Statistical Computing, 36-350
Friday October 14, 2016
lapply(), elements of a list or vectorThe lapply() function takes inputs as in: lapply(x, FUN=my.fun), to apply my.fun() across elements of a list or vector x. The output is a list
my.list = list(nums=seq(0.1,0.6,by=0.1), chars=letters[1:12],
bools=sample(c(TRUE,FALSE), 6, replace=TRUE))
my.list## $nums
## [1] 0.1 0.2 0.3 0.4 0.5 0.6
##
## $chars
## [1] "a" "b" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l"
##
## $bools
## [1] TRUE TRUE TRUE FALSE TRUE TRUElapply(my.list, FUN=mean) # Warning, because mean() can't be applied to chars## Warning in mean.default(X[[i]], ...): argument is not numeric or logical:
## returning NA## $nums
## [1] 0.35
##
## $chars
## [1] NA
##
## $bools
## [1] 0.8333333With lapply() and say, FUN=my.fun, we’ll get back a list, no matter the output type of my.fun()
lapply(my.list, FUN=summary)## $nums
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.100 0.225 0.350 0.350 0.475 0.600
##
## $chars
## Length Class Mode
## 12 character character
##
## $bools
## Mode FALSE TRUE NA's
## logical 1 5 0These all work like they do in apply(). E.g., can use: lapply(x, FUN=my.fun, extra.arg.1, extra.arg.2), for two extra arguments extra.arg.1, extra.arg.2 to be passed to my.fun()
# How to compute the leave-one-out means, also called jackknife means,
# without a for() loop?
mean.omitting.one = function(i, vec) { return(mean(vec[-i])) }
my.vec = state.x77[,"Frost"] # Frost variable in the states data
n = length(my.vec)
my.vec.jack = lapply(1:n, FUN=mean.omitting.one, vec=my.vec)
head(my.vec.jack) # It's a list, and here are the first 5 elements## [[1]]
## [1] 106.1837
##
## [[2]]
## [1] 103.4898
##
## [[3]]
## [1] 106.2857
##
## [[4]]
## [1] 105.2653
##
## [[5]]
## [1] 106.1837
##
## [[6]]
## [1] 103.2041sapply(), elements of a list or vectorThe sapply() function works just like lapply(), but tries to simplify the return value whenever possible. E.g., most common is the conversion from a list to a vector
my.vec.jack = sapply(1:n, FUN=mean.omitting.one, vec=my.vec)
head(my.vec.jack) # Now this is a vector, with same elements as before## [1] 106.1837 103.4898 106.2857 105.2653 106.1837 103.2041sqrt((n-1)^2/n) * sd(my.vec.jack) # Jackknife standard error## [1] 7.351202sd(my.vec)/sqrt(n) # Compare to "usual" standard error of the mean## [1] 7.351202tapply(), levels of a factor vectorThe function tapply() takes inputs as in: tapply(x, INDEX=my.index, FUN=my.fun), to apply my.fun() to subsets of entries in x that share a common level in my.index
head(state.x77) # Matrix of states data, 50 states x 8 variables## Population Income Illiteracy Life Exp Murder HS Grad Frost
## Alabama 3615 3624 2.1 69.05 15.1 41.3 20
## Alaska 365 6315 1.5 69.31 11.3 66.7 152
## Arizona 2212 4530 1.8 70.55 7.8 58.1 15
## Arkansas 2110 3378 1.9 70.66 10.1 39.9 65
## California 21198 5114 1.1 71.71 10.3 62.6 20
## Colorado 2541 4884 0.7 72.06 6.8 63.9 166
## Area
## Alabama 50708
## Alaska 566432
## Arizona 113417
## Arkansas 51945
## California 156361
## Colorado 103766head(state.region) # Factor of regions for the 50 states## [1] South West West South West West
## Levels: Northeast South North Central West# Now, let's average the Frost variable, within in each region
tapply(state.x77[,"Frost"], INDEX=state.region, FUN=mean)## Northeast South North Central West
## 132.7778 64.6250 138.8333 102.1538split(), split by levels of a factorSometimes we want to split up the rows of a data frame or entries of a vector by levels of a factor. The function split() does this, as in: split(x, f=my.index) to split a data frame or vector x according to levels of my.index
# Let's split up the state.x77 matrix according to region
state.x77.df = data.frame(state.x77) # First cast into a data frame
state.x77.by.reg = split(state.x77.df, f=state.region) # Now split
class(state.x77.by.reg) # The result is a list## [1] "list"names(state.x77.by.reg) # This has 4 elements for the 4 regions## [1] "Northeast" "South" "North Central" "West"class(state.x77.by.reg[[1]]) # Each element is a data frame## [1] "data.frame"# For each region, display the first 3 rows of the data frame
lapply(state.x77.by.reg, FUN=head, 3) ## $Northeast
## Population Income Illiteracy Life.Exp Murder HS.Grad Frost
## Connecticut 3100 5348 1.1 72.48 3.1 56.0 139
## Maine 1058 3694 0.7 70.39 2.7 54.7 161
## Massachusetts 5814 4755 1.1 71.83 3.3 58.5 103
## Area
## Connecticut 4862
## Maine 30920
## Massachusetts 7826
##
## $South
## Population Income Illiteracy Life.Exp Murder HS.Grad Frost Area
## Alabama 3615 3624 2.1 69.05 15.1 41.3 20 50708
## Arkansas 2110 3378 1.9 70.66 10.1 39.9 65 51945
## Delaware 579 4809 0.9 70.06 6.2 54.6 103 1982
##
## $`North Central`
## Population Income Illiteracy Life.Exp Murder HS.Grad Frost Area
## Illinois 11197 5107 0.9 70.14 10.3 52.6 127 55748
## Indiana 5313 4458 0.7 70.88 7.1 52.9 122 36097
## Iowa 2861 4628 0.5 72.56 2.3 59.0 140 55941
##
## $West
## Population Income Illiteracy Life.Exp Murder HS.Grad Frost
## Alaska 365 6315 1.5 69.31 11.3 66.7 152
## Arizona 2212 4530 1.8 70.55 7.8 58.1 15
## California 21198 5114 1.1 71.71 10.3 62.6 20
## Area
## Alaska 566432
## Arizona 113417
## California 156361# For each region, average each of the 8 numeric variables
lapply(state.x77.by.reg, FUN=function(df) { apply(df, MARGIN=2, mean) })## $Northeast
## Population Income Illiteracy Life.Exp Murder
## 5495.111111 4570.222222 1.000000 71.264444 4.722222
## HS.Grad Frost Area
## 53.966667 132.777778 18141.000000
##
## $South
## Population Income Illiteracy Life.Exp Murder HS.Grad
## 4208.12500 4011.93750 1.73750 69.70625 10.58125 44.34375
## Frost Area
## 64.62500 54605.12500
##
## $`North Central`
## Population Income Illiteracy Life.Exp Murder HS.Grad
## 4803.00000 4611.08333 0.70000 71.76667 5.27500 54.51667
## Frost Area
## 138.83333 62652.00000
##
## $West
## Population Income Illiteracy Life.Exp Murder
## 2.915308e+03 4.702615e+03 1.023077e+00 7.123462e+01 7.215385e+00
## HS.Grad Frost Area
## 6.200000e+01 1.021538e+02 1.344630e+05