I have a list entitled SET1Bearing1slope with nine numbers, and each number has at least 10 decimal places. When I use the mean() function on the list I get an arithmetic mean
.
Yet if I list the numbers individually and then use the mean() function, I get a different output
I know that this is caused by a rounding and that the second mean is more accurate. Is there a way to avoid this issue? What method can I use to avoid rounding errors when calculating the mean?
In R, mean() expects a vector of values, not multiple values. It is also a generic function so it is tolerant of additional parameters it doesn't understand (but doesn't warn you about them). See
mean(c(1,5,6))
# [1] 4
mean(1, 5, 6) #only "1" is used here, 5 and 6 are ignored.
# [1] 1
So in your example there are no rounding errors, you are just calling the function incorrectly.
Look at the difference in the way you're calling the function:
mean(c(1,2,5))
[1] 2.666667
mean(1,2,5)
[1] 1
As pointed by MrFlick, in the first case you're passing a vector of numbers (the correct way); in the second, you're passing a list of arguments, and just the first one is considered.
As for the number of digits, you can specify it using options():
options(digits = 10)
x <- runif(10)
x
[1] 0.49957540398 0.71266139182 0.07266473584 0.90541790240 0.41799820261
[6] 0.59809536533 0.88133668737 0.17078919476 0.92475634208 0.48827998806
mean(x)
[1] 0.5671575214
But remember that a greater number of digits is not necessarily better. There's a reason why R and others limits the number os digits. Check this topic: https://en.wikipedia.org/wiki/Significance_arithmetic
Related
There is an option in R to get control over digit display. For example:
options(digits=10)
is supposed to give the calculation results in 10 digits till the end of R session. In the help file of R, the definition for digits parameter is as follows:
digits: controls the number of digits
to print when printing numeric values.
It is a suggestion only. Valid values
are 1...22 with default 7
So, it says this is a suggestion only. What if I like to always display 10 digits, not more or less?
My second question is, what if I like to display more than 22 digits, i.e. for more precise calculations like 100 digits? Is it possible with base R, or do I need an additional package/function for that?
Edit: Thanks to jmoy's suggestion, I tried sprintf("%.100f",pi) and it gave
[1] "3.1415926535897931159979634685441851615905761718750000000000000000000000000000000000000000000000000000"
which has 48 decimals. Is this the maximum limit R can handle?
The reason it is only a suggestion is that you could quite easily write a print function that ignored the options value. The built-in printing and formatting functions do use the options value as a default.
As to the second question, since R uses finite precision arithmetic, your answers aren't accurate beyond 15 or 16 decimal places, so in general, more aren't required. The gmp and rcdd packages deal with multiple precision arithmetic (via an interace to the gmp library), but this is mostly related to big integers rather than more decimal places for your doubles.
Mathematica or Maple will allow you to give as many decimal places as your heart desires.
EDIT:
It might be useful to think about the difference between decimal places and significant figures. If you are doing statistical tests that rely on differences beyond the 15th significant figure, then your analysis is almost certainly junk.
On the other hand, if you are just dealing with very small numbers, that is less of a problem, since R can handle number as small as .Machine$double.xmin (usually 2e-308).
Compare these two analyses.
x1 <- rnorm(50, 1, 1e-15)
y1 <- rnorm(50, 1 + 1e-15, 1e-15)
t.test(x1, y1) #Should throw an error
x2 <- rnorm(50, 0, 1e-15)
y2 <- rnorm(50, 1e-15, 1e-15)
t.test(x2, y2) #ok
In the first case, differences between numbers only occur after many significant figures, so the data are "nearly constant". In the second case, Although the size of the differences between numbers are the same, compared to the magnitude of the numbers themselves they are large.
As mentioned by e3bo, you can use multiple-precision floating point numbers using the Rmpfr package.
mpfr("3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825")
These are slower and more memory intensive to use than regular (double precision) numeric vectors, but can be useful if you have a poorly conditioned problem or unstable algorithm.
If you are producing the entire output yourself, you can use sprintf(), e.g.
> sprintf("%.10f",0.25)
[1] "0.2500000000"
specifies that you want to format a floating point number with ten decimal points (in %.10f the f is for float and the .10 specifies ten decimal points).
I don't know of any way of forcing R's higher level functions to print an exact number of digits.
Displaying 100 digits does not make sense if you are printing R's usual numbers, since the best accuracy you can get using 64-bit doubles is around 16 decimal digits (look at .Machine$double.eps on your system). The remaining digits will just be junk.
