my data frame look like this:
A S1 S2 S3 S4
1 ex1 1 0 0 0
2 ex2 0 1 0 0
3 ex3 0 0 1 0
4 ex4 1 0 0 0
5 ex5 0 0 0 1
6 ex6 0 1 0 0
7 ex7 1 0 0 0
8 ex8 0 1 0 0
9 ex9 0 0 1 0
10 ex10 1 0 0 0
i need to have it as a single factor list like:
A Type
ex1 S1
ex2 S2
ex3 S3
ex4 S1
ex5 S4
ex6 S2
ex7 S1
ex8 S2
ex9 S3
ex10 S1
anybody help my problem?
Assuming d is the data, the new column could be obtained with
d$type <- names(d[-1])[apply(d[-1] == 1, 1, which)]
d[c(1, 6)]
# A type
# 1 ex1 S1
# 2 ex2 S2
# 3 ex3 S3
# 4 ex4 S1
# 5 ex5 S4
# 6 ex6 S2
# 7 ex7 S1
# 8 ex8 S2
# 9 ex9 S3
# 10 ex10 S1
apply in your answer?
sapply and apply in there to make it faster.sapply instead of apply (where possible) is faster, but what do you mean by using apply to make it faster? Not sure I understand.[ and ]You can use apply and check the max in the columns 2-5 and then return the corresponding column name:
df$Type <- apply(df[2:5], 1, function(x) names(df)[which.max(x)+1] )
Afterwards, you can delete the columns you don't need anymore:
df <- df[,-c(2:5)]
logical or numeric or integer? As long as the input is the same type, there should be no problems with larger data sets.Could also do (if dat is your data set)
library(reshape2)
dat <- melt(dat, id = "A")
dat[dat$value > 0, 1:2]
You could try:
If df is the dataframe
data.frame(A=df$A, Type=rep(names(df)[-1], nrow(df))[!!t(df[,-1])])
A Type
1 ex1 S1
2 ex2 S2
3 ex3 S3
4 ex4 S1
5 ex5 S4
6 ex6 S2
7 ex7 S1
8 ex8 S2
9 ex9 S3
10 ex10 S1
Also:
names(df)[-1][t(df[-1])*seq_len(ncol(df)-1)]
[1] "S1" "S2" "S3" "S1" "S4" "S2" "S1" "S2" "S3" "S1"
