基于R中的频率采样 [英] sampling based on frequency in R

问题描述

我想根据每个值的大小从相当大的数据中提取20000个样本，以填充NA值: 所以我使用直方图的输出，但是没有成功，并给我一个错误，如何避免呢?

I want to make 20000 sample from a data which is quite big,based on the each value size in order to fill the NA values: so I use the output of histogram, but it wasn't successful, and get me an error, how to avoid it ?

y=hist(maindata,col="red",breaks=length(unique(maindata))
for(k in 1:20000){
data=maindata
for(i in 1:nrow(data)){
if (data[i]="Na"){
 data[i]=sample(y$breaks,size=1,replace=FALSE,prob=y$density)}}}

我收到此错误:

Error in sample.int(length(x), size, replace, prob) : 
  incorrect number of probabilities

我检查了length(y$breaks)和length(y$density)，length(y$breaks)还有一个单位，我该如何解决?

and I check the length(y$breaks) and length(y$density),length(y$breaks) was one unit more, how should I fixed it ?

先谢谢您

编辑:

    structure(list(breaks = c(15, 16, 17, 18, 19, 20, 21, 22, 23, 
24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 
40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 
56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 
72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 
88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 
103, 104, 105, 106, 107, 108, 109), counts = c(27L, 17L, 31L, 
83L, 118L, 144L, 211L, 279L, 354L, 312L, 300L, 377L, 407L, 443L, 
481L, 351L, 302L, 236L, 248L, 178L, 141L, 101L, 77L, 80L, 63L, 
44L, 64L, 44L, 60L, 46L, 24L, 29L, 15L, 28L, 21L, 13L, 19L, 10L, 
30L, 11L, 12L, 12L, 7L, 12L, 12L, 11L, 11L, 7L, 7L, 4L, 4L, 4L, 
1L, 2L, 3L, 6L, 1L, 1L, 3L, 3L, 0L, 0L, 0L, 1L, 1L, 0L, 0L, 0L, 
1L, 0L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L, 1L, 0L, 0L, 1L, 1L, 0L, 
0L, 0L, 0L, 0L, 3L, 0L, 0L, 0L, 1L, 2L), density = c(0.00453172205438067, 
0.00285330647868412, 0.00520308828465928, 0.0139308492782813, 
0.0198053037932192, 0.0241691842900302, 0.035414568647197, 0.0468277945619335, 
0.0594159113796576, 0.0523665659617321, 0.0503524672708963, 0.0632762672037596, 
0.0683115139308493, 0.0743538100033568, 0.0807317891910037, 0.0589123867069486, 
0.0506881503860356, 0.0396106075864384, 0.0416247062772743, 0.0298757972473985, 
0.0236656596173212, 0.0169519973145351, 0.0129237999328634, 0.0134273246055723, 
0.0105740181268882, 0.00738502853306479, 0.0107418596844579, 
0.00738502853306479, 0.0100704934541793, 0.0077207116482041, 
0.0040281973816717, 0.00486740516951997, 0.00251762336354481, 
0.00469956361195032, 0.00352467270896274, 0.00218194024840551, 
0.00318898959382343, 0.00167841557569654, 0.00503524672708963, 
0.0018462571332662, 0.00201409869083585, 0.00201409869083585, 
0.00117489090298758, 0.00201409869083585, 0.00201409869083585, 
0.0018462571332662, 0.0018462571332662, 0.00117489090298758, 
0.00117489090298758, 0.000671366230278617, 0.000671366230278617, 
0.000671366230278617, 0.000167841557569654, 0.000335683115139308, 
0.000503524672708963, 0.00100704934541793, 0.000167841557569654, 
0.000167841557569654, 0.000503524672708963, 0.000503524672708963, 
0, 0, 0, 0.000167841557569654, 0.000167841557569654, 0, 0, 0, 
0.000167841557569654, 0, 0, 0.000167841557569654, 0, 0.000167841557569654, 
0, 0.000167841557569654, 0, 0.000167841557569654, 0.000167841557569654, 
0, 0, 0.000167841557569654, 0.000167841557569654, 0, 0, 0, 0, 
0, 0.000503524672708963, 0, 0, 0, 0.000167841557569654, 0.000335683115139308
), mids = c(15.5, 16.5, 17.5, 18.5, 19.5, 20.5, 21.5, 22.5, 23.5, 
24.5, 25.5, 26.5, 27.5, 28.5, 29.5, 30.5, 31.5, 32.5, 33.5, 34.5, 
35.5, 36.5, 37.5, 38.5, 39.5, 40.5, 41.5, 42.5, 43.5, 44.5, 45.5, 
46.5, 47.5, 48.5, 49.5, 50.5, 51.5, 52.5, 53.5, 54.5, 55.5, 56.5, 
57.5, 58.5, 59.5, 60.5, 61.5, 62.5, 63.5, 64.5, 65.5, 66.5, 67.5, 
68.5, 69.5, 70.5, 71.5, 72.5, 73.5, 74.5, 75.5, 76.5, 77.5, 78.5, 
79.5, 80.5, 81.5, 82.5, 83.5, 84.5, 85.5, 86.5, 87.5, 88.5, 89.5, 
90.5, 91.5, 92.5, 93.5, 94.5, 95.5, 96.5, 97.5, 98.5, 99.5, 100.5, 
101.5, 102.5, 103.5, 104.5, 105.5, 106.5, 107.5, 108.5), xname = "b", 
    equidist = TRUE), .Names = c("breaks", "counts", "density", 
"mids", "xname", "equidist"), class = "histogram")

数据信息:

> head(maindata)
[1] 30 44 -1 32 30 34
> is.numeric(maindata)
[1] TRUE
> is.vector(maindata)
[1] TRUE
> length(maindata)
[1] 36203

推荐答案

您是否只希望从不丢失数据的分布中获得20,000个样本?如果是这样，解决此问题的另一种方法是直接从非缺失数据中直接计算内核密度估计值，然后从中进行采样.例如，使用伪造数据:

Do you just want 20,000 samples from the distribution of the non-missing data? If so, another way to approach this would be to just calculate a kernel density estimate directly from the non-missing data and then sample from that. For example, using fake data:

# Fake data with some missing values
set.seed(31)
dat = rnorm(30000, 20, 10)
dat[sample(1:30000, 5000)] = NA

# Create kernel density estimate from the data
# n is the number of grid points used in the esimate (should always be a power of 2)
dat.dens = density(dat[!is.na(dat)], n=2^10)

sim.sample = sample(dat.dens$x, 2e4, replace=TRUE, prob=dat.dens$y)

plot(dat.dens)
lines(density(sim.sample), col="red")

如果我误解了您要做什么，请告诉我.

Please let me know if I've misunderstood what you're trying to do.

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