11 - Algorítmico: apply
François Rebaudo, IRD francois.rebaudo@ird.fr
Marzo 2019 ; PUCE-Quito-Ecuador http://myrbooksp.netlify.com/
CC BY-NC-ND 3.0
apply
apply
Aplicar una function a todos los elementos de un matrix o array.
Argumentos:
- X: an array, including a matrix.
- MARGIN: a vector giving the subscripts which the function will be applied over. E.g., for a matrix 1 indicates rows, 2 indicates columns, c(1, 2) indicates rows and columns.
- FUN: the function to be applied.
apply
bdd <- matrix(rnorm(200), ncol = 20)
apply(bdd, MARGIN = 2, FUN = mean)## [1] -0.172688675 0.230224263 0.162718357 0.025784095 -0.008130307
## [6] 0.467153737 -0.070431767 -0.204683442 0.153504785 -0.842461722
## [11] -0.622626753 0.058949880 -0.096573100 0.094711924 -0.005039150
## [16] -0.132252628 0.068651924 0.895776443 0.113150713 -0.346795796apply
apply(bdd, MARGIN = 1, FUN = median)## [1] -0.051987590 0.004223318 -0.141736878 -0.167981389 0.064074971
## [6] 0.335765724 -0.140736765 0.351855180 0.055748881 -0.127539494apply
bdd <- matrix(sample(c(1:20, NA), size = 200, replace = TRUE),
ncol = 20)
apply(bdd, MARGIN = 2, FUN = mean)## [1] NA 9.6 NA 15.1 13.0 10.9 9.1 NA 12.9 10.0 NA NA 9.5 NA
## [15] NA 8.3 NA 11.9 NA 7.2apply
apply(bdd, MARGIN = 2, FUN = mean, na.rm = TRUE)## [1] 10.444444 9.600000 11.555556 15.100000 13.000000 10.900000 9.100000
## [8] 15.111111 12.900000 10.000000 11.555556 12.375000 9.500000 12.444444
## [15] 8.444444 8.300000 9.750000 11.900000 10.000000 7.200000apply
apply(bdd, MARGIN = 2, FUN = function(i){
mean(i, na.rm = TRUE)
})## [1] 10.444444 9.600000 11.555556 15.100000 13.000000 10.900000 9.100000
## [8] 15.111111 12.900000 10.000000 11.555556 12.375000 9.500000 12.444444
## [15] 8.444444 8.300000 9.750000 11.900000 10.000000 7.200000lapply
lapply
Aplicar una function a todos los elementos de una list (data.frame es una list).
lapply
myList <- list(
a = sample(1:100, size = 10),
b = sample(1:100, size = 10),
c = sample(1:100, size = 10),
d = sample(1:100, size = 10),
e = sample(1:100, size = 10)
)
print(myList)## $a
## [1] 73 68 96 60 56 80 52 62 95 89
##
## $b
## [1] 71 40 74 90 32 14 68 1 34 67
##
## $c
## [1] 85 33 73 98 99 25 43 67 21 100
##
## $d
## [1] 78 84 63 3 77 46 93 11 2 98
##
## $e
## [1] 30 5 33 26 7 9 49 62 89 48lapply
lapply(myList, FUN = mean)## $a
## [1] 73.1
##
## $b
## [1] 49.1
##
## $c
## [1] 64.4
##
## $d
## [1] 55.5
##
## $e
## [1] 35.8lapply
myList <- list(
a = sample(c(1:5, NA), size = 10, replace = TRUE),
b = sample(c(1:5, NA), size = 10, replace = TRUE),
c = sample(c(1:5, NA), size = 10, replace = TRUE),
d = sample(c(1:5, NA), size = 10, replace = TRUE),
e = sample(c(1:5, NA), size = 10, replace = TRUE)
)
print(myList)## $a
## [1] 4 NA 1 3 3 NA NA 5 2 4
##
## $b
## [1] 4 NA 3 4 3 4 1 3 2 3
##
## $c
## [1] 3 3 5 4 2 2 5 4 2 5
##
## $d
## [1] 3 1 3 NA 1 4 2 5 4 2
##
## $e
## [1] NA 3 5 3 NA 2 4 2 4 1lapply
lapply(myList, FUN = mean)## $a
## [1] NA
##
## $b
## [1] NA
##
## $c
## [1] 3.5
##
## $d
## [1] NA
##
## $e
## [1] NAlapply
lapply(myList, FUN = mean, na.rm = TRUE)## $a
## [1] 3.142857
##
## $b
## [1] 3
##
## $c
## [1] 3.5
##
## $d
## [1] 2.777778
##
## $e
## [1] 3lapply
lapply(myList, FUN = function(i){
mean(i, na.rm = TRUE)
})## $a
## [1] 3.142857
##
## $b
## [1] 3
##
## $c
## [1] 3.5
##
## $d
## [1] 2.777778
##
## $e
## [1] 3lapply
lapply(myList, FUN = function(i){
m <- mean(i, na.rm = TRUE)
if(m > 3){
return("grande")
}else{
return("pequeño")
}
})## $a
## [1] "grande"
##
## $b
## [1] "pequeño"
##
## $c
## [1] "grande"
##
## $d
## [1] "pequeño"
##
## $e
## [1] "pequeño"lapply
Numero de datos faltantes:
lapply(myList, FUN = function(i){
sum(is.na(i))
})## $a
## [1] 3
##
## $b
## [1] 1
##
## $c
## [1] 0
##
## $d
## [1] 1
##
## $e
## [1] 2sapply
sapply
sapply es una forma de lapply con intentos para simplificar el resultado cuando posible (por ejemplo devolver un vector en lugar de un list cuando posible).
