np2000 - 29 days ago 10

R Question

I have a list with various symbols for which I want to create a column and rank a certain column relative to the rest of the list.

For example, I have a list

`x`

`SPY`

`IWM`

`rsi`

`rsi`

`SPY`

`IWM`

I always get a rank of 1, which cannot be correct, so something has to be wrong in my code. As I said, I need the rank of

`rsi`

`library(quantmod)`

stockData <- new.env()

symbols = c("IWM","SPY")

getSymbols(symbols, src='yahoo',from = "2016-10-01",to = Sys.Date())

x <- list()

for (i in 1:length(symbols)) {

x[[i]] <- get(symbols[i], pos=stockData) # get data from stockData environment

x[[i]]$rsi <-RSI(Cl(x[[i]]),14)

x[[i]]$rank <- NA

x[[i]]$rank<-apply(-x[[i]]$rsi,1,rank)

}

Answer

```
library(quantmod)
stockData <- new.env()
symbols = c("IWM","SPY")
getSymbols(symbols, src='yahoo',from = "2016-10-01",to = Sys.Date())
fulldata <- lapply(symbols, get, pos = stockData)
closedata <- lapply(fulldata, Cl)
rsi <- lapply(closedata, RSI, n = 14) # or e.g. n = 2, if RSI based on two periods
```

In order to use rank later on, we need to transform the data to a data.frame, because `rank()`

doesn't like the class of the output that `RSI()`

gives.

```
rsi <- lapply(rsi, as.data.frame)
```

`RSI()`

depends on a moving average of 14 periods, causing the outcome for the first 14 periods to be `NA`

, because no moving average could be calculated for them.

There are a couple of options to deal with the `NA`

-values while doing the ranking. The option you find most suitable will depend on what you are going to use your data for later on:

You can choose to replace all

`NA`

-values in`rsi`

with zeros for the ranking:`for(i in 1:length(rsi)){ rsi[[i]][is.na(rsi[[i]])] <- 0 } ranks <- lapply(rsi, rank)`

You can ignore all

`NA`

-value and simply remove them before ranking`ranks <- lapply(rsi, rank, na.last = NA)`

Rank the

`NA`

-values as either the lowest or highest ranks.`# If NA be put last, use "na.last = TRUE". # If NA be put first, use "na.last = FALSE" ranks <- lapply(rsi, rank, na.last = TRUE)`

I would combine the lists into one data frame and then calculate the row-wise rank:

```
rsiDF <- data.frame(rsi)
rsiDF <- cbind(rsiDF, t(apply(rsiDF, 1, rank)))
```

Note that here you can, again, decide how to deal with tied values and `NA`

-values in your rank calculation (as described above and in `?rank`

)

If you wish to turn it back into lists again:

```
k <- length(symbols)
interranks <- list()
for(i in 1:k){
interranks[[i]] <- rsiDF[,c(i, i+k)]
}
```

Source (Stackoverflow)

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