String matching and manipulating in R

Question

I make progress in cleaning data like this:

df1 <- data.frame(ID=(c("18.1010-2.570322","171114-238509","140808-3481906  
","18055656193","180625-378224","190903-2793831 / -9311442 / -6810125","190808-625-6692","190 807 - 7941125","1807298087721Roland","19060881t1676")),
                  True_ID = c("181010-2570322","171114-2385039","190808-4381906","180556-5619343","180625-3782242", "190903-2793831 190903-9311442 
                              190903-6810125", "190808-6256692","190807-7941125","180729-8087721","190608-8112676"))

The true value is like this: 190312-4184811. So there is a pattern first six integers are a date like 19 = 2019 03 = March and 12 = Day. And the other seven numbers are random. I cleaned a lot of non informative patterns, but here I dont know exactly how to deal with this many different.

I tried something like, but I think there can be a better way:

a = str_extract(data_file$IP_P,"(^|[ ])[:digit:]{6}\\-[:digit:]{7}([ ]|$)")
b = str_extract(data_file$IP_P,"(^|[ ])[:digit:]{5}\\-[:digit:]{7}([ ]|$)")
c = str_extract(data_file$IP_P,"(^|[ ])[:digit:]{4}\\-[:digit:]{7}([ ]|$)")
d = str_extract(data_file$IP_P,"(^|[ ])[:digit:]{6}\\-[:digit:]{6}([ ]|$)")
e = str_extract(data_file$IP_P,"(^|[ ])[:digit:]{6}\\-[:digit:]{5}([ ]|$)")
f = str_extract(data_file$IP_P,"(^|[ ])[:digit:]{6}\\-[:digit:]{4}([ ]|$)")
g = str_extract(data_file$IP_P,"(^|[ ])[:digit:]{6}\\-[:digit:]{8}([ ]|$)")
h = str_extract(data_file$IP_P,"(^|[ ])[:digit:]{6}\\-[:digit:]{9}([ ]|$)")

data_file["Extracted_i"] = NA

data1 <- data.frame(a,b,c,d,e,f,g,h) 
data1 <- data1 %>% unite("z", a:h, remove = FALSE)
data_file["Extracted_i"] =gsub("[^0-9\\.\\-]", "", data1$z)

score 1 · Accepted Answer · answered Sep 13 '20 at 14:37

Couldn't you just strip out all the non-digit characters to give a string of all the digits, then paste together the first 6 and second 6 with a joining "-"?

paste(substr(gsub("\\D", "", df1$ID), 1, 6), 
      substr(gsub("\\D", "", df1$ID), 7, 12), 
      sep = "-")
#> [1] "181010-257032" "171114-238509" "140808-348190" "180556-56193"  
#> [5] "180625-378224" "190903-279383" "190808-625669" "190807-794112" 
#> [9] "180729-808772" "190608-811676"

score 1 · Answer 2 · answered Sep 13 '20 at 17:51

We could also use gsub to capture the characters as a group and in the replacement specify the backreference (\\1, \\2) of the captured group

gsub("^(.{1,6})(.{1,6}).*", "\\1-\\2", gsub("\\D+", "", df1$ID))
#[1] "181010-257032" "171114-238509" "140808-348190" "180556-56193"  
#[5] "180625-378224" "190903-279383" "190808-625669" "190807-794112"
#[9] "180729-808772" "190608-811676"

String matching and manipulating in R

2 Answers2