I am facing a strange behavior of the round() function:
for i in range(1, 15, 2):
n = i / 2
print(n, "=>", round(n))
This code prints:
0.5 => 0
1.5 => 2
2.5 => 2
3.5 => 4
4.5 => 4
5.5 => 6
6.5 => 6
I expected the floating values to be always rounded up, but instead, it is rounded to the nearest even number.
Why such behavior, and what is the best way to get the correct result?
I tried to use the fractions but the result is the same.
The Numeric Types section documents this behaviour explicitly:
round(x[, n])
x rounded to n digits, rounding half to even. If n is omitted, it defaults to 0.
Note the rounding half to even. This is also called bankers rounding; instead of always rounding up or down (compounding rounding errors), by rounding to the nearest even number you average out rounding errors.
If you need more control over the rounding behaviour, use the decimal module, which lets you specify exactly what rounding strategy should be used.
For example, to round up from half:
>>> from decimal import localcontext, Decimal, ROUND_HALF_UP
>>> with localcontext() as ctx:
... ctx.rounding = ROUND_HALF_UP
... for i in range(1, 15, 2):
... n = Decimal(i) / 2
... print(n, '=>', n.to_integral_value())
...
0.5 => 1
1.5 => 2
2.5 => 3
3.5 => 4
4.5 => 5
5.5 => 6
6.5 => 7
For example:
from decimal import Decimal, ROUND_HALF_UP
Decimal(1.5).quantize(0, ROUND_HALF_UP)
# This also works for rounding to the integer part:
Decimal(1.5).to_integral_value(rounding=ROUND_HALF_UP)
You can use this:
import math
def normal_round(n):
if n - math.floor(n) < 0.5:
return math.floor(n)
return math.ceil(n)
It will round number up or down properly.
round() will round either up or down, depending on if the number is even or odd. A simple way to only round up is:
int(num + 0.5)
If you want this to work properly for negative numbers use:
((num > 0) - (num < 0)) * int(abs(num) + 0.5)
Note, this can mess up for large numbers or really precise numbers like 5000000000000001.0 and 0.49999999999999994.
Love the fedor2612 answer. I expanded it with an optional "decimals" argument for those who want to use this function to round any number of decimals (say for example if you want to round a currency $26.455 to $26.46).
import math
def normal_round(n, decimals=0):
expoN = n * 10 ** decimals
if abs(expoN) - abs(math.floor(expoN)) < 0.5:
return math.floor(expoN) / 10 ** decimals
return math.ceil(expoN) / 10 ** decimals
oldRounding = round(26.455,2)
newRounding = normal_round(26.455,2)
print(oldRounding)
print(newRounding)
Output:
26.45
26.46
The behavior you are seeing is typical IEEE 754 rounding behavior. If it has to choose between two numbers that are equally different from the input, it always picks the even one. The advantage of this behavior is that the average rounding effect is zero - equally many numbers round up and down. If you round the half way numbers in a consistent direction the rounding will affect the expected value.
The behavior you are seeing is correct if the objective is fair rounding, but that is not always what is needed.
One trick to get the type of rounding you want is to add 0.5 and then take the floor. For example, adding 0.5 to 2.5 gives 3, with floor 3.
Why make it so complicated? (Only works for positive numbers)
def HalfRoundUp(value):
return int(value + 0.5)
You could of course make it into a lambda which would be:
HalfRoundUp = lambda value: int(value + 0.5)
Unfortunately, this simple answer doesn't work with negative numbers, but it can be fixed with the floor function from math: (This works for both positive and negative numbers too)
from math import floor
def HalfRoundUp(value):
floor(value + 0.5)
Short version: use the decimal module. It can represent numbers like 2.675 precisely, unlike Python floats where 2.675 is really 2.67499999999999982236431605997495353221893310546875 (exactly). And you can specify the rounding you desire: ROUND_CEILING, ROUND_DOWN, ROUND_FLOOR, ROUND_HALF_DOWN, ROUND_HALF_EVEN, ROUND_HALF_UP, ROUND_UP, and ROUND_05UP are all options.
In the question this is basically an issue when dividing a positive integer by 2. The easisest way is int(n + 0.5) for individual numbers.
However we cannot apply this to series, therefore what we then can do for example for a pandas dataframe, and without going into loops, is:
import numpy as np
df['rounded_division'] = np.where(df['some_integer'] % 2 == 0, round(df['some_integer']/2,0), round((df['some_integer']+1)/2,0))
A small addition as the rounding half up with some of the solutions might not work as expected in some cases.
