根据另一个数据集创建分层样本

编程问答 2022-06-07

问题描述

我有一个如下数据库：

DT <- structure(list(Year = c(2005,2005,2006,2007,2007),Type = c(1,1,2,3,3),Value = c(0.504376244608734,0.544791523560323,0.536356351248399,0.186754489979335,0.0145059662169885,0.552467068108315,0.728991908748136,0.0782701833265232,0.0770140143185365,0.745720346755096,0.182549844851049,0.0037854136407528,0.892426526130476,0.670307075099745,0.0787676704471466,0.243642889274613,0.61622932816441,0.773909954748003,0.0368627127466908,0.864836276200213,0.363247130858897,0.170719500081567,0.458862115912474,0.764369844834086,0.22138732039061,0.950217140815184,0.119026355092504,0.806698643902745,0.809697143416323,0.0161168403745759,0.56149794546334,0.0663374185634651,0.851044662622003,0.144127493261805,0.646129610173195,0.180326314861961,0.346305710081752,0.689186084156133,0.0902438913162577,0.493067567084055,0.829728867159447,0.212655417404949,0.873112880345332,0.57019799015934,0.666924788035991,0.421470848297274,0.137822577124685,0.646797965126931,0.00186628356193685,0.220630784144145,0.636097250212043,0.337161167241577,0.763014675300797,0.0290609945874959,0.179775595422681,0.926270372245386,0.14413707866326,0.308460218540821,0.505730133160804,0.92831463570813,0.2406601397661,0.469013177711661,0.0514836845684897,0.8773477591701,0.988870207825279,0.0409427390691713,0.345261503182235,0.457678159145652,0.928521904779235,0.981654149874765,0.165376851871405,0.657749413049735,0.645610554242246,0.288901032482677,0.903464871012278,0.91288926903878,0.331819964874993,0.451775254733976,0.561567931867726,0.934770693643712,0.0515071551015609,0.0772762108900331,0.233674539049138,0.636764452840065,0.673165028674493,0.806944576060158,0.763410488346345,0.661058275398286,0.275215831961986,0.821051953775588),Value2 = c(0.898973133700585,0.0043728119746469,0.90370150590114,0.664255277142381,0.478255150030532,0.428181937562552,0.0547471373342867,0.382060484866744,0.467990590870777,0.44613758335896,0.767317422802576,0.378150639908367,0.490578474103678,0.677901331005272,0.287571260541928,0.201396158908221,0.504989505596871,0.854550423135574,0.545208640791417,0.951248990134053,0.958420479001103,0.916437669811835,0.299402641214852,0.966388390213139,0.511359402704707,0.0867219533353825,0.88481040004275,0.158676351804193,0.0723357399252373,0.605048894989562,0.60104443547608,0.608164723564692,0.309073275149768,0.183031315824665,0.495737621177827,0.981936843144856,0.601436476710344,0.442362735422709,0.497899316486054,0.0545162134700136,0.572666465987199,0.0134330483790179,0.494252845049882,0.752561338910785,0.269231150235318,0.580397043886635,0.00438648885146109,0.974859546601355,0.964309270817873,0.740961468264743,0.966289928060099,0.165450408579171,0.457088887715921,0.725271665700556,0.611801886877621,0.693114823445831,0.509441044895801,0.668642268489104,0.0769213109282016,0.0106313240133811,0.653738670103508,0.515077318720933,0.0355798295524966,0.916849288357794,0.489540407953311,0.355080030655249,0.0584185346727107,0.117505910926226,0.840486642923002,0.0919621689925281,0.513293731647231,0.813987689492758,0.520895630669219,0.417642884334403,0.549898208275446,0.190152036926942,0.730222922437507,0.247328458018061,0.587109508511267,0.850096530635719,0.929032051736368,0.929910983683225,0.461558252621238,0.106247873795127,0.177666580357953,0.85962988262837,0.531897323076434,0.105528819826748,0.0349104003049517,0.180758384726269),Value3 = c(0.728747048185938,0.136214396563203,0.0552254916905935,0.888943411458351,0.593186561829418,0.142192475897417,0.397839605231809,0.128332683559321,0.818143628566787,0.675081193031822,0.267554700398382,0.289692778583473,0.395043380675461,0.582592369450023,0.999361780203229,0.421977850130829,0.723404859329269,0.333410997686596,0.545945290276875,0.510878802866974,0.746682101648222,0.625853669469718,0.0366957172106372,0.417685335838607,0.106323486037796,0.0127310987059773,0.291264331038641,0.690392584005106,0.0367947033685097,0.287721087095362,0.389582158765541,0.179954765659721,0.688980485242488,0.492296704771236,0.177765364735501,0.311877860895471,0.402659917512069,0.579307427105039,0.588566648357923,0.741057591300206,0.111932877257211,0.515443723005798,0.679584351614947,0.0197622696399569,0.0326379476305644,0.736148474541639,0.0115696238487739,0.0530159587501624,0.710708890129421,0.537042840144158,0.0277825198238522,0.851349803530179,0.448963399024373,0.42841165712813,0.0615511042450435,0.210541933956987,0.983517611560273,0.533691182135933,0.61993895519575,0.136074538018663,0.716185070081669,0.67982888131481,0.186059692566576,0.0129160598675656,0.832257317305668,0.0269936347869698,0.579065014243438,0.857987264303428,0.270050217297758,0.606374993010002,0.565105220120649,0.977264711860796,0.14241840012272,0.942496958955904,0.652070963472916,0.912867524689929,0.0249357414986835,0.87704909395977,0.72849611059358,0.525707690655331,0.290223239565496,0.992723233891769,0.178173444691217,0.0292681960925434,0.65696953770876,0.452973377851251,0.471917712361899,0.117830393053313,0.126107861454795,0.0848074010166607
)),row.names = c(NA,-90L),class = c("tbl_df","tbl","data.frame"
))

