R Extract partially matching string - r

Question

I have a question about extracting a part of a string from several files that has these rows:
units = specified
- name 0 = prDM: Pressure, Digiquartz [db]
- name 1 = t090C: Temperature [ITS-90, deg C]
- name 2 = c0S/m: Conductivity [S/m]
- name 3 = t190C:Temperature, 2 [ITS-90, deg C]
- name 4 = c1S/m: Conductivity, 2 [S/m]
- name 5 = flSP: Fluorescence, Seapoint
- name 6 = sbeox0ML/L: Oxygen, SBE 43 [ml/l]
- name 7 = altM: Altimeter [m]
- name 8 = sal00: Salinity, Practical [PSU]
- name 9 = sal11: Salinity, Practical, 2 [PSU]
- span 0 = 1.000, 42.000
I need to extract only the information of the columns that start with "name" and extract everything between = and: .
For example, in the row "name 0 = prDM: Pressure, Digiquartz [db]" the desired result will be prDM.
Some files have different number of "name"rows (i.e. this example has 13 rows but other files has 16, and the number varies), so I want it to be as general as I can so I can allways extract the right strings independently the number of rows.Rows starts with # and a space before name.
I have tried this code but it only extract the first row. Can you please help me with this? Many thanks!
CNV<-NULL
for (i in 1:nro.files){
x <- readLines(all.files[i])
name.col<-grep("^\\# name", x)
df <- data.table::fread(text = x[name.col])
CNV[[i]]<-df
}

Answer 1

using stringr and the regex pattern "name \\d+ = (.*?):" which means in words "name followed by one or more digits followed by an equals sign followed by a space followed by a captured group containing any character (the period) zero or more times (the *) followed by a colon".
library(stringr)
strings <- c("name 0 = prDM: Pressure, Digiquartz [db]",
"name 1 = t090C: Temperature [ITS-90, deg C]",
"name 2 = c0S/m: Conductivity [S/m]",
"name 3 = t190C:Temperature, 2 [ITS-90, deg C]",
"name 4 = c1S/m: Conductivity, 2 [S/m]",
"name 5 = flSP: Fluorescence, Seapoint",
"name 6 = sbeox0ML/L: Oxygen, SBE 43 [ml/l]",
"name 7 = altM: Altimeter [m]",
"name 8 = sal00: Salinity, Practical [PSU]",
"name 9 = sal11: Salinity, Practical, 2 [PSU]")
result <- str_match(strings, "name \\d+ = (.*):")
result[,2]
[1] "prDM" "t090C" "c0S/m" "t190C" "c1S/m" "flSP" "sbeox0ML/L"
[8] "altM" "sal00" "sal11"
Or if you prefer base
pattern = "name \\d+ = (.*):"
result <- regmatches(strings, regexec(pattern, strings))
sapply(result, "[[", 2)
[1] "prDM" "t090C" "c0S/m" "t190C" "c1S/m" "flSP" "sbeox0ML/L"
[8] "altM" "sal00" "sal11"

Answer 2

Use str_extract from package stringr and positive lookahead and lookbehind:
str <- "name 0 = prDM: Pressure, Digiquartz [db]"
str_extract(str, "(?<== ).*(?=:)")
[1] "prDM"
Explanation:
(?<== )if you see =followed by white space on the left (lookbehind)
.* match anything until ...
(?=:)... you see a colon on the right (lookahead)

Answer 3

In Base R
test <- c("name 0 = prDM: Pressure, Digiquartz [db]","name 1 = t090C: Temperature [ITS-90, deg C]")
gsub("^name [0-9]+ = (.+):.+","\\1",test)
[1] "prDM" "t090C"
explanation
^name [0-9]+ Searches for a the beginning of a string ^ with name folowed by any length of number
= (.+): any length + of any character . found between = and : are stored ( ) to be later recalled by \\1