Can you help me to round like the following?
10 -> 12
21 -> 22
22 -> 22
23 -> 32
34 -> 42
I tried answers from below, but all of them are rounding to next multiplier of a number:
-
1What have you tried so far? Please see: Why is “Can someone help me?” not an actual question?Brian McCutchon– Brian McCutchon2021-01-02 23:02:18 +00:00Commented Jan 2, 2021 at 23:02
Add a comment
|
4 Answers
You can compare the number mod 10 to 2; if less than or equal to 2, add 2 - num % 10 else add 12 - num % 10 to get to the nearest higher (or equal) number ending in 2:
def raise_to_two(num):
if num % 10 <= 2:
return num + 2 - num % 10
return num + 12 - num % 10
print(raise_to_two(10))
print(raise_to_two(21))
print(raise_to_two(22))
print(raise_to_two(23))
print(raise_to_two(34))
Output:
12
22
22
32
42
Note (thanks to @MarkDickinson for pointing this out) that because the python modulo operator always returns a positive number if the second argument (to the right of % is positive), you can simplify the above to
def raise_to_two(num):
return num + (2 - num) % 10
The output remains the same
4 Comments
Michael Hofmann
Just compare mod 2. If it is zero add 2, if it is one add 1.
Nick
@MichaelHofmann it's not rounding to the next multiple of 2, it's rounding to the next higher number which ends in 2.
Mark Dickinson
You could get rid of the branching here and just return
num + (2 - num) % 10Nick
@MarkDickinson thanks for that - I should have tested it because I can never seem to remember which languages return negative results for modulo when one input is negative. I've updated the answer.
import math
x = [10, 21, 22, 23, 34]
for n in x:
print((math.ceil((n-2)/10)*10)+2)
outputs:
12
22
22
32
42
4 Comments
Brian McCutchon
The parentheses around
math.ceil(...) * 10 are unnecessary.
couka
@BrianMcCutchon Yes, but improve readability in my opinion
wjandrea
@couka For best readability, you could use a separate variable and add spaces, like
out = math.ceil((n-2)/10)*10 + 2; print(out)couka
Well, readability vs shortness can be a trade-off and that's where the optimum is for me.
this should also work.
arr = [10, 21, 22, 23, 34]
for a in arr:
b = ((a-3) // 10) * 10 + 12
print(f"{a} -> {b}")
Comments
This is the code:
def round_by_two(number):
unitOfDecena = number // 10
residual = number % 10
if residual == 2:
requiredNumber = number
elif residual > 2:
requiredNumber = (unitOfDecena + 1)*10 + 2
else:
requiredNumber = unitOfDecena*10 + 2
return requiredNumber
