R:随机抽样各种类别的观测值 [英] R: Random sampling an even number of observations from a range of categories

问题描述

我以前从我的数据帧中随机抽取了一个邮政编码样本，然后意识到我并没有在所有更高级别的统计单位中进行抽样.我有大约一百万个邮政编码和7000个中间输出统计单位.我希望样本中每个统计单位的邮政编码数量大致相同.

I previously took a random sample of postcodes from my dataframe and then realised that I wasn't sampling across all higher level statistical units. I have around 1 million postcodes and 7000 middle output statistical units. I want the sample to have roughly the same number of postcodes from each statistical unit.

如何从每个较高级别的统计单位中随机抽取35个邮政编码?

How do I randomly sample 35 postcodes from each higher level statistical unit?

我以前使用以下代码随机采样250,000个邮政编码:

I used the following code previously to randomly sample 250,000 postcodes:

total.sample <- total[sample(1:nrow(total), 250000,
                           replace=FALSE),] 

如何根据另一个列变量(例如较高级别的统计单位(请参见下面的数据框结构中的msoa.rank))指定邮政编码的随机样本配额?

How do I specify a random sample quota of postcodes based on another column variable (e.g. such as the higher level statistical unit (see msoa.rank in the dataframe structure below))?

数据库结构:

'data.frame':   1096289 obs. of  25 variables:
$ pcd                : Factor w/ 986055 levels "AL100AB","AL100AD",..: 282268 282258 
$ mbps2              : int  0 1 0 0 0 1 0 0 0 0 ...
$ averagesp          : num  16 7.8 7.8 9.5 9.4 3.2 11.1 19.4 10.5 11.8 ...
$ mediansp           : num  18.2 8 7.8 8.1 8.5 3.2 8.1 18.7 9.7 8.9 ...
$ nga                : int  0 0 0 0 0 0 0 0 0 0 ...
$ x                  : int  533432 532192 533416 533223 532866 531394 532899 532744 
$ total.dps          : int  11 91 10 7 9 10 3 5 21 12 ...
$ connections.density: num  7.909 0.747 3.1 7.714 1.889 ...
$ urban              : int  1 1 1 1 1 1 1 1 1 1 ...
$ gross.pay          : num  36607 36607 36607 36607 36607 ...
$ p.tert             : num  98.8 98.8 98.8 98.8 98.8 ...
$ p.kibs             : num  70.3 70.3 70.3 70.3 70.3 ...
$ density            : num  25.5 25.5 25.5 25.5 25.5 25.5 25.5 25.5 25.5 25.5 ...
$ p_m_s              : num  93.5 93.5 93.5 93.5 93.5 ...
$ p_m_l              : num  6.52 6.52 6.52 6.52 6.52 ...
$ p.edu              : num  62.6 62.6 62.6 62.6 62.6 ...
$ p.claim            : num  1.58 1.58 1.58 1.58 1.58 ...
$ p.non.white        : num  21.4 21.4 21.4 21.4 21.4 21.4 21.4 21.4 21.4 21.4 ...
$ msoa.rank          : int  2 2 2 2 2 2 2 2 2 2 ...
$ oslaua.rank        : int  321 321 321 321 321 321 321 321 321 321 ...
$ nuts2.rank         : int  22 22 22 22 22 22 22 22 22 22 ...
$ gor.rank           : int  8 8 8 8 8 8 8 8 8 8 ...
$ cons               : int  1 1 1 1 1 1 1 1 1 1 ...

pcd =邮政编码

msoa.rank =每个中间输出统计单元的序数变量

msoa.rank = the ordinal variable of each middle output statistical unit

推荐答案

每个msoa.rank是否至少具有35个邮政编码? data.table

Does every msoa.rank have at least 35 postcodes? This will be fast with data.table

#Create a data.table object
require(data.table)
total <- data.table(total)

#Sample by each msoa.rank group (take a sample that is size min(35,total size of msoa grp)
total.sample <- total[ , .SD[sample(1:.N,min(35,.N))], by=msoa.rank]

因此，这是使用经典iris数据集的示例工作方式.

So here is how to example would work using the classic iris dataset.

iris < data.table(iris)
set.seed(2014)
iris.sample <- iris[ , .SD[sample(1:.N,min(10,.N))], by=Species]
summary(iris.sample$Sepal.Length)

Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
4.400   5.000   5.850   5.797   6.525   7.200 

这是另一个示例和摘要，以查看区别

Here is another sample and summary to see the difference

iris.sample2 <- iris[ , .SD[sample(1:.N,min(10,.N))], by=Species]
summary(iris.sample2$Sepal.Length)

Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
4.400   5.100   5.850   5.743   6.275   7.300 

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