I need to round a float to be displayed in a UI. e.g, to one significant figure:
1234 -> 1000
0.12 -> 0.1
0.012 -> 0.01
0.062 -> 0.06
6253 -> 6000
1999 -> 2000
Is there a nice way to do this using the Python library, or do I have to write it myself?
To directly answer the question, here's my version using naming from the R function:
import math
def signif(x, digits=6):
if x == 0 or not math.isfinite(x):
return x
digits -= math.ceil(math.log10(abs(x)))
return round(x, digits)
My main reason for posting this answer are the comments complaining that "0.075" rounds to 0.07 rather than 0.08. This is due, as pointed out by "Novice C", to a combination of floating point arithmetic having both finite precision and a base-2 representation. The nearest number to 0.075 that can actually be represented is slightly smaller, hence rounding comes out differently than you might naively expect.
Also note that this applies to any use of non-decimal floating point arithmetic, e.g. C and Java both have the same issue.
To show in more detail, we ask Python to format the number in "hex" format:
0.075.hex()
which gives us: 0x1.3333333333333p-4. The reason for doing this is that the normal decimal representation often involves rounding and hence is not how the computer actually "sees" the number. If you're not used to this format, a couple of useful references are the Python docs and the C standard.
To show how these numbers work a bit, we can get back to our starting point by doing:
0x13333333333333 / 16**13 * 2**-4
which should should print out 0.075. 16**13 is because there are 13 hexadecimal digits after the decimal point, and 2**-4 is because hex exponents are base-2.
Now we have some idea of how floats are represented we can use the decimal module to give us some more precision, showing us what's going on:
from decimal import Decimal
Decimal(0x13333333333333) / 16**13 / 2**4
giving: 0.07499999999999999722444243844 and hopefully explaining why round(0.075, 2) evaluates to 0.07
To round an integer to 1 significant figure the basic idea is to convert it to a floating point with 1 digit before the point and round that, then convert it back to its original integer size.
To do this we need to know the largest power of 10 less than the integer. We can use floor of the log 10 function for this.
from math import log10, floor def round_int(i,places): if i == 0: return 0 isign = i/abs(i) i = abs(i) if i < 1: return 0 max10exp = floor(log10(i)) if max10exp+1 < places: return i sig10pow = 10**(max10exp-places+1) floated = i*1.0/sig10pow defloated = round(floated)*sig10pow return int(defloated*isign)
The posted answer was the best available when given, but it has a number of limitations and does not produce technically correct significant figures.
numpy.format_float_positional supports the desired behaviour directly. The following fragment returns the float x formatted to 4 significant figures, with scientific notation suppressed.
import numpy as np
x=12345.6
np.format_float_positional(x, precision=4, unique=False, fractional=False, trim='k')
> 12340.
The sigfig package/library covers this. After installing you can do the following:
>>> from sigfig import round
>>> round(1234, 1)
1000
>>> round(0.12, 1)
0.1
>>> round(0.012, 1)
0.01
>>> round(0.062, 1)
0.06
>>> round(6253, 1)
6000
>>> round(1999, 1)
2000
Using python 2.6+ new-style formatting (as %-style is deprecated):
>>> "{0}".format(float("{0:.1g}".format(1216)))
'1000.0'
>>> "{0}".format(float("{0:.1g}".format(0.00356)))
'0.004'
In python 2.7+ you can omit the leading 0s.
I adapted one of the answers. I like this:
def sigfiground(number:float, ndigits=3)->float:
return float(f"{number:.{ndigits}g}")
I use it when I still want a float (I do formatting elsewhere).
I ran into this as well but I needed control over the rounding type. Thus, I wrote a quick function (see code below) that can take value, rounding type, and desired significant digits into account.
import decimal
from math import log10, floor
def myrounding(value , roundstyle='ROUND_HALF_UP',sig = 3):
roundstyles = [ 'ROUND_05UP','ROUND_DOWN','ROUND_HALF_DOWN','ROUND_HALF_UP','ROUND_CEILING','ROUND_FLOOR','ROUND_HALF_EVEN','ROUND_UP']
power = -1 * floor(log10(abs(value)))
value = '{0:f}'.format(value) #format value to string to prevent float conversion issues
divided = Decimal(value) * (Decimal('10.0')**power)
roundto = Decimal('10.0')**(-sig+1)
if roundstyle not in roundstyles:
print('roundstyle must be in list:', roundstyles) ## Could thrown an exception here if you want.
