How to create vector (10^16 + 1, 10^16 + 2, … , 10^16 + 1000) using R?
up vote
0
down vote
favorite
1
I would like to create the following vector (10^16 + 1, 10^16 + 2, ... , 10^16 + 1000).
But if I try it like this:
daten <- rep(1e+16,1000)
daten + 1:1000
It gives me just an approximation:
[1] 10000000000000000 10000000000000002 10000000000000004 10000000000000004
[5] 10000000000000004 10000000000000006 10000000000000008 10000000000000008
[9] 10000000000000008 10000000000000010 10000000000000012 10000000000000012
[13] 10000000000000012 10000000000000014 10000000000000016 10000000000000016
......
How can I get R to calculate more precise ?
Thank you
r precision mantissa
edited Nov 10 at 16:12
Harro Cyranka
896313
asked Nov 10 at 16:03
Joshua
34
New contributor
Joshua is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
1
I would like to create the following vector (10^16 + 1, 10^16 + 2, ... , 10^16 + 1000).
But if I try it like this:
daten <- rep(1e+16,1000)
daten + 1:1000
It gives me just an approximation:
[1] 10000000000000000 10000000000000002 10000000000000004 10000000000000004
[5] 10000000000000004 10000000000000006 10000000000000008 10000000000000008
[9] 10000000000000008 10000000000000010 10000000000000012 10000000000000012
[13] 10000000000000012 10000000000000014 10000000000000016 10000000000000016
......
How can I get R to calculate more precise ?
Thank you
r precision mantissa
edited Nov 10 at 16:12
Harro Cyranka
896313
asked Nov 10 at 16:03
Joshua
34
New contributor
Joshua is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How much more precise?
– Rich Scriven
Nov 10 at 16:20
R is not (at least out of the box) an arbitrary precision calculator. In your example you are dealing with floating point arithmetic. en.wikipedia.org/wiki/IEEE_754#Basic_formats
– jogo
Nov 10 at 16:21
@Rich Scriven: so that the last digit is correct
– Joshua
Nov 10 at 16:31
add a comment |
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
I would like to create the following vector (10^16 + 1, 10^16 + 2, ... , 10^16 + 1000).
But if I try it like this:
daten <- rep(1e+16,1000)
daten + 1:1000
It gives me just an approximation:
[1] 10000000000000000 10000000000000002 10000000000000004 10000000000000004
[5] 10000000000000004 10000000000000006 10000000000000008 10000000000000008
[9] 10000000000000008 10000000000000010 10000000000000012 10000000000000012
[13] 10000000000000012 10000000000000014 10000000000000016 10000000000000016
......
How can I get R to calculate more precise ?
Thank you
r precision mantissa
edited Nov 10 at 16:12
Harro Cyranka
896313
asked Nov 10 at 16:03
Joshua
34
New contributor
Joshua is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I would like to create the following vector (10^16 + 1, 10^16 + 2, ... , 10^16 + 1000).
But if I try it like this:
daten <- rep(1e+16,1000)
daten + 1:1000
It gives me just an approximation:
[1] 10000000000000000 10000000000000002 10000000000000004 10000000000000004
[5] 10000000000000004 10000000000000006 10000000000000008 10000000000000008
[9] 10000000000000008 10000000000000010 10000000000000012 10000000000000012
[13] 10000000000000012 10000000000000014 10000000000000016 10000000000000016
......
How can I get R to calculate more precise ?
Thank you
r precision mantissa
r precision mantissa
edited Nov 10 at 16:12
Harro Cyranka
896313
asked Nov 10 at 16:03
Joshua
34
New contributor
Joshua is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 10 at 16:12
Harro Cyranka
896313
asked Nov 10 at 16:03
Joshua
34
New contributor
Joshua is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 10 at 16:12
Harro Cyranka
896313
edited Nov 10 at 16:12
Harro Cyranka
896313
edited Nov 10 at 16:12
Harro Cyranka
896313
896313
asked Nov 10 at 16:03
Joshua
34
New contributor
Joshua is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 10 at 16:03
Joshua
34
asked Nov 10 at 16:03
Joshua
34
34
New contributor
Joshua is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Joshua is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Joshua is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How much more precise?
– Rich Scriven
Nov 10 at 16:20
R is not (at least out of the box) an arbitrary precision calculator. In your example you are dealing with floating point arithmetic. en.wikipedia.org/wiki/IEEE_754#Basic_formats
– jogo
Nov 10 at 16:21
@Rich Scriven: so that the last digit is correct
– Joshua
Nov 10 at 16:31
add a comment |
How much more precise?
– Rich Scriven
Nov 10 at 16:20
R is not (at least out of the box) an arbitrary precision calculator. In your example you are dealing with floating point arithmetic. en.wikipedia.org/wiki/IEEE_754#Basic_formats
– jogo
Nov 10 at 16:21
@Rich Scriven: so that the last digit is correct
– Joshua
Nov 10 at 16:31
How much more precise?
