exec returns an object with a index property: var match = /bar/.exec(“foobar”); if (match) { console.log(“match found at ” + match.index); } And for multiple matches: var re = /bar/g, str = “foobarfoobar”; while ((match = re.exec(str)) != null) { console.log(“match found at ” + match.index); }
use match(), assuming values in df1 are unique. df1 <- data.frame(x1=1:4,x2=letters[1:4],stringsAsFactors=FALSE) df2 <- data.frame(x1=2:3,x2=c(“zz”,”qq”),stringsAsFactors=FALSE) df1$x2[match(df2$x1,df1$x1)] <- df2$x2 > df1 x1 x2 1 1 a 2 2 zz 3 3 qq 4 4 d If the values aren’t unique, use : for(id in 1:nrow(df2)){ df1$x2[df1$x1 %in% df2$x1[id]] <- df2$x2[id] }
I notice that you mention a function %like% in your current approach. I don’t know if that’s a reference to the %like% from “data.table”, but if it is, you can definitely use it as follows. Note that the object does not have to be a data.table (but also remember that subsetting approaches for data.frames and … Read more
Use regex.test() if all you want is a boolean result: console.log(/^([a-z0-9]{5,})$/.test(‘abc1’)); // false console.log(/^([a-z0-9]{5,})$/.test(‘abc12’)); // true console.log(/^([a-z0-9]{5,})$/.test(‘abc123′)); // true …and you could remove the () from your regexp since you’ve no need for a capture.
The function match works on vectors: x <- sample(1:10) x # [1] 4 5 9 3 8 1 6 10 7 2 match(c(4,8),x) # [1] 1 5 match only returns the first encounter of a match, as you requested. It returns the position in the second argument of the values in the first argument. For … Read more
One option is just to use the regex | character to try to match each of the substrings in the words in your Series s (still using str.contains). You can construct the regex by joining the words in searchfor with |: >>> searchfor = [‘og’, ‘at’] >>> s[s.str.contains(‘|’.join(searchfor))] 0 cat 1 hat 2 dog 3 … Read more
re.match is anchored at the beginning of the string. That has nothing to do with newlines, so it is not the same as using ^ in the pattern. As the re.match documentation says: If zero or more characters at the beginning of string match the regular expression pattern, return a corresponding MatchObject instance. Return None … Read more
You misunderstand how boolean expressions work; they don’t work like an English sentence and guess that you are talking about the same comparison for all names here. You are looking for: if x == 1 or y == 1 or z == 1: x and y are otherwise evaluated on their own (False if 0, … Read more