Using the function from above for instance:
from decimal import Decimal, ROUND_HALF_UP
def round_half_up(x: float, num_decimals: int) -> float:
if num_decimals < 0:
raise ValueError("Num decimals needs to be at least 0.")
target_precision = "1." + "0" * num_decimals
rounded_x = float(Decimal(x).quantize(Decimal(target_precision), ROUND_HALF_UP))
return rounded_x
round_half_up(1.35, 1)
1.4
round_half_up(4.35, 1)
4.3
Where I was expecting 4.4. What did the trick for me was converting x into a string first.
from decimal import Decimal, ROUND_HALF_UP
def round_half_up(x: float, num_decimals: int) -> float:
if num_decimals < 0:
raise ValueError("Num decimals needs to be at least 0.")
target_precision = "1." + "0" * num_decimals
rounded_x = float(Decimal(str(x)).quantize(Decimal(target_precision), ROUND_HALF_UP))
return rounded_x
round_half_up(4.35, 1)
4.4
Rounding to the nearest even number has become common practice in numerical disciplines. "Rounding up" produces a slight bias towards larger results.
So, from the perspective of the scientific establishment, round has the correct behavior.
Here is another solution.
It will work as normal rounding in excel.
from decimal import Decimal, getcontext, ROUND_HALF_UP
round_context = getcontext()
round_context.rounding = ROUND_HALF_UP
def c_round(x, digits, precision=5):
tmp = round(Decimal(x), precision)
return float(tmp.__round__(digits))
c_round(0.15, 1) -> 0.2, c_round(0.5, 0) -> 1
The following solution achieved "school fashion rounding" without using the decimal module (which turns out to be slow).
def school_round(a_in,n_in):
''' python uses "banking round; while this round 0.05 up" '''
if (a_in * 10 ** (n_in + 1)) % 10 == 5:
return round(a_in + 1 / 10 ** (n_in + 1), n_in)
else:
return round(a_in, n_in)
e.g.
print(round(0.005,2)) # 0
print(school_round(0.005,2)) #0.01
So just to make sure there is a crystal clear working example here, I wrote a small convenience function
def round_half_up(x: float, num_decimals: int) -> float:
"""Use explicit ROUND HALF UP. See references, for an explanation.
This is the proper way to round, as taught in school.
Args:
x:
num_decimals:
Returns:
https://stackoverflow.com/questions/33019698/how-to-properly-round-up-half-float-numbers-in-python
"""
if num_decimals < 0:
raise ValueError("Num decimals needs to be at least 0.")
target_precision = "1." + "0" * num_decimals
rounded_x = float(Decimal(x).quantize(Decimal(target_precision), ROUND_HALF_UP))
return rounded_x
And an appropriate set of test cases
def test_round_half_up():
x = 1.5
y = round_half_up(x, 0)
assert y == 2.0
y = round_half_up(x, 1)
assert y == 1.5
x = 1.25
y = round_half_up(x, 1)
assert y == 1.3
y = round_half_up(x, 2)
assert y == 1.25
This is a function that takes the number of decimal places as an argument.
It also rounds up half decimal.
import math
def normal_round(n, decimal_places):
if int((str(n)[-1])) < 5:
return round(n, decimal_places)
return round(n + 10**(-1 * (decimal_places+1)), decimal_places)
Test cases:
>>> normal_round(5.12465, 4)
5.1247
>>> normal_round(5.12464, 4)
5.1246
>>> normal_round(5.12467, 4)
5.1247
>>> normal_round(5.12463, 4)
5.1246
>>> normal_round(5.1241, 4)
5.1241
>>> normal_round(5.1248, 4)
5.1248
>>> normal_round(5.1248, 3)
5.125
>>> normal_round(5.1242, 3)
5.124
You can use:
from decimal import Decimal, ROUND_HALF_UP
for i in range(1, 15, 2):
n = i / 2
print(n, "=>", Decimal(str(n)).quantize(Decimal("1"), rounding=ROUND_HALF_UP))
A classical mathematical rounding without any libraries
def rd(x,y=0):
''' A classical mathematical rounding by Voznica '''
m = int('1'+'0'*y) # multiplier - how many positions to the right
q = x*m # shift to the right by multiplier
c = int(q) # new number
i = int( (q-c)*10 ) # indicator number on the right
if i >= 5:
c += 1
return c/m
Compare:
print( round(0.49), round(0.51), round(0.5), round(1.5), round(2.5), round(0.15,1)) # 0 1 0 2 2 0.1
print( rd(0.49), rd(0.51), rd(0.5), rd(1.5), rd(2.5), rd(0.15,1)) # 0 1 1 2 3 0.2
Knowing that round(9.99,0) rounds to int=10 and int(9.99) rounds to int=9 brings success:
Goal: Provide lower and higher round number depending on value
def get_half_round_numers(self, value):
"""
Returns dict with upper_half_rn and lower_half_rn
:param value:
:return:
"""
hrns = {}
if not isinstance(value, float):
print("Error>Input is not a float. None return.")
