R：重新採樣1到帶有循環的nrow（y）

Question

我有n個變量，每個變量有100行。爲了從1重新取樣到nrows，下面的代碼給出了預期的結果，但是它是單調乏味和不切實際的。要重現的情況下，讓該y具有5行suposse：

y<-rnorm(n=5, mean=10, sd=2) 
R=1000 #number of resamplings 
boot.means = numeric(R) 
for (i in 1:R) { boot.sample = sample(y, 1, replace=T) 
boot.means[i] = mean(boot.sample) } 
m1<-mean(boot.means) 
d1<-sd(boot.means) 
cv1 =(d1*100)/m1 

R=1000 #number of resamplings 
boot.means = numeric(R) 
for (i in 1:R) { boot.sample = sample(y, 2, replace=T) 
boot.means[i] = mean(boot.sample) } 
m2<-mean(boot.means) 
d2<-sd(boot.means) 
cv2 =(d2*100)/m2 

R=1000 #number of resamplings 
boot.means = numeric(R) 
for (i in 1:R) { boot.sample = sample(y, 3, replace=T) 
boot.means[i] = mean(boot.sample) } 
m3<-mean(boot.means) 
d3<-sd(boot.means) 
cv3 =(d3*100)/m3 


R=1000 #number of resamplings 
boot.means = numeric(R) 
for (i in 1:R) { boot.sample = sample(y, 4, replace=T) 
boot.means[i] = mean(boot.sample) } 
m4<-mean(boot.means) 
d4<-sd(boot.means) 
cv4 =(d4*100)/m4 


R=1000 #number of resamplings 
boot.means = numeric(R) 
for (i in 1:R) { boot.sample = sample(y, 5, replace=T) 
boot.means[i] = mean(boot.sample) } 
m5<-mean(boot.means) 
d5<-sd(boot.means) 
cv5 =(d5*100)/m5 

CV.OK<-(c(cv1,cv2,cv3,cv4,cv5)) 
plot(CV.OK) 

我想用類似下面的代碼，但它給出了意想不到的效果。請，有人可以幫助我。謝謝。

R = 1000 #number of resamplings 
boot.sample=seq(1,5, by=1) 
boot.means = numeric(R) 
boot.sd = numeric(R) 
m = 5 
d = 5 
for (i in 1:5) { 
    for (j in 1:R) { 
    boot.sample[i] = sample(y, i, replace=T) 
    boot.means[j] = mean(boot.sample[i]) 
    boot.sd[j] = sd(boot.sample[i]) 
    m[i]=mean(boot.means[j]) 
    d[i]=mean(boot.sd[j]) 
    } 
} 
CV.Fail<-(d*100)/m 

Answer 1

我想你想要這樣的：

y<-rnorm(n=5, mean=10, sd=2) 
R = 1000 #number of resamplings 
CVs <- numeric(5) 
for (i in 1:5) { 
    boot.means = numeric(R) 
    for (j in 1:R) { 
    boot.sample = sample(y, i, replace=T) 
    boot.means[j] = mean(boot.sample) 
    } 
    m=mean(boot.means) 
    d=sd(boot.means) 
    CVs[i] = (d*100)/m 
} 
plot(CVs) 

Answer 2

在R，你應該儘量避免環路，因爲它們是相當緩慢的。 我希望我能夠正確地理解這個問題，並且寫一點功能，讓你從不同的角度開始。

library(plyr) 
library(dplyr) 

# dummy data set 
data_set = data.frame(value = runif(200), group = rep(c("a", "b"), each=100)) 

# create a function that takes the sample size as an argument 
iterative_sample = function(sample_size, data){ 
# group the data (your 'n' equals the number of groups- 
# here thats 'a' and 'b' 
    sample_temp = dplyr::group_by(data, group) %>% 
    # take x (sample size) samples from each group 
    sample_n(sample_size, replace=T) %>% 
    # compute summary stats for each group 
    summarize(mean = mean(value), sd = sd(value)) %>% 
    # attach the sample size to keep track 
    mutate(sample_size = sample_size) 
    # we must return a dataframe to uses ldply later on 
    return(sample_temp) 
} 

# thats the vector we are going to iterate over using ldply 
sample_vect = c(1:2) 

# ldplyr (plyr package) takes a list or vector and returns a dataframe and our custom 
# function -checkout the manpage 
# ?ldply 

# ... 
# 
# 
# .data: list to be processed 
# 
#  .fun: function to apply to each piece 
# 
#  ...: other arguments passed on to ‘.fun’ 
# 
# ... 
# 

ldply(.data = sample_vect, .fun = iterative_sample, data_set)