How to round number to 10, 100, 1000, 10000, 100000

Question

I want a function that will round the number to 10 if it is less than 10, round to 100 if it is between 10 to 1000 and so on.

this will help number concatenation problems, so I don't have to turn number to str and back.

def roundup(n):
    if n < 0: return 0
    if n < 10: return 10
    if n < 100: return 100
    if n < 1000: return 1000
    if n < 10000: return 10000

I want to know if there is a natural way to do this without just stacking 'if's together

Possible duplicate of [Rounding to a power of 10](https://stackoverflow.com/questions/6769406/rounding-to-a-power-of-10). Actually this is the wrong dupe target, but the logic is the same. (`10**floor(log10(x))`) — pault, Jan 09 '19 at 16:34
Yes, you are right. but i got the answer I want here, a naive python way — Matt Cao, Jan 09 '19 at 16:57

score 6 · Answer 1 · edited Jan 09 '19 at 16:47

6

This is more of a math problem than a python specific problem but you are effectively taking a floor of the log of a number.

def round_base_10(x):
    if x < 0:
        return 0
    elif x == 0:
        return 10
    return 10**math.ceil(math.log10(x))

edited Jan 09 '19 at 16:47

Gerges

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answered Jan 09 '19 at 16:37

Victor 'Chris' Cabral

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`round_base_10(11) = 10` which is not correct, it should be 100. – Gerges Jan 09 '19 at 16:39
1

You should use `math.ceil` instead, and it will work. Also, need to take care of case when x = 0 – Gerges Jan 09 '19 at 16:41
1

Thanks for the feedback. I added changed ceil as well as added the n<0 case – Victor 'Chris' Cabral Jan 09 '19 at 16:42
still needs case for x < 1 and also if x == 10, answer should be 100, not 10, etc.... – Gerges Jan 09 '19 at 16:47

score 0 · Answer 2 · answered Jan 09 '19 at 16:41

0

Another option (without math):

def roundup(n):
    return 10**(len(str(n))-1)

answered Jan 09 '19 at 16:41

Tarifazo

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score 0 · Accepted Answer · answered Jan 09 '19 at 16:41

0

You could use the naive approach:

def roundup(n):
    top = 1
    while True:
        if n <= top:
            return top
        top *= 10

You would need top = 1L in Python2, but integers are long by default in Python3.

answered Jan 09 '19 at 16:41

Serge Ballesta

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This is what I am looking for – Matt Cao Jan 09 '19 at 16:49

score 0 · Answer 4 · answered Jan 09 '19 at 16:41

0

def roundup(n):
    if n < 0:
        return 0
    size = len(str(n))
    return int('1'.ljust(size+1, '0'))

answered Jan 09 '19 at 16:41

Victor Outtes

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How to round number to 10, 100, 1000, 10000, 100000

4 Answers4