Python: how can I round a number downward to next 1000

Question

In python, there's a builtin function round(),
it rounds a number like this:

round(1900, -3) == 2000

is there a builtin function that can round a number downward, like this:

function(1900, -3) == 1000

[`math.floor()`](https://docs.python.org/3/library/math.html#math.floor)? — dhke, Aug 23 '18 at 05:40
How much do you want to round it downwards? I mean is not the same to go on units, decimal, hunderds, thousands. — jalazbe, Aug 23 '18 at 05:45
This should help: https://stackoverflow.com/q/34030509/2988730 — Mad Physicist, Aug 23 '18 at 05:50
Possible duplicate of [Python Rounding Down to Custom Step](https://stackoverflow.com/questions/51895370/python-rounding-down-to-custom-step) — tafaust, Aug 23 '18 at 06:33

jpp · Answer 1 · 2018-08-23T08:01:09.937

4

You can use floor division:

def round_down(x, k=3):
    n = 10**k
    return x // n * n

res = round_down(1900)  # 1000

math.floor will also work, but with a drop in performance, see Python integer division operator vs math.floor.

edited Aug 23 '18 at 08:01

answered Aug 23 '18 at 07:44

jpp

score 2 · Answer 2 · edited Aug 23 '18 at 05:53

2

Maybe you can try it this way

import math
math.floor(1900 / 100) * 100

edited Aug 23 '18 at 05:53

Jeru Luke

answered Aug 23 '18 at 05:46

moonscar

Looks good (+1). The idea from user 侯月源 is to bring it to the precision level you want to round for (change the order of magnitude) and then do the inverse operation after rounding. 侯月源 the only thing left is to put it into a function to suite the OP's needs. :-) – tafaust Aug 23 '18 at 06:42

S-Man · Answer 3 · 2018-08-24T14:34:59.183

0

math.floor([field]) rounds down to next integer

math.ceil([field]/1000)*1000 rounds down to next 1000

Maybe you could make an int cast after that.

if you like your syntax with the exponent parameter you could define your own function:

import math

def floorTo10ths(number, exp):
     return int(math.floor(number/10**exp) * 10**exp) 

floorTo10ths(1900, 3)

edited Aug 24 '18 at 14:34

answered Aug 23 '18 at 05:48

S-Man

score 0 · Answer 4 · answered May 06 '22 at 15:14

0

def arroundSup(number, power) :
   nbr = number / (1000*power)
   return ceil(nbr) * (1000*power)


arroundSup(1900, 3) # will return 2000

---------------------------------------------------

def arroundLow(number, power) :
   nbr = number / (1000*power)
   return floor(nbr) * (1000*power)


arroundLow(1900, 3) # will return 1000

answered May 06 '22 at 15:14

KALAGA Serge Eric

"will return 1000" No, it returns 0. Please test code before submitting it. – General Grievance Feb 08 '23 at 14:54