I have a data.frame like this
z <- structure(list(ID = c("R-HSA-977606", "R-HSA-977443", "R-HSA-166658",
"R-HSA-166663", "R-HSA-1236394", "R-HSA-390522", "R-HSA-3232118",
"R-HSA-1630316", "R-HSA-112315", "R-HSA-112314"), GeneRatio = c("6/189",
"6/189", "6/189", "4/189", "5/189", "4/189", "3/189", "7/189",
"11/189", "9/189")), row.names = c("R-HSA-977606", "R-HSA-977443",
"R-HSA-166658", "R-HSA-166663", "R-HSA-1236394", "R-HSA-390522",
"R-HSA-3232118", "R-HSA-1630316", "R-HSA-112315", "R-HSA-112314"
), class = "data.frame")
Is it possible to add a 3rd column with the ratio from the 2nd column calculated? i.e. 6/189=0.0317. So in the third column I should have 0.0317.
CodePudding user response:
As it is a string expression, we can use eval/parse
z$newColumn <- sapply(z$GeneRatio, function(x) eval(parse(text = x)))
-output
> z
ID GeneRatio newColumn
R-HSA-977606 R-HSA-977606 6/189 0.03174603
R-HSA-977443 R-HSA-977443 6/189 0.03174603
R-HSA-166658 R-HSA-166658 6/189 0.03174603
R-HSA-166663 R-HSA-166663 4/189 0.02116402
R-HSA-1236394 R-HSA-1236394 5/189 0.02645503
R-HSA-390522 R-HSA-390522 4/189 0.02116402
R-HSA-3232118 R-HSA-3232118 3/189 0.01587302
R-HSA-1630316 R-HSA-1630316 7/189 0.03703704
R-HSA-112315 R-HSA-112315 11/189 0.05820106
R-HSA-112314 R-HSA-112314 9/189 0.04761905
Or a faster option would be to split by / (or use read.table to create two columns and then divide (assuming the expression includes only division)
z$newColumn <- Reduce(`/`, read.table(text = z$GeneRatio,
header = FALSE, sep = "/"))
CodePudding user response:
This code could be refined but it will work with the eval function
# 1- Creating empty column
z$GeneRatioNum <- NA
# 2- Filling it with eval function
for(i in 1:nrow(z)){z$GeneRatioNum[i] <- (eval(parse(text = z$GeneRatio[i])))}