使用DT数据库，我想基于以下数据创建分层的子样本：

Ratios <- structure(list(State = c("A","A","B","C","C"),Year = c(2005,ratio = c(0.2,0.2,0.6,0.45,0.35,0.1,0.8,0.05,0.5,0.3,0.6
)),-27L),"data.frame"
))

图1. Ratios

strong text

我正在设法弄清楚该问题的解决方法。为了获得最大可能的样本量，我需要根据比率找出最大可能的子样本。图1. Ratios显示了2005年A国的比率。

我首先必须确定DT中的比率：

# This is just to determine the ratio of `Type` in the data `DT`
DT_1 <- DT %>%  count(Year,Type) %>% group_by(Year)
DT_2 <- DT %>%  count(Year,Type) %>% group_by(Year) %>% mutate(n = n/sum(n))
DT_RAT <- cbind(DT_1,DT_2)
setDT(DT_RAT)[,Year1:=NULL]
setDT(DT_RAT)[,Type1:=NULL]
rm(DT_1,DT_2)

现在，我应该利用这些信息来创建样本。我必须从最大份额开始（对于状态A，这是Type==1。）因此，对于样本DT_A，一半样本应为Type =1。从{{1 }}我知道这相当于DT_RAT的13个观测值（类型1观测值的最大数量）。我想做类似的事情：

Type 1

但是我不允许这样做。谁能帮助我实现这一目标？

编辑：

DT_A <- DT[,.SD[sample(.N,min(13,.N))],by = Type==1] （状态A的DT_A的子样本）的期望结果如下。

  • 2005年DT_A中的0.5应该由DT组成（见图1）
  • 2005年DT_A的0.45应该由Type 1
    • 组成
  • 2005年DT_A中的0.05应该由Type 2
    • 组成

DT对2005年的Type 3有13个观测值，因此我们将所有这些都考虑在内。因为这13个应该是样本的一半。总样本为26个观测值 2005年还对Type 1进行了13次观测，但由于Type 2仅应由0.45组成，因此我们取12。Type 2应当只有5％，因此剩下的1观测值。

  • 13 = 2005年DT_A的0.5应该由Type 3组成（见图1）
  • 12 = 2005年DT_A的0.45应包含Type 1
  • 1 = 2005年DT_A的0.05应包含Type 2

解决方法

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data.tablerrsample