R grepl: combine conditions

Question

I'm new in R, and I have to check, if string contains only digits, . and space symbols. My version of code is:

grepl("\\d+.", x)

But it doesn't work. I need following result:

str <- "0. 365" 
str1 <- "S12"

grepl("\\d+.", str) #True
grepl("\\d+.", str1) #False

How can I combine my conditions for grepl correctly?

Answer 1

Try this regex.

str <- c("0. 365", "S12")
grepl("^[ \\.[:digit:]]*$", str)
#[1]  TRUE FALSE

Answer 2

You say you want to check "if string contains only digits, . and space symbols". You do not mention in which order these character types should appear. I take it that the order you are looking for is fixed: first digit, then period, then space (to judge by the fact that you've accepted @Rui's answer, which works for fixed order). If, however, the order is variable rather than fixed, then this regex would work:

Data:

str <- c("0. 365", ".0&0 ", "abc.12 .2", "000. ", "123", "ab.c")

Solution:

grepl("^(?=.*\\s)(?=.*\\d)(?=.*\\.)[\\s\\d\\.]+$", str, perl = T)
[1]  TRUE FALSE FALSE  TRUE FALSE FALSE

This works by asserting that (i) a white space char exists, (ii) a digit exists, and (iii) a period exists, and by allowing only the three characters to appear between start of string ^and end of string $. (For a similar case see How to match strings that only contain a character set in any order and any numbers?)