I want to create a list which is eight times the vector c(2,6), i.e. a list of 8 vectors.
WRONG: object = as.list(rep(c(2,6),8)) results instead in a list of 16 single numbers: 2 6 2 6 2 6 2 6 ...
I tried drop=0 but that didn't help, and I can't get lapply to work.
(My context:
I have a function in which a subfunction will call a numbered list object.
The number will be in a loop and therefore change, and the number and loop size is dependent on user values so I don't know what it'll be. If the user doesn't enter a list of vector values for one of the variables, I need to set a default.
The subfunction is expecting e.g. c(2,6)
The subfunction is currently looping 8 times so I need a list which is eight times c(2,6).
rep(list(c(2,6)),8) is the answer - thanks to Nicola in comments.
Related
I have a large and messy character vector html.
I also have a data frame url_characters:
start end
1 35288 35333
2 36723 36768
3 38168 38213
4 39647 39692
5 41091 41136
6 42549 42594
How can I create a loop in R that would perform substr function on character vector html using each row of the data frame for start and end arguments and store the results in a new character vector with each extraction as a separate item of the vector?
I am an absolute beginner so pardon me if the solution is an obvious one or already exists somewhere, I couldn't find it.
I watched video on YouTube re finding mode in R from list of numerics. When I enter commands they do not work. R does not even give an error message. The vector is
X <- c(1,2,2,2,3,4,5,6,7,8,9)
Then instructor says use
temp <- table(as.vector(x))
to basically sort all unique values in list. R should give me from this command 1,2,3,4,5,6,7,8,9 but nothing happens except when the instructor does it this list is given. Then he says to use command,
names(temp)[temp--max(temp)]
which basically should give me this: 1,3,1,1,1,1,1,1,1 where 3 shows that the mode is 2 because it is repeated 3 times in list. I would like to stay with these commands as far as is possible as the instructor explains them in detail. Am I doing a typo or something?
You're kind of confused.
X <- c(1,2,2,2,3,4,5,6,7,8,9) ## define vector
temp <- table(as.vector(X))
to basically sort all unique values in list.
That's not exactly what this command does (sort(unique(X)) would give a sorted vector of the unique values; note that in R, lists and vectors are different kinds of objects, it's best not to use the words interchangeably). What table() does is to count the number of instances of each unique value (in sorted order); also, as.vector() is redundant.
R should give me from this command 1,2,3,4,5,6,7,8,9 but nothing happens except when the instructor does it this list is given.
If you assign results to a variable, R doesn't print anything. If you want to see the value of a variable, type the variable's name by itself:
temp
you should see
1 2 3 4 5 6 7 8 9
1 3 1 1 1 1 1 1 1
the first row is the labels (unique values), the second is the counts.
Then he says to use command, names(temp)[temp--max(temp)] which basically should give me this: 1,3,1,1,1,1,1,1,1 where 3 shows that the mode is 2 because it is repeated 3 times in list.
No. You already have the sequence of counts stored in temp. You should have typed
names(temp)[temp==max(temp)]
(note =, not -) which should print
[1] "2"
i.e., this is the mode. The logic here is that temp==max(temp) gives you a logical vector (a vector of TRUE and FALSE values) that's only TRUE for the elements of temp that are equal to the maximum value; names(temp)[temp==max(temp)] selects the elements of the names vector (the first row shown in the printout of temp above) that correspond to TRUE values ...
I must first apologize as I have no programming background, so please forgive me if this question is overly simplistic or if it has been addressed repeatedly. I would be very willing to help clarify my issue if it is not clear from my explanation.
I have two sets of data matrices. "A":
[Ac1] [Ac2] ... [Ac500]
[Ac1] 25 30 ... 15
[Ar2] 7 54 ... 41
...
[cr25000]
and
"B" which is similar in the number of columns, but not the number of rows
[Bc1] [Bc2] ... [Bc500]
[Br1] 25 30 ... 15
[Br2] 7 54 ... 41
...
[Br20000]
I'm running an module ("npSeq") in R that uses the matrix A consistently as an input value, a horizontal vector that includes all of the values from a row in matrix B, ex [1]. The module returns a separate list of values. I will need to run the analysis independently for all of the rows in matrix B saving all of the returned lists which I will then need to combine.
However I would like to know if there is a way to automate the process so that the module runs using a vector derived from row [Br1], saves the returned list, and then runs the process again using the vector derived from row [Br2]. Repeating the process until [Br20000].
Again I'm sorry that this is worded so poorly. I wish I understood enough of the terminology to state my problem more clearly.