return_val = decimal.Decimal(divided).quantize(roundto,rounding=roundstyle)*(decimal.Decimal(10.0)**-power)
nozero = ('{0:f}'.format(return_val)).rstrip('0').rstrip('.') # strips out trailing 0 and .
return decimal.Decimal(nozero)
for x in list(map(float, '-1.234 1.2345 0.03 -90.25 90.34543 9123.3 111'.split())):
print (x, 'rounded UP: ',myrounding(x,'ROUND_UP',3))
print (x, 'rounded normal: ',myrounding(x,sig=3))
This function does a normal round if the number is bigger than 10**(-decimal_positions), otherwise adds more decimal until the number of meaningful decimal positions is reached:
def smart_round(x, decimal_positions):
dp = - int(math.log10(abs(x))) if x != 0.0 else int(0)
return round(float(x), decimal_positions + dp if dp > 0 else decimal_positions)
Hope it helps.
https://stackoverflow.com/users/1391441/gabriel, does the following address your concern about rnd(.075, 1)? Caveat: returns value as a float
def round_to_n(x, n):
fmt = '{:1.' + str(n) + 'e}' # gives 1.n figures
p = fmt.format(x).split('e') # get mantissa and exponent
# round "extra" figure off mantissa
p[0] = str(round(float(p[0]) * 10**(n-1)) / 10**(n-1))
return float(p[0] + 'e' + p[1]) # convert str to float
>>> round_to_n(750, 2)
750.0
>>> round_to_n(750, 1)
800.0
>>> round_to_n(.0750, 2)
0.075
>>> round_to_n(.0750, 1)
0.08
>>> math.pi
3.141592653589793
>>> round_to_n(math.pi, 7)
3.141593
This returns a string, so that results without fractional parts, and small values which would otherwise appear in E notation are shown correctly:
def sigfig(x, num_sigfig):
num_decplace = num_sigfig - int(math.floor(math.log10(abs(x)))) - 1
return '%.*f' % (num_decplace, round(x, num_decplace))
Given a question so thoroughly answered why not add another
This suits my aesthetic a little better, though many of the above are comparable
import numpy as np
number=-456.789
significantFigures=4
roundingFactor=significantFigures - int(np.floor(np.log10(np.abs(number)))) - 1
rounded=np.round(number, roundingFactor)
string=rounded.astype(str)
print(string)
This works for individual numbers and numpy arrays, and should function fine for negative numbers.
There's one additional step we might add - np.round() returns a decimal number even if rounded is an integer (i.e. for significantFigures=2 we might expect to get back -460 but instead we get -460.0). We can add this step to correct for that:
if roundingFactor<=0:
rounded=rounded.astype(int)
Unfortunately, this final step won't work for an array of numbers - I'll leave that to you dear reader to figure out if you need.
import math
def sig_dig(x, n_sig_dig):
num_of_digits = len(str(x).replace(".", ""))
if n_sig_dig >= num_of_digits:
return x
n = math.floor(math.log10(x) + 1 - n_sig_dig)
result = round(10 ** -n * x) * 10 ** n
return float(str(result)[: n_sig_dig + 1])
>>> sig_dig(1234243, 3)
>>> sig_dig(243.3576, 5)
1230.0
243.36
Most of these answers involve the math, decimal and/or numpy imports or output values as strings. Here is a simple solution in base python that handles both large and small numbers and outputs a float:
def sig_fig_round(number, digits=3):
power = "{:e}".format(number).split('e')[1]
return round(number, -(int(power) - digits))
A simple variant using the standard decimal library
from decimal import Decimal
def to_significant_figures(v: float, n_figures: int) -> str:
d = Decimal(v)
d = d.quantize(Decimal((0, (), d.adjusted() - n_figures + 1)))
return str(d.quantize(Decimal(1)) if d == d.to_integral() else d.normalize())
Testing it
>>> to_significant_figures(1.234567, 3)
'1.23'
>>> to_significant_figures(1234567, 3)
'1230000'
>>> to_significant_figures(1.23, 7)
'1.23'
>>> to_significant_figures(123, 7)
'123'
This function takes both positive and negative numbers and does the proper significant digit rounding.
