16

I'm trying to check if a specific value is anywhere in a data frame.

I know the %in% operator should allow me to do this, but it doesn't seem to work the way I would expect when applying to a whole data frame:

A = data.frame(B=c(1,2,3,4), C=c(5,6,7,8))
1 %in% A

[1] FALSE

But if I apply this to the specific column the value is in it works the way I expect:

1 %in% A$C

[1] TRUE

What is the proper way of checking if a value is anywhere in a data frame?

Kewl
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    `1 %in% as.matrix(A)` ? Which, of course, makes sense if all the columns of A are numbers... – digEmAll Apr 05 '17 at 14:17
  • @digEmAll, `"1" %in% as.character(as.matrix(A))` may work in all cases but I don't know how much better that is. – d.b Apr 05 '17 at 14:46

6 Answers6

25

You could do:

any(A==1)
#[1] TRUE

OR with Reduce:

Reduce("|", A==1)

OR

length(which(A==1))>0

OR

is.element(1,unlist(A))
989
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    Although all these work, only the first of them makes it clear what the code does. –  Apr 06 '17 at 10:31
10

To find the location of that value you can do f.ex:

which(A == 1, arr.ind=TRUE)
#     row col
#[1,]   1   1
AkselA
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8

Or simply

sum(A == 1) > 0
#[1] TRUE
Sotos
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6

Loop through the variables with sapply, then use any.

any(sapply(A, function(x) 1 %in% x))
[1] TRUE

or following digEmAll's comment, you could use unlist, which takes a list (data.frame) and returns a vector.

1 %in% unlist(A)
[1] TRUE
lmo
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3

The trick to understanding why your first attempt doesn't work, really comes down to understanding what a data frame is - namely a list of vectors of equal length. What you're trying to do here is not check if that list of vectors matches your condition, but checking if the values in those vectors matches the condition.

WHoekstra
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1

Try:

any(A == 1)

Returns FALSE or TRUE

gonzalez.ivan90
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