I am trying to create variables based on a list I have.
The list looks something like this:
color = c("blue","green","yellow")
Each string in the list will become a variable. The variable should take the values based on another column (for example, usercolorlist). Here is the pseudocode:
For each row in usercolorlist
if usercolorlist contains blue
then blue = 1
else 0
Ultimately, the output would be:
usercolorlist blue
"blue/red/green" 1
"red/green" 0
"blue/red" 1
I want to implement this as cleanly as possible. I mainly use python and have been told that for loops are not as efficient in R.
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
I found an interesting data-set from a psychology study (data-set is called WearingTShirt), and I would like to replicate the results. I would need to summarize two variables into a single variable. This is what I have written:
Create empty variable
PinkAndRed = 0
Count instances of people wearing both pink and red and add 1
for i in WearingTShirt:
PinkAndRed+1 if:
WearingTShirt$PINKSHIRT==1 OR WearingTShirt$REDSHIRT==1
Add variable to dataset
WearingTShirt$PinkAndRed
I have not much R experience (I wrote mostly in Python).
Your code is more in python than in R. The equivalent code in R for what you want to do is:
PinkAndRed = rep(0,dim(WearingTShirt)[1])
for(i in 1:dim(WearingTShirt)[1]){
if((WearingTShirt$PINKSHIRT[i]==1) || (WearingTShirt$REDSHIRT[i]==1))
{
PinkAndRed[i] = 1
}
}
WearingTShirt=cbind(WearingTShirt,PinkAndRed)
You need to review basics on R. There are countless small difference between R and python, such as parenthesis in loops or conditions, set the length of a loop (in the above code with dim you calculate the dimension of the dataset and by doing [1] you indicate that you want the number of rows)...
Update:
thanks to the comments i've realized that is not clear if you want a cumulative sum of the individuals with pink and red shirts or a variable which is 1 with the shirt is pink or red, and 0 in other case.
The code above is for a varaible that includes pink and red shirts in one variable.
If you want the sum you must use cumsum function as it's said in the comments
I would not choose to loop, but:
WearingTShirt$PinkAndRed <- ifelse(WearingTShirt$PINKSHIRT==1 |
WearingTShirt$REDSHIRT==1,1,0)
PinkAndRed sounds more like PinkOrRed based on example given.
I'm new here and diving into R, and I'm encountering a problem while trying to solve a knapsack problem.
For optimization purposes I wrote a dynamic program in R, however, now that I am at the point of returning the items, which I succeeded in, I only get the binary numbers saying whether the item has been selected or not (1 = yes). Like this:
Select
[1] 1 0 0 1
However, now I would like the Select function to return the names of values instead of these binary values. Underneath I created an example of what my problem looks like.
This would be the data and a related data frame.
items <- c("Glasses","gloves","shoes")
grams <- c(4,2,3)
value <- c(100,20,50)
data <- data.frame(items,grams,value)
Now, I created various functions, with the final one clarifying whether a product has been selected by 1 (yes) or 0 (no). Like above. However, I would really like for it to return the related name of the item. Is there a manner to go around this by linking back to the dataframe created?
So that it would say instead of (in case all products are selected)
Select
[1] 1 1 1
Select
[1] Glasses gloves shoes
I believe I would have to create a new function. But as I mentioned, is there a good way to refer back to the data frame to take related values from another column in the data frame in case of a 1 (yes)?
I really hope my question is more clear now and someone can direct me in the right direction.
Best, Berber
Lets say your binary vector is
idx <- [1, 0, 1, 0, 1]
just use,
items[as.logical(idx)]
will give you the name for selected items, and
items[!as.logical(idx)]
will give you name for unselected items
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 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.