Loop over list of strings printing string starting with x

Question

I have a dataframe (df) with a column of type strings such as

code     country
DK       Denmark
DE       Germany
BEB      Belgium
B        Brazil

And I wont to write a for loop that prints the countries starting with B. This is how I approach it but I get an error:

for (i in unique(df$country)) {
  if (i == grep("^B", unique(df$country), value = TRUE)) {
    print(i)
  }
}

Error:

Error in if (i == grep("^B", unique(df$country), value = TRUE)) { : 
  the condition has length > 1

Does this answer your question? [how do I grep in R?](https://stackoverflow.com/questions/4220631/how-do-i-grep-in-r) — Andre Wildberg, Oct 30 '22 at 17:22

score 1 · Accepted Answer · answered Oct 30 '22 at 17:20

We don't need a loop here as grep is vectorized

unique(subset(df, grepl("^B", country), select = country))

-output

  country
3 Belgium
4  Brazil

Regarding the issue in the loop, it is just that == does elementwise comparison and returns the logical vector with the length of the second vector as the first one is just of length 1.

'Belgium' == grep("^B", unique(df$country), value = TRUE)
[1]  TRUE FALSE

and this triggers the length warning as if/else expects a vector of length 1 as input

If we use %in%, it returns a single TRUE/FALSE

for (i in unique(df$country)) {
  if (i %in% grep("^B", unique(df$country), value = TRUE)) {
    print(i)
  }
}

-output

[1] "Belgium"
[1] "Brazil"

data

df <- structure(list(code = c("DK", "DE", "BEB", "B"), country = c("Denmark", 
"Germany", "Belgium", "Brazil")), class = "data.frame",
 row.names = c(NA, 
-4L))

Loop over list of strings printing string starting with x

1 Answers1

data