Series.div

Return Floating division of series and other, element-wise (binary operator truediv).

danfo.Series.div(other)  [source]
Parameters
Type
Description
Default
other
Series|int|
values 
​
round
bool
specify if to round off the floating number
true
Return: Series
Example
divide with values of another series
Node
Node
const dfd = require("danfojs-node")
​
let data1 = [30, 40, 3, 5]
let data2 = [1, 2, 3, 4]
let sf1 = new dfd.Series(data1)
let sf2 = new dfd.Series(data2)
sf1.div(sf2).print()
Output
Output
╔═══╤══════════════════════╗
║   │ 0                    ║
╟───┼──────────────────────╢
║ 0 │ 30                   ║
╟───┼──────────────────────╢
║ 1 │ 20                   ║
╟───┼──────────────────────╢
║ 2 │ 1                    ║
╟───┼──────────────────────╢
║ 3 │ 1.25                 ║
╚═══╧══════════════════════╝
divide with a value
Node
Node
const dfd = require("danfojs-node")
​
let data1 = [1, 2, 3, 4, 5]
let sf1 = new dfd.Series(data1)
​
sf1.div(2).print()
Output
Output
╔═══╤══════════════════════╗
║   │ 0                    ║
╟───┼──────────────────────╢
║ 0 │ 0.5                  ║
╟───┼──────────────────────╢
║ 1 │ 1                    ║
╟───┼──────────────────────╢
║ 2 │ 1.5                  ║
╟───┼──────────────────────╢
║ 3 │ 2                    ║
╟───┼──────────────────────╢
║ 4 │ 2.5                  ║
╚═══╧══════════════════════╝

Series.pow

Series.mul

Last updated 11 months ago

Parameters	Type	Description	Default
other	Series\|int\|	values
round	bool	specify if to round off the floating number	true