在R语言中从字符向量提取有效数字的正则表达式解决方案

Hey there, let's get this sorted for you! The issue with your initial regex attempt is probably the syntax differences in how R handles regular expressions—we don't use the leading/trailing / like in JavaScript, and escape characters need double backslashes.

Here's exactly what you need to pull out those valid numbers:

The Correct Regular Expression #

We need a regex that matches full strings representing valid numbers, with no extra spaces or non-digit characters. Here's the pattern tailored for R:

"^-?(0|[1-9]\\d*)(\\.\\d+)?$"

Let's break down what each part does:

• ^ and $: Anchor the match to the start and end of the string, so we only match entirely valid numbers (no extra characters like "4h" slip through)
• -?: Matches an optional leading negative sign
• 0|[1-9]\\d*: Matches a valid integer part—either a single 0, or a non-zero digit followed by any number of digits (prevents leading zeros like "04" if that's a concern)
• (\\.\\d+)?: Matches an optional decimal part—requires a dot immediately followed by one or more digits, which excludes "7. 5" (since there's a space after the dot)

Applying the Regex in R #

You can use either base R or the tidyverse's stringr package to filter your vector.

Base R Approach #

# Your original vector
vec <- c("hi", "4", "-21", "6.5", "7. 5", "-2.2", "4h")

# Filter using grepl()
valid_numbers <- vec[grepl("^-?(0|[1-9]\\d*)(\\.\\d+)?$", vec)]

# Check the result
valid_numbers
# Output: [1] "4"    "-21"  "6.5"  "-2.2"

Tidyverse (stringr) Approach #

If you prefer cleaner syntax, str_subset() does the filtering in one step:

library(stringr)

valid_numbers <- str_subset(vec, "^-?(0|[1-9]\\d*)(\\.\\d+)?$")
valid_numbers
# Same result as above

This will exactly give you the output you're looking for, excluding invalid entries like "hi", "7. 5", and "4h".

内容的提问来源于stack exchange，提问作者Williams86