I have this df. I'd like to replace every instance of "1.1", "2.1" "3.2", etc. based on a lookup table.
Lookup table
Desired output
I looked at Replace values in a dataframe based on lookup table it doesn't solve my problem. Thanks in advance!
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An option would be gsubfn. We match one or more digits with a dot ([0-9.]+) as the pattern and in the replacement, pass a list of key/value pairs created from the second dataset ('df2'). For the matching pattern with the 'keys', it replace the corresponding 'value' within the string
library(gsubfn)
df1$Node <- gsubfn("([0-9.]+)", as.list(setNames(df2$Label,df2$Node)), df1$Node)
df1$Node
#[1] "One one > Two one > Three two" "One two > Two two > Three one" "One one > Two two > Three one" "One two > Two one > Three two"
#[5] "One one > Two two > Three two"
data
df1 <- data.frame(ID = 1:5, Node = c("1.1 > 2.1 > 3.2", "1.2 > 2.2 > 3.1", "1.1 > 2.2 > 3.1", "1.2 > 2.1 > 3.2", "1.1 > 2.2 > 3.2"), stringsAsFactors = FALSE)
df2 <- data.frame(Label = c("One one", "One two", "Two one", "Two two", "Three one", "Three two"), Node = c("1.1", "1.2", "2.1", "2.2", "3.1", "3.2"), stringsAsFactors = FALSE)
akrun
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That works. Thanks! One question. Why do you use "([0-9.]+)" as pattern? – Ketty May 23 '19 at 19:22
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@Ketty Because it is decimal number `0-9` and `.` – akrun May 23 '19 at 19:24
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Ah I see...because df1$Node and df2$Node are numbers and `.`. Thanks again! – Ketty May 23 '19 at 19:35
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@Ketty No problem. I added some explanation – akrun May 23 '19 at 19:36