python fixed digits rounding

Question

Is there an function in python 2 that can do this?

1234 -> round(1234, 2) = 1200
1234 -> round(1234, 3) = 1230
12.34 -> round(12.34, 3) = 12.3

Basically the second number says the precision of the number and the everything behind should be rounded.

Based on the comment i came up with this:

def round_to_precision(x, precision):
    return int(round(x / float(10 ** precision))) * 10 ** precision

But this is still wrong, because i don't know the size of the numbers.

check [this](https://stackoverflow.com/questions/8866046/python-round-up-integer-to-next-hundred) site. This algorithm applies to all numbers — snatchysquid, Jun 02 '20 at 10:21
thank you. I made an edit, but there is still something wrong. — user3680510, Jun 02 '20 at 10:29

score 1 · Accepted Answer · answered Jun 02 '20 at 10:49

Here's a solution (written step by step for clarity).

import math

num_digits = lambda x: int((math.log(x, 10)) + 1)

def round(x, precision): 
    digits = num_digits(x) 
    gap = precision - digits
    x = x * (10 ** gap)
    x = int(x) 
    x = x / (10 ** gap)
    return x

results:

round(1234, 2) # 1200
round(1234, 3) # 1230
round(12.34, 3) # 12.3

score 0 · Answer 2 · answered Jun 02 '20 at 10:42

0

I found a solution:

def round_to_precision(x, precision):
    fmt_string = '{:.' + str(precision) + 'g}'
    return float(fmt_string.format(x))


print round_to_precision(1234, 2)
print round_to_precision(1234, 3)
print round_to_precision(12.34, 3)

answered Jun 02 '20 at 10:42

user3680510

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python fixed digits rounding

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