sapply
lapply(myList, FUN = function(i){
sum(is.na(i))
})## $a
## [1] 3
##
## $b
## [1] 1
##
## $c
## [1] 0
##
## $d
## [1] 1
##
## $e
## [1] 2sapply
sapply(myList, FUN = function(i){
sum(is.na(i))
})## a b c d e
## 3 1 0 1 2sapply
Sacar el elemento “n” de una list:
sapply(myList, FUN = '[[', 2)## a b c d e
## NA NA 3 1 3sapply
myDF <- data.frame(
a = sample(c(1:5, NA), size = 10, replace = TRUE),
b = sample(c(1:5, NA), size = 10, replace = TRUE),
c = sample(c(1:5, NA), size = 10, replace = TRUE),
d = sample(c(1:5, NA), size = 10, replace = TRUE),
e = sample(c(1:5, NA), size = 10, replace = TRUE)
)
print(myDF)## a b c d e
## 1 1 5 1 4 4
## 2 2 3 2 5 3
## 3 1 2 1 5 2
## 4 3 3 3 1 3
## 5 5 3 3 5 2
## 6 1 2 3 5 5
## 7 1 1 NA 3 4
## 8 5 3 NA 2 4
## 9 NA 2 4 3 2
## 10 5 4 3 4 2sapply
sapply(myDF, FUN = function(i){
sum(is.na(i))
})## a b c d e
## 1 0 2 0 0sapply
pruebaTiempo <- lapply(
seq(from = 1000, to = 30000, by = 1000),
function(sampleSize){
sapply(1:30, function(repet){
startTime <- Sys.time()
guessNumber(mySample = 1:sampleSize)
return(Sys.time() - startTime)
})
})sapply
tapply
tapply
col0 <- sample(LETTERS[1:5], size = 1000, replace = TRUE)
col1 <- rnorm(n = 1000, mean = 10, sd = 0.5)
col2 <- rlnorm(n = 1000, meanlog = 10, sdlog = 0.5)
col3 <- rgamma(n = 1000, shape = 10, rate = 0.5)
dfCol <- data.frame(col0, col1, col2, col3)
print(head(dfCol, n = 10))## col0 col1 col2 col3
## 1 E 10.214841 47273.13 19.53360
## 2 B 10.161125 18761.21 21.31726
## 3 A 9.963536 23280.13 31.25191
## 4 E 10.136935 54161.22 13.70548
## 5 B 10.446320 6520.05 23.58053
## 6 A 9.582148 27207.82 17.60784
## 7 E 9.447795 18704.42 19.38604
## 8 B 10.419673 30796.01 28.34310
## 9 A 10.074750 26959.06 24.33891
## 10 E 10.813097 21645.99 20.52869tapply
tapply(dfCol$col1, INDEX = dfCol$col0, FUN = summary)## $A
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 8.522 9.700 9.983 10.003 10.270 11.247
##
## $B
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 8.841 9.664 9.994 10.004 10.356 11.746
##
## $C
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 8.284 9.644 9.962 9.960 10.332 11.487
##
## $D
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 8.813 9.662 10.040 10.048 10.379 11.446
##
## $E
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 8.671 9.613 9.986 9.971 10.317 11.260tapply
sapply(2:4, FUN = function(i){
tapply(dfCol[,i], INDEX = dfCol$col0, FUN = mean)
})## [,1] [,2] [,3]
## A 10.002878 25369.50 19.91896
## B 10.003580 24616.70 20.44072
## C 9.960449 25622.16 20.35900
## D 10.048424 24673.44 19.14632
## E 9.971356 25213.84 20.27397SIGUIENTE
1
11 - Algorítmico: apply
François Rebaudo, IRD francois.rebaudo@ird.fr
Marzo 2019 ; PUCE-Quito-Ecuador http://myrbooksp.netlify.com/
CC BY-NC-ND 3.0