4,5,6,7],[8,9,10])

$ cut(a, 2 7);
([3,4,5,6,7],[8,9,10])

$ m=matrix(1 5 9, 12 20 23, 25 29 32)
$ cut(m,2)

(#0 #1
-- --
1  12
5  20
9  23
,#0
--
25
29
32
)

$ cut(m, 1 2)
(#0
--
12
20
23
,#0
--
25
29
32
)

The cut function can be a convenient tool in time-series data analysis. In the example below, we use the cut function to calculate an aggregate measure between two events.

$ incomes=table(2016.07.31 - 10..1 as date, rand(100,10) as income);
$ incomes;

date

income

2016.07.21

78

2016.07.22

61

2016.07.23

79

2016.07.24

15

2016.07.25

78

2016.07.26

22

2016.07.27

30

2016.07.28

81

2016.07.29

17

2016.07.30

52

$ eventdates = [2016.07.22, 2016.07.25, 2016.07.29];

$ x = incomes.date.binsrch(eventdates);
$ x;
[1,4,8]

$ incomes.date.cut(x);
([2016.07.22,2016.07.23,2016.07.24],[2016.07.25,2016.07.26,2016.07.27,2016.07.28],[2016.07.29,2016.07.30])

$ table(eventdates as startDate, each(last,incomes.date.cut(x)) as endDate, each(sum,incomes.income.cut(x)) as incomeSum);

startDate

endDate

incomeSum

2016.07.22

2016.07.24

155

2016.07.25

2016.07.28

211

2016.07.29

2016.07.30

69