Python: how can I round a number downward to next 1000
Python: how can I round a number downward to next 1000
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()
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 at 5:45
This should help: stackoverflow.com/q/34030509/2988730
– Mad Physicist
Aug 23 at 5:50
Possible duplicate of Python Rounding Down to Custom Step
– Thomas
Aug 23 at 6:33
Did an answer below help? If so, please consider accepting (green tick on the left).
– jpp
Aug 26 at 11:45
3 Answers
3
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.
math.floor
Maybe you can try it this way
import math
math.floor(1900 / 100) * 100
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. :-)
– Thomas
Aug 23 at 6:42
math.floor([field]) rounds down to next integer
math.floor([field])
math.ceil([field]/1000)*1000 rounds down to next 1000
math.ceil([field]/1000)*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)
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math.floor()?– dhke
Aug 23 at 5:40