You can use lapply to loop over B's row indices:
result.list <- lapply(1:nrow(B), function(i) npSeq(A, B[i, ]))
Note that this is not going to be much (any?) faster than using a for loop. It is just a short and clean equivalent. 20,000 iterations does sound like a lot so it may take a while depending on how slow the function is.
I have a vector of binary variables which state whether a product is on promotion in the period. I'm trying to work out how to calculate the duration of each promotion and the duration between promotions.
promo.flag = c(1,1,0,1,0,0,1,1,1,0,1,1,0))
So in other words: if promo.flag is same as previous period then running.total + 1, else running.total is reset to 1
I've tried playing with apply functions and cumsum but can't manage to get the conditional reset of running total working :-(
The output I need is:
promo.flag = c(1,1,0,1,0,0,1,1,1,0,1,1,0)
rolling.sum = c(1,2,1,1,1,2,1,2,3,1,1,2,0)
Can anybody shed any light on how to achieve this in R?
It sounds like you need run length encoding (via the rle command in base R).
unlist(sapply(rle(promo.flag)$lengths,seq))
Gives you a vector 1 2 1 1 1 2 1 2 3 1 1 2 1. Not sure what you're going for with the zero at the end, but I assume it's a terminal condition and easy to change after the fact.
This works because rle() returns a list of two, one of which is named lengths and contains a compact sequence of how many times each is repeated. Then seq when fed a single integer gives you a sequence from 1 to that number. Then apply repeatedly calls seq with the single numbers in rle()$lengths, generating a list of the mini sequences. unlist then turns that list into a vector.
I try to compare multiple vectors of Entrez IDs (integer vectors) by using Reduce(intersect,...). The vectors are selected from a database using "DISTINCT" so a single vector does not contain duplicates.
length(factor(c(l1$entrez)))
gives the same length (and the same IDs w/o the length function) as
length(c(l1$entrez))
When I compare multiple vectors with
length(Reduce(intersect,list(c(l1$entrez),c(l2$entrez),c(l3$entrez),c(l4$entrez))))
or
length(Reduce(intersect,list(c(factor(l1$entrez)),c(factor(l2$entrez)),c(factor(l3$entrez)),c(factor(l4$entrez)))))
the result is not the same. I know that factor!=originalVector but I cannot understand why the result differs although the length and the levels of the initial factors/vectors are the same.
Could somebody please explain the different behaviour of the intersect function on vectors and factors? Is it that the intersect of two factor lists are again factorlists and then duplicates are treated differently?
Edit - Example:
> head(l1)
entrez
1 1
2 503538
3 29974
4 87769
5 2
6 144568
> head(l2)
entrez
1 1743
2 1188
3 8915
4 7412
5 51082
6 5538
The lists contain around 500 to 20K Entrez IDs. So the vectors contain pure integer and should give the intersect among all tested vectors.
> length(Reduce(intersect,list(c(factor(l1$entrez)),c(factor(l2$entrez)),c(factor(l3$entrez)),c(factor(l4$entrez)))))
[1] 514
> length(Reduce(intersect,list(c(l1$entrez),c(l2$entrez),c(l3$entrez),c(l4$entrez))))
[1] 338
> length(Reduce(intersect,list(l1$entrez,l2$entrez,l3$entrez,l4$entrez)))
[1] 494
I have to apologize profusely. The different behaviour of the intersect function may be caused by a problem with the data. I have found fields in the dataset containing comma seperated Entrez IDs (22038, 23207, ...). I should have had a more detailed look at the data first. Thank you for the answers and your time. Although I do not understand the different results yet, I am sure that this is the cause of the different behaviour. Can somebody confirm that?
As Roman says, an example would be very helpful.
Nevertheless, one possibility is that your variables l1$entrez, l2$entrez etc have the same levels but in different orders.
intersect converts its arguments via as.vector, which turns factors into character variables. This is usually the right thing to do, as it means that varying level order doesn't make any difference to the result.
Passing factor(l1$entrez) as an argument to intersect also removes the impact of varying level order, as it effectively creates a new factor with level ordering set to the default. However, if you pass c(l1$entrez), you strip the factor attributes off your variable and what you're left with is the raw integer codes which will depend on level ordering.
Example:
a <- factor(letters[1:3], levels=letters)
b <- factor(letters[1:3], levels=rev(letters)
# returns 1 2 3
intersect(c(factor(a)), c(factor(b)))
# returns integer(0)
intersect(c(a), c(b))
I don't see any reason why you should use c() in here. Just let R handle factors by itself (although to be fair, there are other scenarios where you do want to step in).