Round number to nearest integer

Question

I ve been trying to round long float numbers like   32 268907563  32 268907563  31 2396694215  33 6206896552        With no success so far  I tried math ceil x   math floor x   although that would round up or down  which is not what I m looking for  and round x  which didn t work either  still float numbers    What could I do   EDIT  CODE   for i in widthRange      for j in heightRange          r  g  b   rgb im getpixel  i  j           h  s  v   colorsys rgb to hsv r 255 0  g 255 0  b 255 0          h   h   360         int round h           print h

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You can also use numpy assuming if you are using python3 x here is an example  import numpy as np x   2 3 print np rint x    gt  gt  gt  2 0

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For this purpose I would suggest just do the following thing -   int round x     This will give you nearest integer   Hope this helps

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Your solution is calling round without specifying the second argument  number of decimal places    gt  gt  gt  round 0 44  0  gt  gt  gt  round 0 64  1   which is a much better result than   gt  gt  gt  int round 0 44  2   0  gt  gt  gt  int round 0 64  2   0   From the Python documentation at https   docs python org 3 library functions html round     round number   ndigits        Return number rounded to ndigits precision after the decimal point  If   ndigits is omitted or is None  it returns the nearest integer to its   input       Note      The behavior of round   for floats can be surprising  for example    round 2 675  2  gives 2 67 instead of the expected 2 68  This is not a   bug  it   s a result of the fact that most decimal fractions can   t be   represented exactly as a float  See Floating Point Arithmetic  Issues   and Limitations for more information

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I use and may advise the following solution  python3 6    y   int x    x    1 if x  gt   0 else -1      It works fine for half-numbers  positives and negatives  and works even faster than int round x     round methods    lambda x  int round x                      lambda x  int x    x    1 if x  gt   0 else -1                      lambda x  np rint x  astype int                    lambda x  int proper round x     for rm in round methods       timeit rm 112 5  Out  201 ns    3 96 ns per loop  mean    std  dev  of 7 runs  1000000 loops each  159 ns    0 646 ns per loop  mean    std  dev  of 7 runs  10000000 loops each  925 ns    7 66 ns per loop  mean    std  dev  of 7 runs  1000000 loops each  1 18   s    8 66 ns per loop  mean    std  dev  of 7 runs  1000000 loops each   for rm in round methods      print rm 112 4   rm 112 5   rm 112 6       print rm -12 4   rm -12 5   rm -12 6       print       11   Out  112 112 113 -12 -12 -13             112 113 113 -12 -13 -13             112 112 113 -12 -12 -13             112 113 113 -12 -13 -13

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This one is tricky  to be honest  There are many simple ways to do this nevertheless  Using math ceil    round    and math floor    you can get a integer by using for example  n   int round n    If before we used this function n   5 23  we would get returned 5  If you wanted to round to different place values  you could use this function  def Round n k     point     0     str k     f    if k    0      return int point   n    else      return float point   n   If we used n  5 23  again  round it to the nearest tenth  and print the answer to the console  our code would be  Round 5 23 1   Which would return 5 2  Finally  if you wanted to round something to the nearest  let s say  1 2  you can use the code  def Round n k       return k   round n k   If we wanted n to be rounded to 1 2  our code would be  print Round n 1 2    and our result  4 8  Thank you  If you have any questions  please add a comment      Happy Holidays

