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Division remainder (rem, %)

Description

Adds a new column that returns the remainder of the division of the integer values in a field by another field or value.

Given the following example...

rem(x, y)

rem(x, y)

...the Division remainder operation calculates the remainder as follows, rounding the result of the division to the nearest integer (as can be done using the Rounding (round) operation):

x - y * round(x/y)

x - y * round(x/y)

The Modulo (mod, %%) operation also returns the remainder of a division but instead rounds down the result of the division to the largest previous integer. Go to the operation article to learn more.

How does it work in the search window?

Select Create column in the search window toolbar, then select the Division remainder operation. You need to specify two arguments:

Argument

Data type

Argument

Data type

Dividend mandatory

integer

Divisor mandatory

integer

The data type of the values in the new column is integer.

Example

We want to find the remainder of the division of the values in the posNumbers1_int field by 2. To do it, we will create a new field using the Division remainder operation and call it PosNumbers division remainder.

The arguments needed to create the new field are:

  • Dividend -posNumbers1_int column

  • Divisor - Click the pencil icon and enter 2

Click Create field to see the result.

How does it work in LINQ?

Use the operator select... as...  and add the operation syntax to create the new column. These are the valid formats of the Division remainder operation:

  • rem(integer1, integer2)

  • integer1 % integer2

Example

You can copy the following LINQ scripts and try the above example on the my.upload.sample.data table.

from my.upload.sample.data select split(message, ";", 16) as posNumbers1, int(posNumbers1) as posNumbers1_int select posNumbers1_int % 2 as remainder

or

from my.upload.sample.data select split(message, ";", 16) as posNumbers1, int(posNumbers1) as posNumbers1_int select rem(posNumbers1_int, 2) as remainder