from math import floor
def significant_arithmetic_rounding(n, d):
'''
This function takes a floating point number and the no. of significant digit d, perform significant digits
arithmetic rounding and returns the floating point number after rounding
'''
if n == 0:
return 0
else:
# Checking whether the no. is negative or positive. If it is negative we will take the absolute value of it and proceed
neg_flag = 0
if n < 0:
neg_flag = 1
n = abs(n)
n1 = n
# Counting the no. of digits to the left of the decimal point in the no.
ld = 0
while(n1 >= 1):
n1 /= 10
ld += 1
n1 = n
# Counting the no. of zeros to the right of the decimal point and before the first significant digit in the no.
z = 0
if ld == 0:
while(n1 <= 0.1):
n1 *= 10
z += 1
n1 = n
# No. of digits to be considered after decimal for rounding
rd = (d - ld) + z
n1 *= 10**rd
# Increase by 0.5 and take the floor value for rounding
n1 = floor(n1+0.5)
# Placing the decimal point at proper position
n1 /= 10 ** rd
# If the original number is negative then make it negative
if neg_flag == 1:
n1 = 0 - n1
return n1
Testing:
>>> significant_arithmetic_rounding(1234, 3)
1230.0
>>> significant_arithmetic_rounding(123.4, 3)
123.0
>>> significant_arithmetic_rounding(0.0012345, 3)
0.00123
>>> significant_arithmetic_rounding(-0.12345, 3)
-0.123
>>> significant_arithmetic_rounding(-30.15345, 3)
-30.2
Easier to know an answer works for your needs when it includes examples. The following is built on previous solutions, but offers a more general function which can round to 1, 2, 3, 4, or any number of significant digits.
import math
# Given x as float or decimal, returns as string a number rounded to "sig" significant digts
# Return as string in order to control significant digits, could be a float or decimal
def round_sig(x, sig=2):
r = round(x, sig-int(math.floor(math.log10(abs(x))))-1)
floatsig = "%." + str(sig) + "g"
return "%d"%r if abs(r) >= 10**(sig-1) else '%s'%float(floatsig % r)
>>> a = [1234, 123.4, 12.34, 1.234, 0.1234, 0.01234, 0.25, 1999, -3.14, -48.01, 0.75]
>>> [print(i, "->", round_sig(i,1), round_sig(i), round_sig(i,3), round_sig(i,4)) for i in a]
1234 -> 1000 1200 1230 1234
123.4 -> 100 120 123 123.4
12.34 -> 10 12 12.3 12.34
1.234 -> 1 1.2 1.23 1.234
0.1234 -> 0.1 0.12 0.123 0.1234
0.01234 -> 0.01 0.012 0.0123 0.01234
0.25 -> 0.2 0.25 0.25 0.25
1999 -> 2000 2000 2000 1999
-3.14 -> -3 -3.1 -3.14 -3.14
-48.01 -> -50 -48 -48.0 -48.01
0.75 -> 0.8 0.75 0.75 0.75
in very cases, the number of significant is depend on to the evaluated process, e.g. error. I wrote the some codes which returns a number according to it's error (or with some desired digits) and also in string form (which doesn't eliminate right side significant zeros)
import numpy as np
def Sig_Digit(x, *N,):
if abs(x) < 1.0e-15:
return(1)
N = 1 if N ==() else N[0]
