How to add the results of lapply or for loop to specific rows - r

Question

I have a database with 5 variables (columns). I want to subset the data frame, in order to check if a certain value of one specific column is present. If this is the case, then assign 1, else 0. Then paste the result (1 or 0) to a specific column of the restricted data frame, and then continue.
The data frame looks like the following:
## Year Month Product Supermarket Price
## 2015 1 67 1 10
## 2015 1 65 1 11
## 2015 1 69 1 15
## 2015 2 65 2 20
## 2015 2 67 2 25
## 2015 2 67 3 15
## 2015 2 69 3 12
Now I want to restrict for each Year, Month and Supermarket and check if Product = 65 is present. If it is, then to assign 1 for the rows restricted in a new variable (column). If not, to assign 0.
I have tried using lapply:
prueba <- function(x)
ifelse(any(base$Product == 65), 1, 0)
lapply(unique(base$Supermarket) & unique(base$Year) & unique(base$Month),
base$NewVar <- prueba)
but have the following result
Error in rep(value, length.out = nrows) :
attempt to replicate an object of type 'closure'
Next, I try to make a for loop:
for(i in unique(base$Supermarket)) {
for(j in unique(base$Year))
for(h in unique(base$Month)) {
try <- ifelse(any((filter(base, Supermarket == i, Year == j, Month == h))$Product == 65), 1, 0)
base[base$Supermarket == i && base$Year ==j && base$Month == h,]$NewVar <- try
}
}
}
And have the following results:
Error in if (nrow(try) == 0) { : argument has zero lenght
I shall say that the database has 50 million rows, so speed is an issue here (so I try to use lapply instead of for loop)
I do not how to obtain the proper result, which should be like the following:
## Year Month Product Supermarket Price NewVar
## 2015 1 67 1 10 1
## 2015 1 65 1 11 1
## 2015 1 69 1 15 1
## 2015 2 65 2 20 1
## 2015 2 67 2 25 1
## 2015 2 67 3 15 0
## 2015 2 69 3 12 0
Do not know how to solve the whole problem. When using lapply I get the "right" answer, but then could not paste the result to the right rows in the dataframe.
Thanks in advance.

Answer 1

For fast operation, try to use data.table or dplyr. With data.table, you can simply create the new variable with logic check grouped by the Year, Month and Supermarket variables(suppose your original data frame is called df):
library(data.table)
setDT(df)[, NewVar := as.numeric(65 %in% Product), .(Year, Month, Supermarket)]
df
# Year Month Product Supermarket Price NewVar
# 1: 2015 1 67 1 10 1
# 2: 2015 1 65 1 11 1
# 3: 2015 1 69 1 15 1
# 4: 2015 2 65 2 20 1
# 5: 2015 2 67 2 25 1
# 6: 2015 2 67 3 15 0
# 7: 2015 2 69 3 12 0
Or correspondingly using dplyr: df <- df %>% group_by(Year, Month, Supermarket) %>% mutate(NewVar = as.numeric(65 %in% Product))

Answer 2

## read data
base <- c(2015, 1, 67, 1, 10,
2015, 1, 65, 1, 11,
2015, 1, 69, 1, 15,
2015, 2, 65, 2, 20,
2015, 2, 67, 2, 25,
2015, 2, 67, 3, 15,
2015, 2, 69, 3, 12)
base <- data.frame(matrix(base, 7, byrow = TRUE))
names(base) <- c('Year', 'Month', 'Product', 'Supermarket', 'Price')
Made a couple changes to function. I changed the object to match input (x) and specified the third element (since column of interest is column 3)
## create function
prueba <- function(x) ifelse(x[3] == 65, 1, 0)
To apply this function to each row, use the apply() function with 1 (for rows) apply(x, 1, function).
base$new_var <- apply(base, 1, prueba)
base
## Year Month Product Supermarket Price new_var
## 1 2015 1 67 1 10 0
## 2 2015 1 65 1 11 1
## 3 2015 1 69 1 15 0
## 4 2015 2 65 2 20 1
## 5 2015 2 67 2 25 0
## 6 2015 2 67 3 15 0
## 7 2015 2 69 3 12 0
You could also create a new variable and conditionally enter '1' to relevant rows. This is the way I'd do it:
base$new_var <- 0
base$new_var[base$Product == 65] <- 1
base
## Year Month Product Supermarket Price new_var
## 1 2015 1 67 1 10 0
## 2 2015 1 65 1 11 1
## 3 2015 1 69 1 15 0
## 4 2015 2 65 2 20 1
## 5 2015 2 67 2 25 0
## 6 2015 2 67 3 15 0
## 7 2015 2 69 3 12 0

Answer 3

We can do this easily in base R
df1$NewVar <- with(df1, ave(Product, Year, Month, Supermarket,
FUN= function(x) 65 %in% x))
df1$NewVar
#[1] 1 1 1 1 1 0 0