1

I have a string like this:

str1 <- "get all securities in portfolio port1 on date 2010-12-31 where field value of Close on 2010-12-31+10 less than 2000"

I am trying to convert "2010-12-31+10" to "2011-01-10" in str1. I tried str_replace_all method of stringr package But i didnt get the output. 

> str_replace_all(str1,"2010-12-31+10","2011-01-10")
[1] "get all securities in portfolio port1 on date 2010-12-31 where field value of Close on 2010-12-31+10 less than 2000"

What is the reason for that?

Dinoop Nair
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1 Answers1

3

The second argument of str_replace_all is not a character string but a regular expression. So you have to escape the symbols such as + that have a special meaning in regexps :

R> str_replace_all(str1,"2010-12-31\\+10","2011-01-10")
[1] "get all securities in portfolio port1 on date 2010-12-31 where field value of Close on 2011-01-10 less than 2000"

Or you can use the fixed function of stringr to make it match your pattern as a regular string :

R> str_replace_all(str1,fixed("2010-12-31+10"),"2011-01-10")
[1] "get all securities in portfolio port1 on date 2010-12-31 where field value of Close on 2011-01-10 less than 2000"
juba
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