已解决- lmer：qr.default(X, tol = tol, LAPACK = FALSE) 中的错误：外部函数中的 NA/NaN/Inf (arg 1)

如何解决已解决- lmer：qr.default(X, tol = tol, LAPACK = FALSE) 中的错误：外部函数中的 NA/NaN/Inf (arg 1)

我对 R 不是很精通，偶然发现了一个我似乎无法解决的问题。

我在一个经济交易游戏的数据帧（“试验”）中有一些数据，我想对其进行一些预测。我通过以下方式将数据加载到 R：

trials <- read.table("TrialsCG2.txt",sep="\t",header=T)

示例数据，通过 dput(head(trials,18)) 获取前 18 行，其中包括一些Inf值

structure(list(SubjectID = structure(c(86L,86L,86L
),.Label = c("01HPG","01IVW","02BCH","02BLB","02LIM","02LRH","02TBH","03AAG","03bss","04BAR","04BSR","05BEH","05BEL","05FAF","05KBF","05rgf","05SBK","05ssw","06ANG","06HHG","06HSP","06IEF","06IMA","06SBW","07BDI","07BRC","07BUS","07FBU","07SAA","08FAV","08FIF","08IBA","08KMM","08WHH","09HIC","09kaa","09KUR","09OIR","10GKC","10IMK","11IIA","11SEB","11SSM","11WBG","12IAE","12IDG","12IMG","12LNC","12SDM","12SMS","12WCO","13DOR","13GBR","13HKA","13WPA","13ZMS","13ZRE","14HBG","14IMS","15DMC","15IHG","15KTR","15SAC","15ser","15wmk","16BAS","16BTT","16SJS","17fga","17HBV","17ICK","17ILJ","17KRA","18HAK","18HSB","18MMR","19LNC","20BIP","20IBM","20KEK","20MKS","21BAP","21DMW","21FSM","21HAR","21MAT","22BSB","22LDS","22lmr","23ERW","23MSV","23uls","24BFM","24LCF","24SIC","25HCA","25IBG","25SSM","26DMW","26emh","27DGJ","27IDH","27WNV","28BJT","28HAS","28kmh","28MBM","28SGW","29FCJ","29GEJ","29KHF","30BKH","31BRF","31LGR","31LWH"),class = "factor"),lossPercent = c(8.695652174,8.695652174,8.695652174),Neuro_Score = c(3L,3L,3L),Female = c(0L,0L,0L),Education = c(4L,4L,4L),Age = c(27L,27L,27L
),Satisfaction = c(2L,2L,2L),LifeEvent = c(1L,1L,1L),Ment_Score = c(3.67,3.67,3.67),MentSelf_Score = c(3,3,3),MentEmot_Score = c(3.25,3.25,3.25),MentEqui_Score = c(4,4,4),MentAffect_Score = c(4.67,4.67,4.67),LPFS_Score = c(10L,10L,10L),SchemaED_Score = c(5L,5L,5L),SchemaAB_Score = c(11L,11L,11L),SchemaMA_Score = c(10L,10L
),SchemaSI = c(12L,12L,12L),SchemaTotal_Score = c(38L,38L,38L),RejectionTotal_Score = c(8.56,8.56,8.56),UV1_Score = c(4L,UV2_Score = c(3L,UV3_Score = c(4L,UVTotal_Score = c(11L,currentPartner = structure(c(1L,.Label = c("Spieler A","Spieler B"),goodPartner = c(0L,finalBalance = c(10L,13L,14L,15L,16L,17L,19L,21L,23L,22L,24L,25L),goodPartnerBalance = c(10L,29L,30L,31L),badPartnerBalance = c(12L,20L,24L),investment = c(1L,sentBack = c(1L,win = c(0L,-2L,round = 1:18,preCrash = c(TRUE,TRUE,FALSE,FALSE),crash = c(FALSE,initialBalance = c(10L,visibiltyBalance = c(TRUE,TRUE),postCrash = c(FALSE,previousSentBack = c(NA,perInvest = c(10,10,20,8.33333333333333,7.69230769230769,7.14285714285714,6.66666666666667,6.25,11.7647058823529,10.5263157894737,9.52380952380952,8.69565217391304,9.09090909090909,4.34782608695652,4.16666666666667),ReInvestRatio = c(NA,1,2,0.25,0.5,Inf,0.666666666666667,0.333333333333333,0.5)),row.names = c(NA,18L),class = "data.frame")    ``

