If else statement refuses to execute

[Quack]

I created a dictionary of unigrams, bigrams, trigrams, and 4-grams. The dictionary looks like this:

word1                 word2     frequency
will                            5153
like                            5081
get                             4821
let                   know       627
right                 now        575
look                  like       559
next                  week       478
social                media      465
let_us                know       194
let_know              think      173
cinco_de              mayo       172
manufacturer_custom   built      171
custom_built          painted    171

Here is a function to look up words in the dictionary (named dat)

nxtword1<- function(word){dat %>% filter(word1 == word) %>% select(-word1)}

I tested this function, and it worked.

The purpose of the following code is to look up a text string in the dictionary. Messages will be printed to the console, on whether the phrase appears in the dictionary or not.

predict<-nxtword1("new_manufacturer_custom_built")
if (row.names(predict)>0) {print("match found")
} else {print("no match found")}
predict

I knew that the phrase "new_manufacturer_custom_built" did not appear in the dictionary. Apparently R does not like my conditional statement. I was hoping that the else statement would be executed. Instead an error message appears

Error in if (row.names(predict) > 0) { : argument is of length zero

> else {print("no match found")}

> predict}

Apparently R does not like my conditional statement. How to fix this?

[Anantadinath]

Your else command is on a different line and that's the reason

It should have been exactly after the curly braces of if statement like

If (true){
Something
} else {
Something
}

Else should be on the same line

[Quack]

Thanks for the correction. I submitted the corrected code

predict<-nxtword1("new_manufacturer_custom_built")
if (row.names(predict)>0) {print("match found")
}  else {print("no match found")}
Predict

I got a different error message this time:
Error in if (row.names(predict) > 0) { : argument is of length zero

To try to figure out what was wrong, I typed
> predict and got on the console

[1] word2     frequency
<0 rows> (or 0-length row.names)

I wanted R to execute the else statement, but it's still not doing that.

[Kill3rbee]

It will be easier if you provided a reproducible example. I would just do it this way and get rid of nxtword1 function.

 ifelse(word %in% dat$word1, print("match found"), print("match not found"))

That should be able to do it. Again

[Anantadinath]

it means your predict object doesn't have row names.
some data frame just doesn't have row names.
try nrow(predict) instead.

[Quack]

Here is some from the dictionary
word1 word2 frequency
let know 627
right now 575
look like 559
next week 478
social media 465
let_us know 194
let_know think 173
cinco_de mayo 172
manufacturer_custom built 171
custom_built painted 171

Thank you for your code. It worked:) I tried elaborating on it to get it to execute functions I had written before.
checkdic <-function(word) {ifelse(word %in% dat$word1, nxtword1(word), nxtword1(less1gram(word)))
}
When the word appeared in the dictionary, all was well

checkdic("social")
[[1]]
[1] " " "media" "networks" "experience" "economic" "order" "good" "buying"

But if the word did not appear in the dictionary, there was an error message

checkdic("dog_social")

[1] "match found"

[1] "match found"

Show Traceback

Rerun with Debug

Error in word1 == word :

comparison (1) is possible only for atomic and list types

Thanks so much for your attention. What's going on now?

[woodward]

You really do have to pay attention to data types in R. Dataframes, lists, vectors, logical etc.

row.names(predict) is a vector of strings (possibly with length 0), so how can you compare it to > 0? nrow(predict) is probably what you want, as mentioned above.

The double square bracket shows that the nxtword1 function returns a list, and the first element [[1]] is a vector of strings.

However the less1gram function fails for words that are not in dat$word1 (which is a vector). You need to fix it to use %in% like you did with nxtword1, I suspect.

[Kill3rbee]

You seem to have multiple post. here is something I put together based on my understanding of your problem.

dat<- data.frame(word1=c( "will", "like","get", "look", "next", "social",
                          "cinco_de","manufacturer_custom", "custom_built"), 
                 word2=c(" ", " ", " ", "like","week", "media", "mayo", "built", "painted"), 
                 frequency = c( 5153, 5081, 4821, 559, 478,465, 172,171,171 ) )

word_exist <- function(word, df){
  ifelse(word %in% df$word1, "match found", "match not found")
}

word_exist("dog_social", dat)

[Quack]

Thank you for your attention. After reading your example, I fixed my code. This time it worked!
following_word<- function(phrase){match <-nxtword1(phrase)
if(nrow(match)>0){print(match)
} else{(nxtword1(less1gram(phrase)))
}
}