Find the number of elements in a list

Question

I am working with a data set that has a column which I have turned into a list that looks like this:

[[997]]
[1] NA

[[998]]
[1] "0.99"  " 0.99" " 0.99" " 0.99" " 0.99" " 2.99"

[[999]]
[1] "19.99"  " 9.99"  " 4.99"  " 79.99" " 15.99" " 14.99" " 99.99" " 7.99" 
[9] " 59.99" " 49.99"

[[1000]]
[1] NA

I need to count how many components are in each element of a list and save these values in a matrix. For example, in 999 the output should be 10 and in 1000 the output should be 0. Can anyone suggest a function that could help? Thank you.

BTW, your title does not reflect your question. You are not counting for character, you want to count for number of elements in a list. — dc37, Dec 01 '19 at 21:03

dc37 · Accepted Answer · 2019-12-01T21:17:08.943

3

Here, using a small example of list v:

v = vector("list",4)
v[[1]] = 1:5
v[[2]] = 1:50
v[[3]] = NA

> v
[[1]]
[1] 1 2 3 4 5

[[2]]
 [1]  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
[37] 37 38 39 40 41 42 43 44 45 46 47 48 49 50

[[3]]
[1] NA

[[4]]
NULL

And you get the count by doing:

l = unlist(lapply(v,length))

and

> l
[1]  5 50  1  0

If you don't want to count for NA

l = unlist(lapply(v,function(x)length(x[!is.na(x)])))

and you get:

> l
[1]  5 50  0  0

EDIT: from @markus and @A5C1D2H2I1M1N2O1R2T1 comments

As mentioned by @markus, you can go much much simpler by doing:

> lengths(v)
[1]  5 50  1  0

And as mentioned by @A5C1D2H2I1M1N2O1R2T1, you can get rid of NA count by doing:

> replace(lengths(v), is.na(v), 0)
[1]  5 50  0  0

edited Dec 01 '19 at 21:17

answered Dec 01 '19 at 20:55

dc37

15,840
4
15
32

Great, Thank you! I was struggling with not counting for NA's. This was very helpful! – H. Minear Dec 01 '19 at 21:03
2

Better use `lengths(v)` – markus Dec 01 '19 at 21:07
OMG !! I didn't know about this function ! Thank you @markus. I will edit my answer to add it. – dc37 Dec 01 '19 at 21:08
as pointed out by @user20650, how you get rid of the count for NA ? – dc37 Dec 01 '19 at 21:12
2

@dc37, `replace(lengths(v), is.na(v), 0)`. – A5C1D2H2I1M1N2O1R2T1 Dec 01 '19 at 21:13
@A5C1D2H2I1M1N2O1R2T1, thanks I'm editing my answer – dc37 Dec 01 '19 at 21:16

ThomasIsCoding · Answer 2 · 2019-12-02T07:04:37.700

2

Additional to the existing solution by @dc37, you can also use the following with lengths()

lengths(v)

If you don't want to take NA into account, then try

lengths(v)*!is.na(v)

edited Dec 02 '19 at 07:04

answered Dec 01 '19 at 21:08

ThomasIsCoding

96,636
9
24
81

but this gives a length of one for the NA values – user20650 Dec 01 '19 at 21:09
@user20650 thanks, please see my updated solution – ThomasIsCoding Dec 01 '19 at 21:14
thanks, I dont think that will work **if** the aim is to count the non NA elements. Try using dc's example but with `v[[1]] = c(1:5, NA)` – user20650 Dec 01 '19 at 21:15
@user20650, I think the OP needs to clarify what they would want in that case. – A5C1D2H2I1M1N2O1R2T1 Dec 01 '19 at 21:17
@A5C1D2H2I1M1N2O1R2T1 ; yes fair point – user20650 Dec 01 '19 at 21:18
@user20650, in this particular case of mix NAs and elements, the `unlist(lapply(v,function(x)length(x[!is.na(x)])))` will work – dc37 Dec 01 '19 at 21:19
@user20650 well....`rapply` does not work, but `lengths` is still available. I am using dc's example – ThomasIsCoding Dec 01 '19 at 21:19

Find the number of elements in a list

2 Answers2