Assign values to matrix based on condition [duplicate] - r

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I want to create a matrix with 3 columns and many rows assigning 1 or 0 if the condition is satisfied.
I have data stored in 3 variables
df1 <- data.frame(names=c("A","B","C","D","E","F"))
df2 <- data.frame(names=c("A","B","C","F"))
df3 <- data.frame(names=c("E","F","H"))
output will be
df1 df2 df3
A 1 1 0
B 1 1 0
C 1 1 0
D 1 0 0
E 1 1 1
F 1 0 1
H 0 0 1
In first row if A is present in dataset then I will assign 1 under each column and 0 if A not present in dataset
Here is what I have tried
DF <- rbind(df1,df2,df3)
for (i in DF) {
for (j in 1:length(df1$names)) {
if(i == df1$names[j]){
A3 <-data.frame(paste0("",i),paste0(1),paste0(0),paste0(0))
names(A3) <- NULL
}
else{
A3 <-data.frame(paste0("",i),paste0(0),paste0(0),paste0(0))
}
}
}
I have written this code only for df1 but its very slow because I have more than 1500 rows in my orignal data set. What would be the fastest way to do it?

Answer 1

Add a grouping variable to each dataframe:
df1 <- data.frame(names=c("A","B","C","D","E","F"),group="df1")
df2 <- data.frame(names=c("A","B","C","F"),group="df2")
df3 <- data.frame(names=c("E","F","H"),group="df3")
DF <- rbind(df1,df2,df3)
Then do this:
res <- table(DF)
> res
group
names df1 df2 df3
A 1 1 0
B 1 1 0
C 1 1 0
D 1 0 0
E 1 0 1
F 1 1 1
H 0 0 1
Or if you want a dataframe:
library(reshape2)
dcast(names~group, data=DF,fun.aggregate = length)

Answer 2

When using the idcol parameter in rbindlist of the data.table package, there is no need of creating a grouping column for each dataframe separately:
library(data.table) # I used v1.9.5 for this
DT <- rbindlist(list(df1, df2, df3), idcol="id")
dcast(DT[, .N , by=.(id,names)], names ~ id, fill=0)
which gives:
names 1 2 3
1: A 1 1 0
2: B 1 1 0
3: C 1 1 0
4: D 1 0 0
5: E 1 0 1
6: F 1 1 1
7: H 0 0 1

Answer 3

%in% operator lets you check if a string is present in a vector of strings. It is also vectorised, so it works quite quick:
x=c(LETTERS[c(1:6,8)])
df=data.frame(x=x,df1=as.numeric(x %in% df1$names),
df2=as.numeric(x %in% df2$names),
df3=as.numeric(x %in% df3$names))
df
If speed is crucial, {data.table} package gives a little speed boost with %chin% operator:
library(data.table)
x=c(LETTERS[c(1:6,8)])
dt=data.table(x=x,df1=as.numeric(x %chin% as.character(df1$names)),
df2=as.numeric(x %chin% as.character(df2$names)),
df3=as.numeric(x %chin% as.character(df3$names)))
dt

Answer 4

The code below is slightly more general than the other answers. Also, I think it's useful to know how to dynamically create commands...
I use the data frames as you prepared them:
df1 <- data.frame( names = c( "A", "B", "C", "D", "E", "F") )
df2 <- data.frame( names = c( "A", "B", "C"," F") )
df3 <- data.frame( names = c( "E", "F", "H") )
DF <- rbind( df1, df2, df3 )
nDF <- unique( DF ) #we don't want to duplicate tests.
Then the main loop is just like this:
n_ <- 3
for( ii in 1 : n_){
nDF[ paste0( "df", ii ) ] <- as.logical( NA ) #dynamically creates a new variable in your data frame
cmnd <- paste0("nDF$names %in% df",ii,"$names") #dynamically creates the appropriate command (in this case you want to test e.g. whether "nDF$names %in% df1$names".
nDF[ paste0("df",ii)] <- eval( parse( text = cmnd ) ) #evaluates the dynamically created command and saves it into the previously created variable.
}
Should be relatively fast. But if you don't have duplicates in your data, then heroka's suggestion to this questions is probably the way to go.