d1 <- data.frame("char"=c("a", "b", "c", "o", "m", "p"))
print("a" %in% d1)

I don't know why this code is "FALSE".

CodePudding user response：

That returns false because d1 is a data frame - and a data frame is a list that contains columns. The %in% method as you've written it "a" %in% d1 is testing if the value "a" is a column of d1, which it is not.

You can use "a" %in% d1$char to look in a specific column. You could use "a" %in% unlist(d1) to look in all the columns. You can use "a" == d1 to compare a to every individual value in d1, and wrap that in any() as any("a" == d1) to also get a single TRUE result.

Things can get a little strange depending of if the data frame has multiple rows and if the vector you're testing has multiple values:

## 1 value in 1-row data frame works
"a" %in% data.frame(x = "a", y = "b")
# [1] TRUE

## With two values to test, we're still looking for individual columns
c("a", "b") %in% data.frame(x = c("a", "b"))
# [1] FALSE FALSE
## This is FALSE FALSE because there isn't a column that the whole
## column is "a", nor is there a column that the whole column is "b"

## Both "a" and "b" are whole columns here:
c("a", "b") %in% data.frame(x = "a", y = "b")
# [1] TRUE TRUE
## So we get TRUE TRUE

## Here's the way to test a whole column with more than 1 value:
list(c("a", "b")) %in% data.frame(x = c("a", "b"))
[1] TRUE

CodePudding user response：

The statement inside your print() function is a logic test.

If you want to print the contents of your "char" column, you could try this:

d1 <- data.frame("char"=c("a", "b", "c", "o", "m", "p"))
print(d1$char)

Extra tips:

• You can change "char"to just char - the dataframe will still name your column char
• you don't need to use print()at all in this instance, just type d1$char