return None
value = round(value,2)
whole = int(value) # Rounds 9.99 to 9
remainder = (value - whole) * 100
if remainder >= 51:
hrns['upper_half_rn'] = round(round(value,0),2) # Rounds 9.99 to 10
hrns['lower_half_rn'] = round(round(value,0) - 0.5,2)
else:
hrns['lower_half_rn'] = round(int(value),2)
hrns['upper_half_rn'] = round(int(value) + 0.5,2)
return hrns
Some testing:
yw
import math
# round tossing n digits from the end
def my_round(n, toss=1):
def normal_round(n):
if isinstance(n, int):
return n
intn, dec = str(n).split(".")
if int(dec[-1]) >= 5:
if len(dec) == 1:
return math.ceil(n)
else:
return float(intn + "." + str(int(dec[:-1]) + 1))
else:
return float(intn + "." + dec[:-1])
while toss >= 1:
n = normal_round(n)
toss -= 1
return n
for n in [1.25, 7.3576, 30.56]:
print(my_round(n, 2))
1.0
7.36
31
import math
def round_half_up(x: float) -> int:
if x < 0:
return math.trunc(x) if -x % 1 < 0.5 else math.floor(x)
else:
return math.trunc(x) if x % 1 < 0.5 else math.ceil(x)
This even works for corner cases like 0.49999999999999994 and 5000000000000001.0.
You can try this
def round(num):
return round(num + 10**(-9))
it will work since num = x.5 will always will be x.5 + 0.00...01 in the process which its closer to x+1 hence the round function will work properly and it will round x.5 to x+1
Let me start with an example -
I have a range of numbers from 1 to 9. And let's say the target number that I want is 29.
In this case the minimum number of operations that are required would be (9*3)+2 = 2 operations. Similarly for 18 the minimum number of operations is 1 (9*2=18).
I can use any of the 4 arithmetic operators - +, -, / and *.
How can I programmatically find out the minimum number of operations required?
Thanks in advance for any help provided.
clarification: integers only, no decimals allowed mid-calculation. i.e. the following is not valid (from comments below): ((9/2) + 1) * 4 == 22
I must admit I didn't think about this thoroughly, but for my purpose it doesn't matter if decimal numbers appear mid-calculation. ((9/2) + 1) * 4 == 22 is valid. Sorry for the confusion.
For the special case where set Y = [1..9] and n > 0:
n <= 9 : 0 operations
n <=18 : 1 operation (+)
otherwise : Remove any divisor found in Y. If this is not enough, do a recursion on the remainder for all offsets -9 .. +9. Offset 0 can be skipped as it has already been tried.
Notice how division is not needed in this case. For other Y this does not hold.
This algorithm is exponential in log(n). The exact analysis is a job for somebody with more knowledge about algebra than I.
For more speed, add pruning to eliminate some of the search for larger numbers.
Sample code:
def findop(n, maxlen=9999):
# Return a short postfix list of numbers and operations
# Simple solution to small numbers
if n<=9: return [n]
if n<=18: return [9,n-9,'+']
# Find direct multiply
x = divlist(n)
if len(x) > 1:
mults = len(x)-1
x[-1:] = findop(x[-1], maxlen-2*mults)
x.extend(['*'] * mults)
return x
shortest = 0
for o in range(1,10) + range(-1,-10,-1):
x = divlist(n-o)
if len(x) == 1: continue
mults = len(x)-1
# We spent len(divlist) + mults + 2 fields for offset.
# The last number is expanded by the recursion, so it doesn't count.
recursion_maxlen = maxlen - len(x) - mults - 2 + 1
if recursion_maxlen < 1: continue
x[-1:] = findop(x[-1], recursion_maxlen)
x.extend(['*'] * mults)
if o > 0:
x.extend([o, '+'])
else:
x.extend([-o, '-'])
if shortest == 0 or len(x) < shortest:
shortest = len(x)
maxlen = shortest - 1
solution = x[:]
if shortest == 0:
# Fake solution, it will be discarded
return '#' * (maxlen+1)
return solution
def divlist(n):
l = []
for d in range(9,1,-1):
while n%d == 0:
l.append(d)
n = n/d
if n>1: l.append(n)
return l
The basic idea is to test all possibilities with k operations, for k starting from 0. Imagine you create a tree of height k that branches for every possible new operation with operand (4*9 branches per level). You need to traverse and evaluate the leaves of the tree for each k before moving to the next k.