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int round x     Will round it and change it to integer  EDIT   You are not assigning int round h   to any variable  When you call int round h    it returns the integer number but does nothing else  you have to change that line for   h   int round h     To assign the new value to h  EDIT 2   As  plowman said in the comments  Python s round   doesn t work as one would normally expect  and that s because the way the number is stored as a variable is usually not the way you see it on screen  There are lots of answers that explain this behavior   round   doesn  39 t seem to be rounding properly  One way to avoid this problem is to use the Decimal as stated by this answer  https   stackoverflow com a 15398691 4345659  In order for this answer to work properly without using extra libraries it would be convenient to use a custom rounding function  After a lot of corrections  I came up with the following solution  that as far as I tested avoided all the storing issues  It is based on using the string representation  obtained with repr    NOT str      It looks hacky but it was the only way I found to solve all the cases  It works with both Python2 and Python3   def proper round num  dec 0       num   str num   str num  index      dec 2      if num -1  gt   5           return float num  -2- not dec   str int num -2- not dec    1       return float num  -1     Tests    gt  gt  gt  print proper round 1 0005 3   1 001  gt  gt  gt  print proper round 2 0005 3   2 001  gt  gt  gt  print proper round 3 0005 3   3 001  gt  gt  gt  print proper round 4 0005 3   4 001  gt  gt  gt  print proper round 5 0005 3   5 001  gt  gt  gt  print proper round 1 005 2   1 01  gt  gt  gt  print proper round 2 005 2   2 01  gt  gt  gt  print proper round 3 005 2   3 01  gt  gt  gt  print proper round 4 005 2   4 01  gt  gt  gt  print proper round 5 005 2   5 01  gt  gt  gt  print proper round 1 05 1   1 1  gt  gt  gt  print proper round 2 05 1   2 1  gt  gt  gt  print proper round 3 05 1   3 1  gt  gt  gt  print proper round 4 05 1   4 1  gt  gt  gt  print proper round 5 05 1   5 1  gt  gt  gt  print proper round 1 5   2 0  gt  gt  gt  print proper round 2 5   3 0  gt  gt  gt  print proper round 3 5   4 0  gt  gt  gt  print proper round 4 5   5 0  gt  gt  gt  print proper round 5 5   6 0  gt  gt  gt    gt  gt  gt  print proper round 1 000499999999 3   1 0  gt  gt  gt  print proper round 2 000499999999 3   2 0  gt  gt  gt  print proper round 3 000499999999 3   3 0  gt  gt  gt  print proper round 4 000499999999 3   4 0  gt  gt  gt  print proper round 5 000499999999 3   5 0  gt  gt  gt  print proper round 1 00499999999 2   1 0  gt  gt  gt  print proper round 2 00499999999 2   2 0  gt  gt  gt  print proper round 3 00499999999 2   3 0  gt  gt  gt  print proper round 4 00499999999 2   4 0  gt  gt  gt  print proper round 5 00499999999 2   5 0  gt  gt  gt  print proper round 1 0499999999 1   1 0  gt  gt  gt  print proper round 2 0499999999 1   2 0  gt  gt  gt  print proper round 3 0499999999 1   3 0  gt  gt  gt  print proper round 4 0499999999 1   4 0  gt  gt  gt  print proper round 5 0499999999 1   5 0  gt  gt  gt  print proper round 1 499999999   1 0  gt  gt  gt  print proper round 2 499999999   2 0  gt  gt  gt  print proper round 3 499999999   3 0  gt  gt  gt  print proper round 4 499999999   4 0  gt  gt  gt  print proper round 5 499999999   5 0   Finally  the corrected answer would be     Having proper round defined as previously stated h   int proper round h     EDIT 3   Tests    gt  gt  gt  proper round 6 39764125  2  6 31   should be 6 4  gt  gt  gt  proper round 6 9764125  1  6 1    should be 7   The gotcha here is that the dec-th decimal can be 9 and if the dec 1-th digit   5 the 9 will become a 0 and a 1 should be carried to the dec-1-th digit    If we take this into consideration  we get   def proper round num  dec 0       num   str num   str num  index      dec 2      if num -1  gt   5         a   num  -2- not dec           integer part       b   int num -2- not dec    1   decimal part       return float a  b   -dec 1  if a and b    10 else float a str b       return float num  -1     In the situation described above b   10 and the previous version would just concatenate a and b which would result in a concatenation of 10 where the trailing 0 would disappear  This version transforms b to the right decimal place based on dec  as a proper carry

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For positives  try  int x   0 5    To make it work for negatives too  try  int x    0 5 if x  gt  0 else -0 5     int   works like a floor function and hence you can exploit this property  This is definitely the fastest way

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If you need  for example  a two digit approximation for A  then  int A 100 0 5  100 0 will do what you are looking for   If you need three digit approximation multiply and divide by 1000 and so on

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Isn t just Python doing round half to even  as prescribed by IEEE 754   Be careful redefining  or using  non-standard  rounding      See also https   stackoverflow com a 33019948 109839

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Use round x  y   It will round up your number up to your desired decimal place   For example    gt  gt  gt  round 32 268907563  3  32 269

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Some thing like this should also work  import numpy as np      def proper round a               given any real number  a  returns an integer closest to  a              a ceil   np ceil a      a floor   np floor a      if np abs a ceil - a   lt  np abs a floor - a           return int a ceil      else          return int a floor

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round value significantDigit  is the ordinary solution  however this does not operate as one would expect from a math perspective when round values ending in 5   If the 5 is in the digit just after the one you re rounded to  these values are only sometimes rounded up as expected  i e  8 005 rounding to two decimal digits gives 8 01    For certain values due to the quirks of floating point math  they are rounded down instead   i e    gt  gt  gt  round 1 0005 3  1 0  gt  gt  gt  round 2 0005 3  2 001  gt  gt  gt  round 3 0005 3  3 001  gt  gt  gt  round 4 0005 3  4 0  gt  gt  gt  round 1 005 2  1 0  gt  gt  gt  round 5 005 2  5 0  gt  gt  gt  round 6 005 2  6 0  gt  gt  gt  round 7 005 2  7 0  gt  gt  gt  round 3 005 2  3 0  gt  gt  gt  round 8 005 2  8 01   Weird   Assuming your intent is to do the traditional rounding for statistics in the sciences  this is a handy wrapper to get the round function working as expected needing to import extra stuff like Decimal    gt  gt  gt  round 0 075 2   0 07   gt  gt  gt  round 0 075 10   -2 5  2   0 08   Aha  So based on this we can make a function     def roundTraditional val digits      return round val 10   -len str val  -1   digits    Basically this adds a value guaranteed to be smaller than the least given digit of the string you re trying to use round on  By adding that small quantity it preserve s round s behavior in most cases  while now ensuring if the digit inferior to the one being rounded to is 5 it rounds up  and if it is 4 it rounds down   The approach of using 10   -len val -1  was deliberate  as it the largest small number you can add to force the shift  while also ensuring that the value you add never changes the rounding even if the decimal   is missing   I could use just 10   -len val   with a condiditional if  val gt 1  to subtract 1 more    but it s simpler to just always subtract the 1 as that won t change much the applicable range of decimal numbers this workaround can properly handle   This approach will fail if your values reaches the limits of the type  this will fail  but for nearly the entire range of valid decimal values it should work   You can also use the decimal library to accomplish this  but the wrapper I propose is simpler and may be preferred in some cases     Edit  Thanks Blckknght for pointing out that the 5 fringe case occurs only for certain values   Also an earlier version of this answer wasn t explicit enough that the odd rounding behavior occurs only when the digit immediately inferior to the digit you re rounding to has a 5

[python] Round number to nearest integer

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