将数字四舍五入至最接近的基数10

Question

I want to get the nearest base 10 number (ex. 10, 100, 1000) called new_n of the input n . For example, 99 gets 100 (not 10 ), 551 gets 1000 (not 100 ).

What I try is using np.log10 to extract the power of the input number n and use it to power 10

import numpy as np

n = 0.09
new_n = 10**(int(round(np.log10(n))))
print new_n

n = 35
new_n = 10**(int(round(np.log10(n))))
print new_n

n = 999
new_n = 10**(int(round(np.log10(n))))
print new_n

n = 4655
new_n = 10**(int(round(np.log10(n))))
print new_n

> 0.1
> 100
> 1000
> 10000

The problem is the number such as 35 its np.log10(n) (which I expect to use as power) is 1.544068 and its rounding is 2 . So the result is 100 instead of 10 . Or the result of 4655 is 10000 . How to round the number to the nearest base 10 number?

Answer 1

Add a check back in the linear space (with a possible correction) after obtaining the exponent:

n = np.array([0.09,35,549,551,999,4655])
e = np.log10(n).round()
#array([-1.,  2.,  3.,  3.,  3.,  4.])

So, is the number closer to the "rounded" answer or to the previous degree of 10?

new_n = np.where(10**e - n <= n - 10**(e - 1), 10**e, 10**(e - 1))
#array([  1.00000000e-01,   1.00000000e+01,   1.00000000e+02,
#         1.00000000e+03,   1.00000000e+03,   1.00000000e+03])

Answer 2

You can achieve what you want by scaling your numbers before taking the logs. Multiply each value by sqrt(10) / 5.5 and you should get the results you want:

n = np.array([0.09,35,549,551,999,4655]) # borrowed test values from @DYZ
multiplier = np.sqrt(10) / 5.5
results = 10**np.log10(multiplier * n).round()
# array([  1.00000000e-01,   1.00000000e+01,   1.00000000e+02,
#          1.00000000e+03,   1.00000000e+03,   1.00000000e+03])

Answer 3

You can try this one:

import np
new_n = 10**(np.floor(np.log10(n)-np.log10(0.5)))