k = int(round(abs(N)-1))-int(np.floor(np.log10(abs(x))))
return(k);
def Sig_Format(x, *Error,):
if abs(x) < 1.0e-15:
return('{}')
Error = 1 if Error ==() else abs(Error[0])
k = int(np.floor(np.log10(abs(x))))
z = x/10**k
k = -Sig_Digit(Error, 1)
m = 10**k
y = round(x*m)/m
if k < 0:
k = abs(k)
if z >= 9.5:
FMT = '{:'+'{}'.format(1+k)+'.'+'{}'.format(k-1)+'f}'
else:
FMT = '{:'+'{}'.format(2+k)+'.'+'{}'.format(k)+'f}'
elif k == 0:
if z >= 9.5:
FMT = '{:'+'{}'.format(1+k)+'.0e}'
else:
FMT = '{:'+'{}'.format(2+k)+'.0f}'
else:
FMT = '{:'+'{}'.format(2+k)+'.'+'{}'.format(k)+'e}'
return(FMT)
def Sci_Format(x, *N):
if abs(x) < 1.0e-15:
return('{}')
N = 1 if N ==() else N[0]
N = int(round(abs(N)-1))
y = abs(x)
k = int(np.floor(np.log10(y)))
z = x/10**k
k = k-N
m = 10**k
y = round(x/m)*m
if k < 0:
k = abs(k)
if z >= 9.5:
FMT = '{:'+'{}'.format(1+k)+'.'+'{}'.format(k-1)+'f}'
else:
FMT = '{:'+'{}'.format(2+k)+'.'+'{}'.format(k)+'f}'
elif k == 0:
if z >= 9.5:
FMT = '{:'+'{}'.format(1+k)+'.0e}'
else:
FMT = '{:'+'{}'.format(2+k)+'.0f}'
else:
FMT = '{:'+'{}'.format(2+N)+'.'+'{}'.format(N)+'e}'
return(FMT)
def Significant(x, *Error):
N = 0 if Error ==() else Sig_Digit(abs(Error[0]), 1)
m = 10**N
y = round(x*m)/m
return(y)
def Scientific(x, *N):
m = 10**Sig_Digit(x, *N)
y = round(x*m)/m
return(y)
def Scientific_Str(x, *N,):
FMT = Sci_Format(x, *N)
return(FMT.format(x))
def Significant_Str(x, *Error,):
FMT = Sig_Format(x, *Error)
return(FMT.format(x))
test code:
X = [19.03345607, 12.075, 360.108321344, 4325.007605343]
Error = [1.245, 0.1245, 0.0563, 0.01245, 0.001563, 0.0004603]
for x in X:
for error in Error:
print(x,'+/-',error, end=' \t==> ')
print(' (',Significant_Str(x, error), '+/-', Scientific_Str(error),')')
print out:
19.03345607 +/- 1.245 ==> ( 19 +/- 1 )
19.03345607 +/- 0.1245 ==> ( 19.0 +/- 0.1 )
19.03345607 +/- 0.0563 ==> ( 19.03 +/- 0.06 )
19.03345607 +/- 0.01245 ==> ( 19.03 +/- 0.01 )
19.03345607 +/- 0.001563 ==> ( 19.033 +/- 0.002 )
19.03345607 +/- 0.0004603 ==> ( 19.0335 +/- 0.0005 )
12.075 +/- 1.245 ==> ( 12 +/- 1 )
12.075 +/- 0.1245 ==> ( 12.1 +/- 0.1 )
12.075 +/- 0.0563 ==> ( 12.07 +/- 0.06 )
12.075 +/- 0.01245 ==> ( 12.07 +/- 0.01 )
12.075 +/- 0.001563 ==> ( 12.075 +/- 0.002 )
12.075 +/- 0.0004603 ==> ( 12.0750 +/- 0.0005 )
360.108321344 +/- 1.245 ==> ( 360 +/- 1 )
360.108321344 +/- 0.1245 ==> ( 360.1 +/- 0.1 )
360.108321344 +/- 0.0563 ==> ( 360.11 +/- 0.06 )
360.108321344 +/- 0.01245 ==> ( 360.11 +/- 0.01 )
360.108321344 +/- 0.001563 ==> ( 360.108 +/- 0.002 )
360.108321344 +/- 0.0004603 ==> ( 360.1083 +/- 0.0005 )
4325.007605343 +/- 1.245 ==> ( 4325 +/- 1 )
4325.007605343 +/- 0.1245 ==> ( 4325.0 +/- 0.1 )
4325.007605343 +/- 0.0563 ==> ( 4325.01 +/- 0.06 )
4325.007605343 +/- 0.01245 ==> ( 4325.01 +/- 0.01 )
4325.007605343 +/- 0.001563 ==> ( 4325.008 +/- 0.002 )
4325.007605343 +/- 0.0004603 ==> ( 4325.0076 +/- 0.0005 )