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.
Related
I have a code I'm working with which has the following line,
data2 <- apply(data1[,-c(1:(index-1))],2,log)
I understand that this creates a new data frame, from the data1, taking column-wise values log-transformed and some columns are eliminated, but I don't understand how the columns are removed. what does 1:(index-1) do exactly?
The ":" operator creates an integer sequence. Because (1:(index-1) ) is numeric and being used in the second position for the extraction operator"[" applied to a dataframe, it is is referring to column numbers. The person writing the code didn't need the c-function. It could have been more economically written:
data1[,-(1:(index-1))]
# but the outer "("...")"'s are needed so it starts at 1 rather than -1
So it removes the first index-1 columns from the object passed to apply. (As MrFlick points out, index must have been defined before this gets passed to R. There's not default value or interpretation for index in R.
Suppose the index is 5, then index -1 returns 4 so the sequence will be from 1 to 4 i.e. and then we use - implies loop over the columns other than the first 4 columns as MARGIN = 2
The problem
I would like to find a length of a list.
The expected output
I would like to find the length based on a condition.
Example
Suppose that I have a list of 4 elements as follows:
myve <–list(1,2,3,0)
Here I have 4 elements, one of them is zero. How can I find the length by extracting the zero values? Then, if the length is > 1I would like to substruct one. That is:
If the length is 4 then, I would like to have 4-1=3. So, the output should be 3.
Note
Please note that I am working with a problem where the zero values may be changed from one case to another. For example, For the first list may I have only one 0 value, while for the second list may I have 2 or 3 zero values.
The values are always positive or zero.
You just need to apply the condition to each element. This will produce a list of boolean, then you sum it to get the number of True elements (i.e. validation your condition).
In your case:
sum(myve != 0)
In a more complex case, where the confition is expressed by a function f:
sapply(myve, f)
Use sapply to extract the ones different to zeros and sum to count them
sum(sapply(myve, function(x) x!=0))
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.
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)
Hello I am new to R and I can't find the way to do exactly what I want to. I have a vector of x numbers, and what i want to do is order it in increasing order, and then start making subtractions like this (let's say the vecto has 100 numbers for example):
[x(100)-x(99)]+[x(99)-x(98)]+[x(98)-x(97)]+[x(97)-x(96)]+...[x(2)-x(1)]
and then divide all that sum by the number of elements the vector has, in this case 100.
The only thing that I am able to do at the moment is order the vector with:
sort(nameOfTheVector)
Sorry for my bad English.
diff returns suitably lagged and iterated differences. In your case you want the default single lag. sum will return the sum any arguments passed to it, so....
sum(diff(sort(nameOfTheVector))) / length(nameOfTheVector)