I am trying to count the length of occurrances of a value in a vector such as
q <- c(1,1,1,1,1,1,4,4,4,4,4,4,4,4,4,4,4,4,6,6,6,6,6,6,6,6,6,6,1,1,4,4,4)
Actual vectors are longer than this, and are time based. What I would like would be an output for 4 that tells me it occurred for 12 time steps (before the vector changes to 6) and then 3 time steps. (Not that it occurred 15 times total).
Currently my ideas to do this are pretty inefficient (a loop that looks element by element that I can have stop when it doesn't equal the value I specified). Can anyone recommend a more efficient method?
x <- with(rle(q), data.frame(values, lengths)) will pull the information that you want (courtesy of d.b. in the comments).
From the R Documentation: rle is used to "Compute the lengths and values of runs of equal values in a vector – or the reverse operation."
y <- x[x$values == 4, ] will subset the data frame to include only the value of interest (4). You can then see clearly that 4 ran for 12 times and then later for 3.
Modifying the code will let you check whatever value you want.
Related
I'm working with a 4-dimensional matrix (Year, Simulation, Flow, Time instant: 10x5x20x10) in R. I need to remove some values from the matrix. For example, for year 1 I need to remove simulations number 1 and 2; for year 2 I need to remove simulation number 5.
Can anyone suggest me how I can make such changes?
Arrays (which is how R documentation usually refers to higher-dimensional 'matrices') can be indexed with negative values in the same way as matrices or vectors: a negative value removes the corresponding row/column/slice. So if you wanted to remove year 1 completely (for example), you could use a[-1,,,]; to remove simulation 5 completely, a[,-5,,].
However, arrays can't be "ragged", there has to be something in every row/column/slice combination. You could replace the values you want to remove with NAs (and then make sure to account for the NAs appropriately when computing, e.g. using na.rm = TRUE in sum()/min()/max()/median()/etc.): a[1,1:2,,] <- NA or a[2,5,,] <- NA in your examples.
If you knew that all values of Flow and Time would always be present, you could store your data as a list of lists of matrices: e.g.
results <- list(Year1 = list(Simulation1 = matrix(...),
Simulation2 = matrix(...),
...),
Year2 = list(Simulation1 = matrix(...),
Simulation2 = matrix(...),
...))
Then you could easily remove years or simulations within years (by setting them to NULL, but it would make indexing a little bit harder (e.g. "retrieve Simulation1 values for all years" would require an lapply or a loop across years).
cv.uk.df$new.d[2:nrow(cv.uk.df)] <- tail(cv.uk.df$deaths, -1) - head(cv.uk.df$deaths, -1) # this line of code works
I wanted to know why do we -1 in the tail and -1 in head to create this new column.
I made an effort to understand by removing the -1 and "R"(The code is in R studio) throws me this error.
Could anyone shed some light on this? I can't explain how much I would appreciate it.
Look at what is being done. On the left-hand side of the assignment operator, we have:
cv.uk.df$new.d[2:nrow(cv.uk.df)] <-
Let's pick this apart.
cv.uk.df # This is the data.frame
$new.d # a new column to assign or a column to reassign
[2:nrow(cv.uk.df)] # the rows which we are going to assign
Specifically, this line of code will assign a new value all rows of this column except the first. Why would we want to do that? We don't have your data, but from your example, it looks like you want to calculate the change from one line to the next. That calculation is invalid for the first row (no previous row).
Now let's look at the right-hand side.
<- tail(cv.uk.df$deaths, -1) - head(cv.uk.df$deaths, -1)
The cv.uk.df$deaths column has the same number of rows as the data.frame. R gets grouchy when the numbers of elements don't follow sum rules. For data.frames, the right-hand side needs to have the same number of elements, or a number that can be recycled a whole-number of times. For example, if you have 10 rows, you need to have a replacement of 10 values. Or you can have 5 values that R will recycle.
If your data.frame has 100 rows, only 99 are being replaced in this operation. You cannot feed 100 values into an operation that expects 99. We need to trim the data. Let's look at what is happening. The tail() function has the usage tail(x, n), where it returns the last n values of x. If n is a negative integer, tail() returns all values but the first n. The head() function works similarly.
tail(cv.uk.df$deaths, -1) # This returns all values but the first
head(cv.uk.df$deaths, -1) # This returns all values but the last
This makes sense for your calculation. You cannot subtract the number of deaths in the row before the first row from the number in the first row, nor can you subtract the number of deaths in the last row from the number in the row after the last row. There are more intuitive ways to do this thing using functions from other packages, but this gets the job done.
Lets say I have this data. My objective is to extraxt combinations of sequences.
I have one constraint, the time between two events may not be more than 5, lets call this maxGap.
User <- c(rep(1,3)) # One users
Event <- c("C","B","C") # Say this is random events could be anything from LETTERS[1:4]
Time <- c(c(1,12,13)) # This is a timeline
df <- data.frame(User=User,
Event=Event,
Time=Time)
If want to use these sequences as binary explanatory variables for analysis.
Given this dataframe the result should be like this.
res.df <- data.frame(User=1,
C=1,
B=1,
CB=0,
BC=1,
CBC=0)
(CB) and (CBC) will be 0 since the maxGap > 5.
I was trying to write a function for this using many for-loops, but it becomes very complex if the sequence becomes larger and the different number of evets also becomes larger. And also if the number of different User grows to 100 000.
Is it possible of doing this in TraMineR with the help of seqeconstraint?
Here is how you would do that with TraMineR
df.seqe <- seqecreate(id=df$User, timestamp=df$Time, event=df$Event)
constr <- seqeconstraint(maxGap=5)
subseq <- seqefsub(df.seqe, minSupport=0, constraint=constr)
(presence <- seqeapplysub(subseq, method="presence"))
which gives
(B) (B)-(C) (C)
1-(C)-11-(B)-1-(C) 1 1 1
presence is a table with a column for each subsequence that occurs at least once in the data set. So, if you have several individuals (event sequences), the table will have one row per individual and the columns will be the binary variable you are looking for. (See also TraMineR: Can I get the complete sequence if I give an event sub sequence? )
However, be aware that TraMineR works fine only with subsequences of length up to about 4 or 5. We suggest to set maxK=3 or 4 in seqefsub. The number of individuals should not be a problem, nor should the number of different possible events (the alphabet) as long as you restrict the maximal subsequence length you are looking for.
Hope this helps
I would like to create a function that looks at a column of values. from those values look at each value individually, and asses which of the other data points value is closest to that data point.
I'm guessing it could be done by checking the length of the data frame, making a list of the respective length in steps of 1. Then use that list to reference which cell is being analysed against the rest of the column. though I don't know how to implement that.
eg.
data:
20
17
29
33
1) is closest to 2)
2) is closest to 1)
3) is closest to 4)
4) is closest to 3)
I found this example which tests for similarity but id like to know what letter is assigns to.
x=c(1:100)
your.number=5.43
which(abs(x-your.number)==min(abs(x-your.number)))
Also if you know how I could do this, could you expain the parts of the code and what they mean?
I wrote a quick function that does the same thing as the code you provided.
The code you provided takes the absolute value of the difference between your number and each value in the vector, and compares that the minimum value from that vector. This is the same as the which.min function that I use below. I go through my steps below. Hope this helps.
Make up some data
a = 1:100
yourNumber = 6
Where Num is your number, and x is a vector
getClosest=function(x, Num){
return(which.min(abs(x-Num)))
}
Then if you run this command, it should return the index for the value of the vector that corresponds to the closest value to your specified number.
getClosest(x=a, Num=yourNumber)
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.