This question already has answers here:
How can I disable scientific notation?
(4 answers)
Why are these numbers not equal?
(6 answers)
Closed 3 years ago.
I have just started with R and am stuck on this bug.
fuel_efficiency<-c(28.2, 28.3, 28.4, 28.5, 29.0)
mean=28.48
deviation<-(fuel_efficiency-mean)
deviation
sum(deviation)
I have written this code to subtract the mean from the elements of the fuel efficiency vector to get the deviation vector. Then am trying to get the sum of the updated deviation vector.
The sum answer should return 0 but instead gives something like -3.552714e-15
The deviation vector is printed properly as expected.
#[1] -0.28 -0.18 -0.08 0.02 0.52
This are just rounding errors, -3.552714e-15 is a very, very small number:
Computers are notoriously bad at handling decimal numbers, (one could even say they are just not able to do it exactly). To overcome this R provides a function to check for equality:
all.equal(sum(deviation), 0)
This returns:
[1] TRUE
format(sum(deviation), scientific = FALSE)
Your answer is very close to zero. It's a rounding error. R uses IEEE 754 double-precision floating point numbers. Read more here: https://en.wikipedia.org/wiki/Double-precision_floating-point_format
Related
This question already has answers here:
Round up from .5
(7 answers)
Closed 1 year ago.
There are a number of threads about this question. None seems to answer the simple question: why does R round incorrectly and how can I let it round correctly?
Correct rounding to the i-th decimal x considers the i+1-th decimal. If it is 5 or larger then x is is set to x+1. If it is 4 or smaller then x is returned. For example 1.45 is rounded to the first decimal as 1.5. 1.44 is rounded 1.4. However, in R
> round(1.45,1)
[1] 1.4
But
> round(1.46,1)
[1] 1.5
So it changes the convention to 'if the i+1th decimal is 6 or larger, then x is set to x+1'. Why? And how can I change this to the convention I am familiar with?
Most decimal fractions are not exactly representable in binary double precision
Learned here: https://stat.ethz.ch/R-manual/R-devel/library/base/html/Round.html
Section "Warnings":
Rounding to decimal digits in binary arithmetic is non-trivial (when digits != 0) and may be surprising. Be aware that most decimal fractions are not exactly representable in binary double precision. In R 4.0.0, the algorithm for round(x, d), for d > 0, has been improved to measure and round “to nearest even”, contrary to earlier versions of R (or also to sprintf() or format() based rounding).
This question already has answers here:
Why are these numbers not equal?
(6 answers)
Closed 2 years ago.
I came across a strange thing in R programming. When I simulate a sequence and want to judge whether the element is less than 0.5,
t=(1:1440)/1440
x=(t[720]-t[648])/0.1
x
#output:[1] 0.5
x<1/2
#output:[1] TRUE
x=0.5
x<1/2
#output:[1] FALSE
The two results are completely opposite and obviously the second result is what I want. Can anybody help me?
Floating point arithmetic is not exact in R, and the value you expect to be numerically exact to 0.5 may in fact be slightly more (or less). One possible workaround here would be to use rounding:
t <- (1:1440)/1440
x <- (t[720]-t[648]) / 0.1
round(x, 1) < 0.5
This question already has answers here:
Is floating point math broken?
(31 answers)
Closed 3 years ago.
I am using a Math.Round function to round the decimal numbers with a precision of 3 digits, but for some equations it gives a wrong result.The code given below gives me a wrong result.It gives 1.428 but the expected result is 1.429
Math.Round(28.57 * 5.0 / 100, 3, MidpointRounding.AwayFromZero)
I got the right answer by just putting 5.0 / 100 in a bracket.The code given below gives 1.429
Math.Round(28.57 * (5.0 / 100), 3, MidpointRounding.AwayFromZero)
I don't understand why it is happening like this. Can you explain this?
Usually, floating-point numbers may not have an "exact" representation for the number, so they loose precision.
For your example, if you tried to subtract the values in both expressions, you will get a very very small result.
This question already has answers here:
Round up from .5
(7 answers)
Closed 4 years ago.
I saw already a question with very large number of decimal digits R rounding explanation.
round(62.495, digits=2)
gives me 62.49. I would expect already 62.5, but it seems, R (3.4.3, 3.5.0) rounds up only starting at 6, e.g.,
round(62.485, 2) == 62.48
round(62.486, 2) == 62.49.
For other reasons, I am using the option
options(digits.secs=6)
From what I have learnt, one rounds up starting at 5. I tested also with Python and Matlab. Matlab rounds up, Python 3.5.4 down.
How can I change the behaviour or is this definition different, e.g. between Europe and US?
This is a floating point representation issue, 62.495 is actually represented by a slightly smaller number which then gets rounded downwards.
print(62.495,digits=22)
[1] 62.49499999999999744205
R's rounding is statistical rounding, or round half to even. It should round halves up or down to an even number, eg
round(0.5) # rounds the half down to 0
[1] 0
round(1.5) # rounds the half up to 2
[1] 2
This question already has answers here:
Why are these numbers not equal?
(6 answers)
Closed 8 years ago.
Does the last statement in this series of statements make logical sense to anybody else? R seems to give similar results for a small subset of possible sums of decimals under 1. I cannot recall any basic mathematical principles that would make this true, but it seems to be unlikely to be an error.
> 0.4+0.6
[1] 1
> 0.4+0.6==1.0
[1] TRUE
> 0.3+0.6
[1] 0.9
> 0.3+0.6==0.9
[1] FALSE
Try typing 0.3+0.6-0.9, on my system the result is -1.110223e-16 this is because the computer doesn't actually sum them as decimal numbers, it stores binary approximations, and sums those. And none of those numbers can be exactly represented in binary, so there is a small amount of error present in the calculations, and apparently it's small enough not to matter in the first one, but not the second.
Floating point arithmetic is not exact, but the == operator is. Use all.equal to compare two floating point values in R.
isTRUE(all.equal(0.3+0.6, 0.9))
You can also define a tolerance when calling all.equals.
isTRUE(all.equal(0.3+0.6, 0.9, tolerance = 0.001))