One more solution able to control the how many decimal digits to print out based on needs (if you don't want to print redundant zero(s))
For example, if you have a vector as elements and would like to get sum of it
elements <- c(-1e-05, -2e-04, -3e-03, -4e-02, -5e-01, -6e+00, -7e+01, -8e+02)
sum(elements)
## -876.5432
Apparently, the last digital as 1 been truncated, the ideal result should be -876.54321, but if set as fixed printing decimal option, e.g sprintf("%.10f", sum(elements)), redundant zero(s) generate as -876.5432100000
Following the tutorial here: printing decimal numbers, if able to identify how many decimal digits in the certain numeric number, like here in -876.54321, there are 5 decimal digits need to print, then we can set up a parameter for format function as below:
decimal_length <- 5
formatC(sum(elements), format = "f", digits = decimal_length)
## -876.54321
We can change the decimal_length based on each time query, so it can satisfy different decimal printing requirement.
If you work primarily with tibbles, there is a function that enforces digits: num().
Here is an example:
library(tidyverse)
data <- tribble(
~ weight, ~ weight_selfreport,
81.5,81.66969147005445,
72.6,72.59528130671505,
92.9,93.01270417422867,
79.4,79.4010889292196,
94.6,96.64246823956442,
80.2,79.4010889292196,
116.2,113.43012704174228,
95.4,95.73502722323049,
99.5,99.8185117967332
)
data <-
data %>%
mutate(across(where(is.numeric), ~ num(., digits = 3)))
data
#> # A tibble: 9 × 2
#> weight weight_selfreport
#> <num:.3!> <num:.3!>
#> 1 81.500 81.670
#> 2 72.600 72.595
#> 3 92.900 93.013
#> 4 79.400 79.401
#> 5 94.600 96.642
#> 6 80.200 79.401
#> 7 116.200 113.430
#> 8 95.400 95.735
#> 9 99.500 99.819
Thus you can even decide to have different rounding options depending on what your needs are. I find it very helpful and a rather quick solution to printing dfs.
There is an option in R to get control over digit display. For example:
options(digits=10)
is supposed to give the calculation results in 10 digits till the end of R session. In the help file of R, the definition for digits parameter is as follows:
digits: controls the number of digits
to print when printing numeric values.
It is a suggestion only. Valid values
are 1...22 with default 7
So, it says this is a suggestion only. What if I like to always display 10 digits, not more or less?
My second question is, what if I like to display more than 22 digits, i.e. for more precise calculations like 100 digits? Is it possible with base R, or do I need an additional package/function for that?
Edit: Thanks to jmoy's suggestion, I tried sprintf("%.100f",pi) and it gave
[1] "3.1415926535897931159979634685441851615905761718750000000000000000000000000000000000000000000000000000"
which has 48 decimals. Is this the maximum limit R can handle?
The reason it is only a suggestion is that you could quite easily write a print function that ignored the options value. The built-in printing and formatting functions do use the options value as a default.
As to the second question, since R uses finite precision arithmetic, your answers aren't accurate beyond 15 or 16 decimal places, so in general, more aren't required. The gmp and rcdd packages deal with multiple precision arithmetic (via an interace to the gmp library), but this is mostly related to big integers rather than more decimal places for your doubles.
Mathematica or Maple will allow you to give as many decimal places as your heart desires.
EDIT:
It might be useful to think about the difference between decimal places and significant figures. If you are doing statistical tests that rely on differences beyond the 15th significant figure, then your analysis is almost certainly junk.
On the other hand, if you are just dealing with very small numbers, that is less of a problem, since R can handle number as small as .Machine$double.xmin (usually 2e-308).
Compare these two analyses.
x1 <- rnorm(50, 1, 1e-15)
y1 <- rnorm(50, 1 + 1e-15, 1e-15)
t.test(x1, y1) #Should throw an error
x2 <- rnorm(50, 0, 1e-15)
y2 <- rnorm(50, 1e-15, 1e-15)
t.test(x2, y2) #ok
In the first case, differences between numbers only occur after many significant figures, so the data are "nearly constant". In the second case, Although the size of the differences between numbers are the same, compared to the magnitude of the numbers themselves they are large.
As mentioned by e3bo, you can use multiple-precision floating point numbers using the Rmpfr package.
mpfr("3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825")
These are slower and more memory intensive to use than regular (double precision) numeric vectors, but can be useful if you have a poorly conditioned problem or unstable algorithm.
If you are producing the entire output yourself, you can use sprintf(), e.g.
> sprintf("%.10f",0.25)
[1] "0.2500000000"
specifies that you want to format a floating point number with ten decimal points (in %.10f the f is for float and the .10 specifies ten decimal points).
I don't know of any way of forcing R's higher level functions to print an exact number of digits.
Displaying 100 digits does not make sense if you are printing R's usual numbers, since the best accuracy you can get using 64-bit doubles is around 16 decimal digits (look at .Machine$double.eps on your system). The remaining digits will just be junk.
