Remove all leading zeros and turn number into a positive

Question

I'm trying to convert .000278 into 278 in R but all functions I see can only move it to .278. It needs to be a whole, positive number.

Is there a function that will remove the .0+ preceding a number??

`.000278*1000000`? Is the number a character or numeric? – Psidom Jun 17 '16 at 22:01 — Psidom, Jun 17 '16 at 22:01

score 3 · Answer 1 · edited May 23 '17 at 10:29

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I assume you want to apply this to many numbers at once (otherwise I might not understand the question).

a <- c(0.003, 0.0056, 0.000278)#Example data
a*(10^(nchar(a)-2))
[1]   3  56 278

Make sure scientific notation is disabled with scipen, as discussed in this post (e.g., by doing options(scipen=999). This is important because nchar counts the number of characters the numbers have (minus 2 for "0.").

Another approach. Use the package stringr.

a <- c(0.003, 0.0056, 0.000278)#Example data
library(stringr)
as.numeric(str_replace(a, "0.", ""))
[1]   3  56 278

Note that with this method you need to convert the output of str_replace back to numeric using as.numeric (not ideal).

edited May 23 '17 at 10:29

Community

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answered Jun 17 '16 at 22:19

milan

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I think `a*(10^(nchar(a)-2))` is the best approach. You always want to avoid unnecessary coercion. – Rich Scriven Jun 18 '16 at 00:43
Thanks for the comment, I changed the order. – milan Jun 18 '16 at 00:44
@RichardScriven `nchar` approach fails for `a <- c(0.00003, 0.0056, 0.0000278)` – 989 Jun 19 '16 at 14:19
You have to make sure that scientific notation is disabled. Try doing this first: options(scipen=999). – milan Jun 19 '16 at 14:46

Miha · Answer 2 · 2016-06-18T07:21:24.637

2

Or use

substr and regexpr what gives exactly what you wanted

x <- 0.000278
as.numeric(substr(x ,regexpr("[^0.]",x),nchar(x)))
[1] 278

And this also works for different numbers, just set:

options("scipen"=100, "digits"=10) # Force R not to use exponential notation (e.g. e-8)

and you could try for example:

z <- 0.000000588
as.numeric(substr(z ,regexpr("[^0.]",z),nchar(z)))
[1] 588

edited Jun 18 '16 at 07:21

answered Jun 17 '16 at 22:32

Miha

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@Richard Scriven good point, unfortunately I've made a mistake which I corrected (see my edit in a post). – Miha Jun 18 '16 at 07:23
Your solution heavily relies on `options("scipen"=100, "digits"=10)`. One should change R default options to use it. – 989 Jun 19 '16 at 14:27

989 · Answer 3 · 2016-06-17T22:14:33.097

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Try this (len is adjustable):

a <- 0.000278
a*(10^match(T,round(a, 1:(len=10)) == a)) # counts the number of decimals places
# 278

Use as.numeric(a) in case a is of type character.

edited Jun 17 '16 at 22:14

answered Jun 17 '16 at 22:08

989

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score 0 · Answer 4 · answered Jun 18 '16 at 00:23

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>  as.numeric(gsub("^[0.]*", "", paste(0.000278)))
[1] 278

answered Jun 18 '16 at 00:23

Marcelo Ventura

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Remove all leading zeros and turn number into a positive

4 Answers4