I'm struggling to remove a row in a matrix, where this matrix's name is "unknown". What I mean by "unknown" is that there are several matrices, and the last 3 characters of each matrix's name is different.
An example would make this a lot clearer I think.
Say I have 3 matrices, Trades_ABC, Trades_DEF, Trades_HIJ. Each of these matrices has x rows and 5 columns.
I currently have the following code:
for (k in 1:3)
assign(get(paste0("Trades_",sellLeg))[1,1],y)
next k
Where "sellLeg" is one of "ABC","DEF","HIJ"
In this code I am trying to change the value of the first element in each of the three matrices to some number, represented by "1", as an example. In reality, I'm not so much looking to CHANGE a value as I am looking to REMOVE a row, but my main problem is that I don't know how to assign a value to a matrix with an "unknown" name (once I can do this I should be able to remove a row)
Many thanks!
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
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.
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 have a dataframe ma
it has a factor called type
type is comprised of the following factors: I210, I210plus, I210plusc, KV2c, KV2cplus
I'd like to put some of these factors in a vector, say, selected_types
so, selected_types<-c("I210plusc","KV2c")
then, have this command subset the dataframe ma
ma1<-subset(ma, type==selected_types)
such that ma1 would be a subset of ma consisting of only the observations that had
type I210plusc and KV2c
however, when I do this, the number of observations in the resulting dataframe ma1 is less than the sum of the occurrences of the two types in selected_types from the original ma
Any ideas on what I'm doing incorrectly?
Thank you
I originally had this in a comment, but it's a bit lengthy, plus I wanted to add to it. Here some details on what's happening:
what you're doing with == is recycling your two length vector, so that every even row is compared to "KV2c", and every odd one to "I210plusc", so your final result will be the data frame of odd rows that are "KV2c" and even rows that are "I210plusc".
An alternate solution that might make the issue clear is as follows:
subset(ma, type == selected_types[[1]] | type == selected_types[[2]])
Or, more gracefully:
subset(ma, type %in% selected_types)
The %in% operator returns a logical vector of same length as type with TRUE for every position in type that "is in" selected_types (hence the name of the operator).
Given data like this
C1<-c(3,-999.000,4,4,5)
C2<-c(3,7,3,4,5)
C3<-c(5,4,3,6,-999.000)
DF<-data.frame(ID=c("A","B","C","D","E"),C1=C1,C2=C2,C3=C3)
How do I go about removing the -999.000 data in all of the columns
I know this works per column
DF2<-DF[!(DF$C1==-999.000 | DF$C2==-999.000 | DF$C3==-999.000),]
But I'd like to avoid referencing each column. I am thinking there is an easy way to reference all of the columns in a particular data frame aka:
DF3<-DF[!(DF[,]==-999.000),]
or
DF3<-DF[!(DF[,(2:4)]==-999.000),]
but obviously these do not work
And out of curiosity, bonus points if you can me why I need that last comma before the ending square bracket as in:
==-999.000),]
The following may work
DF[!apply(DF==-999,1,sum),]
or if you can have multiple -999 on a row
DF[!(apply(DF==-999,1,sum)>0),]
or
DF[!apply(DF==-999,1,any),]
Based on your code, I'll assume that you want to remove all rows that contain -999.
DF2 <- DF[rowSums(DF == -999) == 0, ]
As for your bonus question: A data frame is a list of vectors, all of which have the same length. If we think of the vectors as columns, then a data frame can be thought of as a matrix where the columns might have different types (numeric, character, etc). R allows you to refer to elements of a data frame much the same way you refer to elements of a matrix; by using row and column indices. So DF[i, j] refers to the ith element in the jth vector of DF, which you can think of as the ith row and jth column. So if you want to retain only some of the rows of the data frame and all columns, you can use a matrix-like notation: DF[row.indices, ].
To address your "bonus" question, if we go to the documentation for ?Extract.data.frame we will find:
Data frames can be indexed in several modes. When [ and [[ are used
with a single index (x[i] or x[[i]]), they index the data frame as if
it were a list. In this usage a drop argument is ignored, with a
warning.
and also:
When [ and [[ are used with two indices (x[i, j] and x[[i, j]]) they
act like indexing a matrix: [[ can only be used to select one element.
Note that for each selected column, xj say, typically (if it is not
matrix-like), the resulting column will be xj[i], and hence rely on
the corresponding [ method, see the examples section.
So you need the comma to ensure that R knows you are referring to a row, not a column.
I don't understand if your target is to remove all the rows that contain at least one NA, if this is what you are looking for, then this could be a possible answer:
DF[DF==-999] <- NA
na.omit(DF)
ID C1 C2 C3
1 A 3 3 5
3 C 4 3 3
4 D 4 4 6