Count number of words with 3 or more letters from a string in R

I have a string sentence. I need to find the count of words from the sentence which have more than or equal to 3 letters.

For example:

sentence <- 'I have a string sentence but i do not know how to get three lettered words from it'
# Total words = 18
# 3 or more lettered words = 12

How I do it in Python in a basic way:

words = sentence.split(' ') #this creates a list of words
count = 0
for each in words:
    if len(each) >= 3:
        count = count + 1

print(count)

Alternative way in python (a little crude but):

print(len(list(filter(lambda word: len(word)>= 3, words))))

I tried doing the same thing in R:

words <- strsplit(sentence, split = ' ')
count <- 0
for (word in words) {
    l <- nchar(word)
    if (l >= 3) {
        count <- count + 1
    }
}

print(count)

This results in an error for me:

# Error in if (l >= 3) { : the condition has length > 1
# ERROR!
# Execution halted

When I checked this error on the web, it says that if we provide a vector to the if condition, then this error occurs. But I provided it with a simple numeric variable, so I do not understand what is causing this error.

Can someone please explain and help me out?

P.s.: I do not want to use any external package for this. I am learning R so want to do it with basics.

You can use lengths

sentence <- 'I have a string sentence but i do not know how to get three lettered words from it'

lengths(strsplit(sentence, '\\s+'))
# [1] 18

To count words with min. three chars, we use the first element of the resulting list, test if nchar is >= three and sum.

sum(nchar(el(strsplit(sentence, "\\s+"))) >= 3)
# [1] 12

or using pipes:

strsplit(sentence, '\\s+') |> el() |> nchar() |> base::`>=`(3) |> sum()
# [1] 12

The regex '\\s+', one or more spaces, instead of ' ' cares for (accidentally) multiple whitespaces.

Note:

To clarify lengths vs. length:

length(list(1:3))
# [1] 1

lengths(list(1:3))
# [1] 3

sapply(list(1:3), length)  ## equiv.
# [1] 3