I have a dataset named bwght which contains the variable cigs (cigarattes smoked per day)
When I calculate the mean of cigs in the dataset bwght using:
mean(bwght$cigs), I get a number 2.08.
Only 212 of the 1388 women in the sample smoke (and 1176 does not smoke):
summary(bwght$cigs>0) gives the result:
Mode FALSE TRUE NA's
logical 1176 212 0
I'm asked to find the average of cigs among the women who smoke (the 212).
I'm having a hard time finding the right syntax for excluding the non smokers = 0
I have tried:
mean(bwght$cigs| bwght$cigs>0)
mean(bwght$cigs>0 | bwght$cigs=TRUE)
if (bwght$cigs > 0){
sum(bwght$cigs)
}
x <-as.numeric(bwght$cigs, rm="0");
mean(x)
But nothing seems to work! Can anyone please help me??
If you want to exclude the non-smokers, you have a few options. The easiest is probably this:
mean(bwght[bwght$cigs>0,"cigs"])
With a data frame, the first variable is the row and the next is the column. So, you can subset using dataframe[1,2] to get the first row, second column. You can also use logic in the row selection. By using bwght$cigs>0 as the first element, you are subsetting to only have the rows where cigs is not zero.
Your other ones didn't work for the following reasons:
mean(bwght$cigs| bwght$cigs>0)
This is effectively a logical comparison. You're asking for the TRUE / FALSE result of bwght$cigs OR bwght$cigs>0, and then taking the mean on it. I'm not totally sure, but I think R can't even take data typed as logical for the mean() function.
mean(bwght$cigs>0 | bwght$cigs=TRUE)
Same problem. You use the | sign, which returns a logical, and R is trying to take the mean of logicals.
if(bwght$cigs > 0){sum(bwght$cigs)}
By any chance, were you a SAS programmer originally? This looks like how I used to type at first. Basically, if() doesn't work the same way in R as it does in SAS. In that example, you are using bwght$cigs > 0 as the if condition, which won't work because R will only look at the first element of the vector resulting from bwght$cigs > 0. R handles looping differently from SAS - check out functions like lapply, tapply, and so on.
x <-as.numeric(bwght$cigs, rm="0")
mean(x)
I honestly don't know what this would do. It might work if rm="0" didn't have quotes...?
mean(bwght[bwght$cigs>0,"cigs"])
I found the statement failed, returning "argument is not numeric or logical: returning NA"
Converting to matrix solved this:
mean(data.matrix(bwght[bwght$cigs>0,"cigs"]))
Related
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.
Excuse the horrible title. I don't think I'd be able to summarise this problem in such few words.
So I have a table in R with data whose proportions by row have been calculated using the prop.table function ( prop.table(tab, 1) ). It looks like this:
The row headings (i.e. Q1-00-05, etc.) denote times of the day. The column headings TRUE and FALSE denote whether a particular 999 call was responded to within 10 minutes.
What I need from this table is the proportion of 999 calls responded to efficiently (< 10 mins) between 1800hrs and 0500 hrs.
I tried doing this:
tab2<-table(callouts$daytime=="Q4-18-23"|"Q1-00-05", callouts$tenmins)
but this proved fruitless. I got an error message saying:
operations are possible only for numeric, logical or complex types
I expected the table to come out with TRUE or FALSE as the row headings (for whether the callout time was within this time frame or not) and TRUE or FALSE as the column headings (for whether the response time was sub-10mins)
Any help would be much appreciated. Thanks!
I have a dataframe, and I want to confirm that two columns match for each entry. So I tried:
> nrow(subset(df, col.a!=col.b))
[1] 0
That seemed good to me, but then I tried to compare how many matches there were to the total number of entries in the data frame. It seems like these numbers should be equal but they are not:
nrow(subset(df, col.a==col.b))
[1] 3443
nrow(df)
[1] 3453
Any idea what is going on here? Why does it looked like the subset dropped 10 entries? Thanks so much for your help.
Also, I'm fairly new to this, so please let me know if there is a better way of checking if the two columns match.
subset automatically drops rows where the criterion is NA. It should always (?) be the case that
nrow(d)
and
nrow(subset(d, col.a!=col.b))+
nrow(subset(d, col.a==col.b))+
nrow(subset(d, is.na(col.a) | is.na(col.b)))
should be equal.
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).
I have a dataset named bwght which contains the variable cigs (cigarattes smoked per day)
When I calculate the mean of cigs in the dataset bwght using:
mean(bwght$cigs), I get a number 2.08.
Only 212 of the 1388 women in the sample smoke (and 1176 does not smoke):
summary(bwght$cigs>0) gives the result:
Mode FALSE TRUE NA's
logical 1176 212 0
I'm asked to find the average of cigs among the women who smoke (the 212).
I'm having a hard time finding the right syntax for excluding the non smokers = 0
I have tried:
mean(bwght$cigs| bwght$cigs>0)
mean(bwght$cigs>0 | bwght$cigs=TRUE)
if (bwght$cigs > 0){
sum(bwght$cigs)
}
x <-as.numeric(bwght$cigs, rm="0");
mean(x)
But nothing seems to work! Can anyone please help me??
If you want to exclude the non-smokers, you have a few options. The easiest is probably this:
mean(bwght[bwght$cigs>0,"cigs"])
With a data frame, the first variable is the row and the next is the column. So, you can subset using dataframe[1,2] to get the first row, second column. You can also use logic in the row selection. By using bwght$cigs>0 as the first element, you are subsetting to only have the rows where cigs is not zero.
Your other ones didn't work for the following reasons:
mean(bwght$cigs| bwght$cigs>0)
This is effectively a logical comparison. You're asking for the TRUE / FALSE result of bwght$cigs OR bwght$cigs>0, and then taking the mean on it. I'm not totally sure, but I think R can't even take data typed as logical for the mean() function.
mean(bwght$cigs>0 | bwght$cigs=TRUE)
Same problem. You use the | sign, which returns a logical, and R is trying to take the mean of logicals.
if(bwght$cigs > 0){sum(bwght$cigs)}
By any chance, were you a SAS programmer originally? This looks like how I used to type at first. Basically, if() doesn't work the same way in R as it does in SAS. In that example, you are using bwght$cigs > 0 as the if condition, which won't work because R will only look at the first element of the vector resulting from bwght$cigs > 0. R handles looping differently from SAS - check out functions like lapply, tapply, and so on.
x <-as.numeric(bwght$cigs, rm="0")
mean(x)
I honestly don't know what this would do. It might work if rm="0" didn't have quotes...?
mean(bwght[bwght$cigs>0,"cigs"])
I found the statement failed, returning "argument is not numeric or logical: returning NA"
Converting to matrix solved this:
mean(data.matrix(bwght[bwght$cigs>0,"cigs"]))