I didn't test this pseudo-code:
for every k from 0 to infinity
for every n from 1 to 9
if compute(n,0,k):
return k
boolean compute(n,j,k):
if (j == k):
return (n == target)
else:
for each operator in {+,-,*,/}:
for every i from 1 to 9:
if compute((n operator i),j+1,k):
return true
return false
It doesn't take into account arithmetic operators precedence and braces, that would require some rework.
Really cool question :)
Notice that you can start from the end! From your example (9*3)+2 = 29 is equivalent to saying (29-2)/3=9. That way we can avoid the double loop in cyborg's answer. This suggests the following algorithm for set Y and result r:
nextleaves = {r}
nops = 0
while(true):
nops = nops+1
leaves = nextleaves
nextleaves = {}
for leaf in leaves:
for y in Y:
if (leaf+y) or (leaf-y) or (leaf*y) or (leaf/y) is in X:
return(nops)
else:
add (leaf+y) and (leaf-y) and (leaf*y) and (leaf/y) to nextleaves
This is the basic idea, performance can be certainly be improved, for instance by avoiding "backtracks", such as r+a-a or r*a*b/a.
I guess my idea is similar to the one of Peer Sommerlund:
For big numbers, you advance fast, by multiplication with big ciphers.
Is Y=29 prime? If not, divide it by the maximum divider of (2 to 9).
Else you could subtract a number, to reach a dividable number. 27 is fine, since it is dividable by 9, so
(29-2)/9=3 =>
3*9+2 = 29
So maybe - I didn't think about this to the end: Search the next divisible by 9 number below Y. If you don't reach a number which is a digit, repeat.
The formula is the steps reversed.
(I'll try it for some numbers. :) )
I tried with 2551, which is
echo $((((3*9+4)*9+4)*9+4))
But I didn't test every intermediate result whether it is prime.
But
echo $((8*8*8*5-9))
is 2 operations less. Maybe I can investigate this later.
This is part of a search function on a website. So im trying to find a way to get to the end result as fast as possible.
Have a binary number where digit order matters.
Input Number = 01001
Have a database of other binary numbers all the same length.
01000, 10110, 00000, 11111
I dont know how to write what im doing, so im going to do it more visually below.
// Zeros mean nothing & the location of a 1 matters, not the total number of 1's.
input num > 0 1 0 0 1 = 2 possible matches
number[1] > 0 1 0 0 0 = 1 match = 50% match
number[2] > 1 0 1 1 0 = 0 match = 0% match
number[3] > 0 0 0 0 0 = 0 match = 0% match
number[4] > 1 1 1 1 1 = 2 match = 100% match
Now obviously, you could go digit by digit, number by number and compare it that way (using a loop and what not). But I was hoping there might be an algorithm or something that will help. Mostly because in the above example I only used 5 digit numbers. But im going to be routinely comparing around 100,000 numbers with 200 digits each, that's a lot of calculating.
I usually deal with php and MySQL. But if something spectacular comes up I could always learn.
If it's possible to somehow chop up your bitstrings in integer-size chunks some elementary boolean arithmetic would do, and that kind of instructions is generally pretty fast
$matchmask = ~ ($inputval ^ $tomatch) & $inputval
What this does:
the xor determines the bits that are different in the inputval and tomatch
negation gives a value where all bits that are equal in inputval and tomatch are set
and that with inputval and only the bits that are 1 in both inputval and tomatch remain set.
Then count the number of bits set in the result, look at How to count the number of set bits in a 32-bit integer? for an optimal solution, easily translated into php
Instead of checking each bit, you could pre-process the input and determine which bits need checking. In the worst case, this devolves into processing each bit, but for a normal distribution, you'll save some processing.
That is, for input
01001, iterate over the database and determine if number1[0] & input is non-zero, and (number1[3] >> 8) & input is non-zero, assuming 0 as the index of the LSB. How you get fast bit-shifting and anding with the large numbers is on you, however. If you detect 1s than 0s in the input, you could always invert the input and test for zero to detect coverage.
This will give you modest improvement, but it's at best a constant-time reduction of the problem. If most of your inputs are balanced between 0s and 1s, you'll halve the number of required operations. If it's more biased, you'll get better results.
Well, the first thing I can think of is a simple bitwise AND between the two numbers; you can then analyze the result to get the match percentage:
if( result >= input )
//100% match
else {
result ^= input;
/* The number of 1's in result is the number of 1 of "input"
* that are missing in "result".