One more solution able to control the how many decimal digits to print out based on needs (if you don't want to print redundant zero(s))
For example, if you have a vector as elements and would like to get sum of it
elements <- c(-1e-05, -2e-04, -3e-03, -4e-02, -5e-01, -6e+00, -7e+01, -8e+02)
sum(elements)
## -876.5432
Apparently, the last digital as 1 been truncated, the ideal result should be -876.54321, but if set as fixed printing decimal option, e.g sprintf("%.10f", sum(elements)), redundant zero(s) generate as -876.5432100000
Following the tutorial here: printing decimal numbers, if able to identify how many decimal digits in the certain numeric number, like here in -876.54321, there are 5 decimal digits need to print, then we can set up a parameter for format function as below:
decimal_length <- 5
formatC(sum(elements), format = "f", digits = decimal_length)
## -876.54321
We can change the decimal_length based on each time query, so it can satisfy different decimal printing requirement.
If you work primarily with tibbles, there is a function that enforces digits: num().
Here is an example:
library(tidyverse)
data <- tribble(
~ weight, ~ weight_selfreport,
81.5,81.66969147005445,
72.6,72.59528130671505,
92.9,93.01270417422867,
79.4,79.4010889292196,
94.6,96.64246823956442,
80.2,79.4010889292196,
116.2,113.43012704174228,
95.4,95.73502722323049,
99.5,99.8185117967332
)
data <-
data %>%
mutate(across(where(is.numeric), ~ num(., digits = 3)))
data
#> # A tibble: 9 × 2
#> weight weight_selfreport
#> <num:.3!> <num:.3!>
#> 1 81.500 81.670
#> 2 72.600 72.595
#> 3 92.900 93.013
#> 4 79.400 79.401
#> 5 94.600 96.642
#> 6 80.200 79.401
#> 7 116.200 113.430
#> 8 95.400 95.735
#> 9 99.500 99.819
Thus you can even decide to have different rounding options depending on what your needs are. I find it very helpful and a rather quick solution to printing dfs.
I have a boolean vector in which I want to count the number of occurrences of some patterns.
For example, for the pattern "(1,1)" and the vector "(1,1,1,0,1,1,1)", the answer should be 4.
The only built-in function I found to help is grepRaw, which finds the occurrences of a particular string in a longer string. However, it seems to fail when the sub-strings matching the pattern overlap:
length(grepRaw("11","1110111",all=TRUE))
# [1] 2
Do you have any ideas to obtain the right answer in this case?
Edit 1
I'm afraid that Rich's answer works for the particular example I posted, but fails in a more general setting:
> sum(duplicated(rbind(c(FALSE,FALSE),embed(c(TRUE,TRUE,TRUE,FALSE,TRUE,TRUE,TRUE),2))))
[1] 3
In this other example, the expected answer would be 0.
Using the function rollapply you can apply a moving window of width = 2 summing the values. Then you can sum the records where the result is equal to 2 i.e. sum(c(1,1))
library(zoo)
z <- c(1,1,1,0,1,1,1)
sum(rollapply(z, 2, sum) == 2)
There is an option in R to get control over digit display. For example:
options(digits=10)
is supposed to give the calculation results in 10 digits till the end of R session. In the help file of R, the definition for digits parameter is as follows:
digits: controls the number of digits
to print when printing numeric values.
It is a suggestion only. Valid values
are 1...22 with default 7
So, it says this is a suggestion only. What if I like to always display 10 digits, not more or less?
My second question is, what if I like to display more than 22 digits, i.e. for more precise calculations like 100 digits? Is it possible with base R, or do I need an additional package/function for that?
Edit: Thanks to jmoy's suggestion, I tried sprintf("%.100f",pi) and it gave
[1] "3.1415926535897931159979634685441851615905761718750000000000000000000000000000000000000000000000000000"
which has 48 decimals. Is this the maximum limit R can handle?
The reason it is only a suggestion is that you could quite easily write a print function that ignored the options value. The built-in printing and formatting functions do use the options value as a default.
As to the second question, since R uses finite precision arithmetic, your answers aren't accurate beyond 15 or 16 decimal places, so in general, more aren't required. The gmp and rcdd packages deal with multiple precision arithmetic (via an interace to the gmp library), but this is mostly related to big integers rather than more decimal places for your doubles.
Mathematica or Maple will allow you to give as many decimal places as your heart desires.
EDIT:
It might be useful to think about the difference between decimal places and significant figures. If you are doing statistical tests that rely on differences beyond the 15th significant figure, then your analysis is almost certainly junk.
On the other hand, if you are just dealing with very small numbers, that is less of a problem, since R can handle number as small as .Machine$double.xmin (usually 2e-308).
