I have one data frame that has 6 variables. df is the name of my data frame.
I found the expectation of e (variable in df) using
Ee <- mean(df[["e"]])
How do I find E[e|Z=z] for z in {0,1}?
Similarly, how do I find E[e|X=x] for x in {1...20} using the sapply function?
Here's a thought:
set.seed(42)
sampdata <- data.frame(e = runif(1000), z = sample(0:1, size=1000, replace=TRUE), x = sample(1:20, size=1000, replace=TRUE))
head(sampdata)
# e z x
# 1 0.9148060 1 15
# 2 0.9370754 0 2
# 3 0.2861395 1 13
# 4 0.8304476 1 12
# 5 0.6417455 1 4
# 6 0.5190959 0 7
aggregate(e ~ z, data = sampdata, FUN = mean)
# z e
# 1 0 0.4910876
# 2 1 0.4852118
aggregate(e ~ x, data = sampdata, FUN = mean)
# x e
# 1 1 0.5097038
# 2 2 0.4495141
# 3 3 0.5077897
# 4 4 0.5300375
# 5 5 0.4549345
# 6 6 0.5122537
# 7 7 0.4704425
# 8 8 0.4911532
# 9 9 0.5572367
# 10 10 0.4634067
# 11 11 0.4408758
# 12 12 0.4815633
# 13 13 0.5503166
# 14 14 0.4922317
# 15 15 0.5205427
# 16 16 0.4999023
# 17 17 0.4784551
# 18 18 0.4282990
# 19 19 0.4202285
# 20 20 0.4852303
But if you feel you must use sapply, then this can be equivalent.
sapply(setNames(nm = unique(sampdata$z)), function(Z) mean(sampdata[["e"]][ sampdata[["z"]] == Z ]))
# 1 0
# 0.4852118 0.4910876
sapply(setNames(nm = unique(sampdata$x)), function(X) mean(sampdata[["e"]][ sampdata[["x"]] == X ]))
# 15 2 13 12 4 7 19 16 10 1
# 0.5205427 0.4495141 0.5503166 0.4815633 0.5300375 0.4704425 0.4202285 0.4999023 0.4634067 0.5097038
# 9 3 14 18 11 20 5 8 17 6
# 0.5572367 0.5077897 0.4922317 0.4282990 0.4408758 0.4852303 0.4549345 0.4911532 0.4784551 0.5122537