*/
}
Of course, you'll need to implement your own AND and XOR function (this will work only for 32 bit integers). Note that it works only with unsigned numbers.
Suppose the input number is called A (so in your example A = 01001) and the other number is x. You'll have 100% match when x & A == A. Otherwise, for partial matches, the number of 1 bits will be (taken from hacker's delight):
x = (x & 0x55555555) + ((x >> 1) & 0x55555555);
x = (x & 0x33333333) + ((x >> 2) & 0x33333333);
x = (x & 0x0F0F0F0F) + ((x >> 4) & 0x0F0F0F0F);
x = (x & 0x00FF00FF) + ((x >> 8) & 0x00FF00FF);
x = (x & 0x0000FFFF) + ((x >>16) & 0x0000FFFF);
Note this will work for 32 bits integers.
Let's assume you have a function bit1count, then from what you describe, the "likeness" formula should be:
100.0 / min(bit1count(n1), bit1count(n2)) * bit1count(n1 & n2)
With n1 and n2 being the two numbers and & being the logical and operator.
bit1count can be easily implemented using a loop, or, more elegant, using the algorithm provided in BigBears answer.
There is actually a BIT_COUNT in mysql, so something like this should work:
SELECT 100.0 / IF(BIT_COUNT(n1) < BIT_COUNT(n2), BIT_COUNT(n1), BIT_COUNT(n2)) * BIT_COUNT(n1 & n2) FROM table
Do the below without any conditional or comparison operator.
if (Number <= 0)
{
Print '0';
}
else
{
print Number;
}
thanks..
My original simple solution:
1. print( (abs(Number)+Number) / 2 )
That solution would work in most cases, unless Number is very large (more than half the maximum e.g. Number >= MAX_INT/2) in which case the addition may cause overflow.
The following solution solves the overflow problem:
2. print( (abs(Number)/2) + (Number/2) )
However, there may be a case in which Number is and must remain integer, and the division operator (/) is integer division, so that 7/2=3. In this case solution 2 won't work because if Number=7 it will print 6 (for this case solution 1 will work just fine).
So if we need to deal with both large numbers AND integer arithmetic, the following monstrosity comes to the rescue, adding compensation for the 1 that may be lost in the division by 2 in case of odd integer:
3. print(
( (abs(Number)/2)+(Number/2) ) +
((
(Number-(2*(Number/2))) +
(abs(Number)-(2*(abs(Number)/2)))
) / 2)
)
print max(0, number)
Let's say that number is represented by an 8-bit two's complement integer.
Positive numbers including 0 all have the MSB set to 0.
Negative numbers all have the MSB set to 1.
So we take the complement of the MSB, extend it to the full 8 bits, and bitwise AND it with the original number, e.g.
Positive:
00110101 -> MSB is 0
11111111 -> complement of MSB extended
00110101 -> bitwise AND of above
Negative:
10110101 -> MSB is 1
00000000 -> complement of MSB extended
00000000 -> bitwise AND of above
No comparisons needed - I'm kind of assuming that bitwise AND isn't strictly a comparison.
Also, sorry for the lack of code, but you get the idea.
I haven't seen a solution yet that is valid for the complete domain.
An other solution is to call a function that raises an exception if the input value is 0 or below 0. Then catch the exception and print 0.
Or you can use a function Sign, that returns -1 if the input is <0, 0 if it's 0 and 1 otherwise.
print ((sign(x)+1) * sign(x) / 2) * x.
sign can be -1, 0 or 1, so ((sign(x)+1) * sign(x) / 2) can have the following values:
-1 -> ((-1+1)*-1)/2 = 0
0 -> ((0+1)*0)/2 = 0
1 -> ((1+1) * 1)/2 = 1
Another method is to create a lookup table that maps all non negative numbers to themself and the rest to 0.
But, in my opinion, the original function is much clearer. So why violate the KISS principle.
Similar to the accepted answer. Although acceptable where absolute value is implemented using comparisons, but also more prone to overflow:
print( (sqrt(Number * Number) + x) / 2);
Assuming C or C++:
switch ((unsigned long)Number & ~(unsigned long)LONG_MAX) {
case 0:
printf("%d\n", Number);
break;
default:
printf("0\n", Number);
break;
}
If you consider switch to be a conditional operator, then try this:
unsigned long flag = (unsigned long)Number & ~(unsigned long)LONG_MAX;
flag /= (unsigned long)LONG_MAX + 1;
flag = 1 - flag;
printf("%d\n", Number * flag);