How to round a number to significant figures in Python

Question

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

User · Answer

I modified indgar s solution to handle negative numbers and small numbers  including zero    from math import log10  floor def round sig x  sig 6  small value 1 0e-9       return round x  sig - int floor log10 max abs x   abs small value      - 1

User · Answer

I can t think of anything that would be able to handle this out of the box  But it s fairly well handled for floating point numbers    gt  gt  gt  round 1 2322  2  1 23   Integers are trickier  They re not stored as base 10 in memory  so significant places isn t a natural thing to do  It s fairly trivial to implement once they re a string though   Or for integers    gt  gt  gt  def intround n  sigfigs         n   str n        return n  sigfigs      0     len n - sigfigs      gt  gt  gt  intround 1234  1   1000   gt  gt  gt  intround 1234  2    If you would like to create a function that handles any number  my preference would be to convert them both to strings and look for a decimal place to decide what to do    gt  gt  gt  def roundall1 n  sigfigs         n   str n        try          sigfigs   n index            except ValueError          pass       return intround n  sigfigs    Another option is to check for type  This will be far less flexible  and will probably not play nicely with other numbers such as Decimal objects    gt  gt  gt  def roundall2 n  sigfigs         if type n  is int  return intround n  sigfigs        else  return round n  sigfigs

User · Answer

If you want to round without involving strings  the link I found buried in the comments above   http   code activestate com lists python-tutor 70739   strikes me as best   Then when you print with any string formatting descriptors  you get a reasonable output  and you can use the numeric representation for other calculation purposes   The code at the link is a three liner  def  doc  and return   It has a bug  you need to check for exploding logarithms   That is easy   Compare the input to sys float info min   The complete solution is   import sys math  def tidy x  n      Return  x  rounded to  n  significant digits     y abs x  if y  lt   sys float info min  return 0 0 return round  x  int  n-math ceil math log10 y         It works for any scalar numeric value  and n can be a float if you need to shift the response for some reason   You can actually push the limit to   sys float info min sys float info epsilon   without provoking an error  if for some reason you are working with miniscule values

User · Answer

g in string formatting will format a float rounded to some number of significant figures   It will sometimes use  e  scientific notation  so convert the rounded string back to a float then through  s string formatting    gt  gt  gt    s    float    1g    1234   1000   gt  gt  gt    s    float    1g    0 12   0 1   gt  gt  gt    s    float    1g    0 012   0 01   gt  gt  gt    s    float    1g    0 062   0 06   gt  gt  gt    s    float    1g    6253   6000 0   gt  gt  gt    s    float    1g    1999   2000 0

User · Answer

Using python 2 6  new-style formatting  as  -style is deprecated     gt  gt  gt    0   format float   0  1g   format 1216     1000 0   gt  gt  gt    0   format float   0  1g   format 0 00356     0 004    In python 2 7  you can omit the leading 0s

User · Answer

If you want to have other than 1 significant decimal  otherwise the same as Evgeny     gt  gt  gt  from math import log10  floor  gt  gt  gt  def round sig x  sig 2         return round x  sig-int floor log10 abs x    -1        gt  gt  gt  round sig 0 0232  0 023  gt  gt  gt  round sig 0 0232  1  0 02  gt  gt  gt  round sig 1234243  3  1230000 0

User · Answer

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

User · Answer

I have created the package to-precision that does what you want  It allows you to give your numbers more or less significant figures   It also outputs standard  scientific  and engineering notation with a specified number of significant figures   In the accepted answer there is the line   gt  gt  gt  round to 1 1234243  1000000 0   That actually specifies 8 sig figs  For the number 1234243 my library only displays one significant figure    gt  gt  gt  from to precision import to precision  gt  gt  gt  to precision 1234243  1   std    1000000   gt  gt  gt  to precision 1234243  1   sci    1e6   gt  gt  gt  to precision 1234243  1   eng    1e6    It will also round the last significant figure and can automatically choose what notation to use if a notation isn t specified    gt  gt  gt  to precision 599  2   600   gt  gt  gt  to precision 1164  2   1 2e3

User · Answer

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   gt  gt  gt  round to n 750  2  750 0  gt  gt  gt  round to n 750  1  800 0  gt  gt  gt  round to n  0750  2  0 075  gt  gt  gt  round to n  0750  1  0 08  gt  gt  gt  math pi 3 141592653589793  gt  gt  gt  round to n math pi  7  3 141593

User · Answer

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  lt  1          return 0     max10exp   floor log10 i       if max10exp 1  lt  places          return i     sig10pow   10   max10exp-places 1      floated   i 1 0 sig10pow     defloated   round floated  sig10pow     return int defloated isign

User · Answer

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    quot   e  quot  format number  split  e   1      return round number  - int power  - digits

User · Answer

f  float f  i  1g    g     Or with Python  lt 3 6     g   format float     1g   format i      This solution is different from all of the others because    it exactly solves the OP question it does not need any extra package it does not need any user-defined auxiliary function or mathematical operation   For an arbitrary number n of significant figures  you can use   print    g   format float      p g   format i  p n       Test   a    1234  0 12  0 012  0 062  6253  1999  -3 14  0   -48 01  0 75  b       g   format float     1g   format i    for i in a    b      1000    0 1    0 01    0 06    6000    2000    -3    0    -50    0 8     Note  with this solution  it is not possible to adapt the number of significant figures dynamically from the input because there is no standard way to distinguish numbers with different numbers of trailing zeros  3 14    3 1400   If you need to do so  then non-standard functions like the ones provided in the to-precision package are needed

User · Answer

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

User · Answer

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  gt  0 else decimal positions    Hope it helps

User · Answer

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 lt  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

User · Answer

import math    def sig dig x  n sig dig         num of digits   len str x  replace                 if n sig dig  gt   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          gt  gt  gt  sig dig 1234243  3       gt  gt  gt  sig dig 243 3576  5           1230 0         243 36

User · Answer

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    gt  12340

User · Answer

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

User · Answer

def round to n x  n       if not x  return 0     power   -int math floor math log10 abs x        n - 1      factor    10    power      return round x   factor    factor  round to n 0 075  1         0 08 round to n 0  1             0 round to n -1e15 - 1  16    1000000000000001 0   Hopefully taking the best of all the answers above  minus being able to put it as a one line lambda         Haven t explored yet  feel free to edit this answer   round to n 1e15   1  11     999999999999999 9

User · Answer

You can use negative numbers to round integers    gt  gt  gt  round 1234  -3  1000 0   Thus if you need only most significant digit    gt  gt  gt  from math import log10  floor  gt  gt  gt  def round to 1 x         return round x  -int floor log10 abs x            gt  gt  gt  round to 1 0 0232  0 02  gt  gt  gt  round to 1 1234243  1000000 0  gt  gt  gt  round to 1 13  10 0  gt  gt  gt  round to 1 4  4 0  gt  gt  gt  round to 1 19  20 0   You ll probably have to take care of turning float to integer if it s bigger than 1

User · Answer

The sigfig package library covers this   After installing you can do the following    gt  gt  gt  from sigfig import round  gt  gt  gt  round 1234  1  1000  gt  gt  gt  round 0 12  1  0 1  gt  gt  gt  round 0 012  1  0 01  gt  gt  gt  round 0 062  1  0 06  gt  gt  gt  round 6253  1  6000  gt  gt  gt  round 1999  1  2000

[python] How to round a number to significant figures in Python

Examples related to python

Examples related to math

Examples related to rounding