– Rich Scriven
Nov 10 at 16:20
How much more precise?
– Rich Scriven
Nov 10 at 16:20
R is not (at least out of the box) an arbitrary precision calculator. In your example you are dealing with floating point arithmetic. en.wikipedia.org/wiki/IEEE_754#Basic_formats
– jogo
Nov 10 at 16:21
R is not (at least out of the box) an arbitrary precision calculator. In your example you are dealing with floating point arithmetic. en.wikipedia.org/wiki/IEEE_754#Basic_formats
– jogo
Nov 10 at 16:21
@Rich Scriven: so that the last digit is correct
– Joshua
Nov 10 at 16:31
@Rich Scriven: so that the last digit is correct
– Joshua
Nov 10 at 16:31
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The Rmpfr package allows you as much precision as you need. Create numbers using the mpfr function, specifying the value + the number of bits of precision you need.
library(Rmpfr)
daten <- mpfr(1:1000, 120)
daten <- daten + mpfr(1e16, 120)
print(daten[1:5])
This yields the following:
5 'mpfr' numbers of precision 120 bits
[1] 10000000000000001 10000000000000002 10000000000000003 10000000000000004 10000000000000005
answered Nov 10 at 16:31
Edward Carney
92926
thank you very much
– Joshua
Nov 10 at 16:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The Rmpfr package allows you as much precision as you need. Create numbers using the mpfr function, specifying the value + the number of bits of precision you need.
library(Rmpfr)
daten <- mpfr(1:1000, 120)
daten <- daten + mpfr(1e16, 120)
print(daten[1:5])
This yields the following:
5 'mpfr' numbers of precision 120 bits
[1] 10000000000000001 10000000000000002 10000000000000003 10000000000000004 10000000000000005
answered Nov 10 at 16:31
Edward Carney
92926
thank you very much
– Joshua
Nov 10 at 16:56
add a comment |
up vote
1
down vote
accepted
The Rmpfr package allows you as much precision as you need. Create numbers using the mpfr function, specifying the value + the number of bits of precision you need.
library(Rmpfr)
daten <- mpfr(1:1000, 120)
daten <- daten + mpfr(1e16, 120)
print(daten[1:5])
This yields the following:
5 'mpfr' numbers of precision 120 bits
[1] 10000000000000001 10000000000000002 10000000000000003 10000000000000004 10000000000000005
answered Nov 10 at 16:31
Edward Carney
92926
thank you very much
– Joshua
Nov 10 at 16:56
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The Rmpfr package allows you as much precision as you need. Create numbers using the mpfr function, specifying the value + the number of bits of precision you need.
library(Rmpfr)
daten <- mpfr(1:1000, 120)
daten <- daten + mpfr(1e16, 120)
print(daten[1:5])
This yields the following:
5 'mpfr' numbers of precision 120 bits
[1] 10000000000000001 10000000000000002 10000000000000003 10000000000000004 10000000000000005
answered Nov 10 at 16:31
Edward Carney
92926
The Rmpfr package allows you as much precision as you need. Create numbers using the mpfr function, specifying the value + the number of bits of precision you need.
library(Rmpfr)
daten <- mpfr(1:1000, 120)
daten <- daten + mpfr(1e16, 120)
print(daten[1:5])
This yields the following:
5 'mpfr' numbers of precision 120 bits
[1] 10000000000000001 10000000000000002 10000000000000003 10000000000000004 10000000000000005
answered Nov 10 at 16:31
Edward Carney
92926
answered Nov 10 at 16:31
Edward Carney
92926
answered Nov 10 at 16:31
Edward Carney
92926
answered Nov 10 at 16:31
Edward Carney
92926
92926
thank you very much
– Joshua
Nov 10 at 16:56
add a comment |
thank you very much
– Joshua
Nov 10 at 16:56
thank you very much
– Joshua
Nov 10 at 16:56
thank you very much
– Joshua
Nov 10 at 16:56
add a comment |
Joshua is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
Joshua is a new contributor. Be nice, and check out our Code of Conduct.
Joshua is a new contributor. Be nice, and check out our Code of Conduct.
Joshua is a new contributor. Be nice, and check out our Code of Conduct.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53240770%2fhow-to-create-vector-1016-1-1016-2-1016-1000-using-r%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
How much more precise?
– Rich Scriven
Nov 10 at 16:20
R is not (at least out of the box) an arbitrary precision calculator. In your example you are dealing with floating point arithmetic. en.wikipedia.org/wiki/IEEE_754#Basic_formats
– jogo
Nov 10 at 16:21
@Rich Scriven: so that the last digit is correct
– Joshua
Nov 10 at 16:31