Compare these two analyses.
x1 <- rnorm(50, 1, 1e-15)
y1 <- rnorm(50, 1 + 1e-15, 1e-15)
t.test(x1, y1) #Should throw an error
x2 <- rnorm(50, 0, 1e-15)
y2 <- rnorm(50, 1e-15, 1e-15)
t.test(x2, y2) #ok
In the first case, differences between numbers only occur after many significant figures, so the data are "nearly constant". In the second case, Although the size of the differences between numbers are the same, compared to the magnitude of the numbers themselves they are large.
As mentioned by e3bo, you can use multiple-precision floating point numbers using the Rmpfr package.
mpfr("3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825")
These are slower and more memory intensive to use than regular (double precision) numeric vectors, but can be useful if you have a poorly conditioned problem or unstable algorithm.
If you are producing the entire output yourself, you can use sprintf(), e.g.
> sprintf("%.10f",0.25)
[1] "0.2500000000"
specifies that you want to format a floating point number with ten decimal points (in %.10f the f is for float and the .10 specifies ten decimal points).
I don't know of any way of forcing R's higher level functions to print an exact number of digits.
Displaying 100 digits does not make sense if you are printing R's usual numbers, since the best accuracy you can get using 64-bit doubles is around 16 decimal digits (look at .Machine$double.eps on your system). The remaining digits will just be junk.
One more solution able to control the how many decimal digits to print out based on needs (if you don't want to print redundant zero(s))
For example, if you have a vector as elements and would like to get sum of it
elements <- c(-1e-05, -2e-04, -3e-03, -4e-02, -5e-01, -6e+00, -7e+01, -8e+02)
sum(elements)
## -876.5432
Apparently, the last digital as 1 been truncated, the ideal result should be -876.54321, but if set as fixed printing decimal option, e.g sprintf("%.10f", sum(elements)), redundant zero(s) generate as -876.5432100000
Following the tutorial here: printing decimal numbers, if able to identify how many decimal digits in the certain numeric number, like here in -876.54321, there are 5 decimal digits need to print, then we can set up a parameter for format function as below:
decimal_length <- 5
formatC(sum(elements), format = "f", digits = decimal_length)
## -876.54321
We can change the decimal_length based on each time query, so it can satisfy different decimal printing requirement.
If you work primarily with tibbles, there is a function that enforces digits: num().
Here is an example:
library(tidyverse)
data <- tribble(
~ weight, ~ weight_selfreport,
81.5,81.66969147005445,
72.6,72.59528130671505,
92.9,93.01270417422867,
79.4,79.4010889292196,
94.6,96.64246823956442,
80.2,79.4010889292196,
116.2,113.43012704174228,
95.4,95.73502722323049,
99.5,99.8185117967332
)
data <-
data %>%
mutate(across(where(is.numeric), ~ num(., digits = 3)))
data
#> # A tibble: 9 × 2
#> weight weight_selfreport
#> <num:.3!> <num:.3!>
#> 1 81.500 81.670
#> 2 72.600 72.595
#> 3 92.900 93.013
#> 4 79.400 79.401
#> 5 94.600 96.642
#> 6 80.200 79.401
#> 7 116.200 113.430
#> 8 95.400 95.735
#> 9 99.500 99.819
Thus you can even decide to have different rounding options depending on what your needs are. I find it very helpful and a rather quick solution to printing dfs.
I'm having troubles with
set.seed(1)
sum(abs(rnorm(100)))
set.seed(1)
cumsum(abs(rnorm(100)))
Why does the value of the sum differ from the last value of the cumulative sum with the cumulative sum preserving the all decimal digits and sum rounding 1 digit off.
Also note that this really really is about how values are printed i.e. presented. This does not change the values themselves, e.g. ...
set.seed(1)
d1 <- sum(abs(rnorm(100)))
set.seed(1)
d2 <- cumsum(abs(rnorm(100)))
(d1 == d2)[100]
## [1] TRUE
This is a consequence of the way R prints atomic vectors.
With the default digits option set to 7 as you likely have, any value between -1 and 1 will print with seven decimal places. Because of the way R prints atomic vectors, all other values in the vector will also have seven decimal places. Furthermore, a value of .6264538 with digits option set to 7 must print with eight digits (0.6264538) because it must have a leading zero. There are two of these values in your rnorm() vector.
If you look at cumsum(abs(rnorm(100)))[100] alone and you can see the difference (actually it becomes the same as printed value as sum(abs(rnorm(100))), although not exactly the same value).
sum(abs(rnorm(100)))
# [1] 71.67207
cumsum(abs(rnorm(100)))[100]
# [1] 71.67207
Notice that both of these values have seven digits. Probably the most basic example of this I can think of is as follows
0.123456789
#[1] 0.1234568
1.123456789
#[1] 1.123457
11.123456789
# [1] 11.12346
## and so on ...