`

现在我想用一个lmer来预测玩家在每一轮的投资，比如：

fit1 <- lmer(log(investment + 0.5)~LPFS_Score+Education+visibiltyBalance+RejectionTotal_Score+(1|SubjectID),subset=preCrash==T)

运行良好。

我正在处理一个多轮投资游戏，玩家在每一轮内发送和接收资金。我想使用在给定轮次投资之前一轮返回的总和的比率作为预测值。 像这样计算之前退回的钱：

previousSentBack[round == 1] <- NA
previousSentBack[round == 2] <- sentBack[round == 1]
previousSentBack[round == 3] <- sentBack[round==2]
previousSentBack[round == 4] <- sentBack[round==3]
previousSentBack[round == 5] <- sentBack[round==4]
previousSentBack[round == 6] <- sentBack[round==5]
previousSentBack[round == 7] <- sentBack[round==6]
previousSentBack[round == 8] <- sentBack[round==7]
previousSentBack[round == 9] <- sentBack[round==8]
previousSentBack[round == 10] <- sentBack[round==9]
previousSentBack[round == 11] <- sentBack[round==10]
previousSentBack[round == 12] <- sentBack[round==11]
previousSentBack[round == 13] <- sentBack[round==12]
previousSentBack[round == 14] <- sentBack[round==13]
previousSentBack[round == 15] <- sentBack[round==14]
previousSentBack[round == 16] <- sentBack[round==15]
previousSentBack[round == 17] <- sentBack[round==16]
previousSentBack[round == 18] <- sentBack[round==17]

trials$previousSentBack <- previousSentBack

效果也很好。

现在是令人不安的变量。我这样计算比率：

ReInvestRatio = trials$ReInvestRatio

ReInvestRatio <- investment/previousSentBack

trials$ReInvestRatio <- ReInvestRatio

在上面的 lmer 中使用这个新计算的变量给了我这个错误：

qr.default(X,tol = tol,LAPACK = FALSE) 中的错误：外部函数中的 NA/NaN/Inf (arg 1)

在这里和谷歌上做了一些研究，我发现 NaN 和 Inf 是罪魁祸首，但我似乎无法删除它们。应该注意的是，我假设 NA 对我的 lmer 应该没有太大问题，因为其他一些预测变量也包含 NAs。我更愿意修复变量而不必将所有 NA/NaN/Inf 替换为 0。但是，替换似乎首先是问题所在。

我尝试了以下代码：

trials <- do.call(data.frame,lapply(trials,function(x) replace(x,is.infinite(x),NA)))

还有：

is.na(trials$ReInvestRatio) <- sapply(trials$ReInvestRatio,is.infinite)  

is.na(trials$ReInvestRatio) <- sapply(trials$ReInvestRatio,is.nan) 

都明显地从我的数据集中删除了 NaN 和 Inf，但它们似乎没有“内部”这样做，因为我仍然从 lmer 中得到相同的错误，例如sum(is.infinite(ReInvestRatio)) 仍然列出了数百个案例。

我绝对达到了我的 R 能力极限，非常感谢您的帮助！

亲切的问候， 最大

修复

毕竟没有什么真正的数学：

我使用的是 attach(trials)，因此为什么 lmer 中的所有变量都没有任何 trials$ 序言。我原以为这对 ReInvestRatio 也同样有效，但显然不是。当我将 ReInvestRatio 放入 lmer(log(investment +0.5)~ReInvestRatio +(1|SubjectID)) 时，R 不断从工作区中最初创建的变量中提取值。这个没有（或者至少在我的情况下没有）清除任何 Inf 等。trials$ReInvestRatio 就像 lmer(log(investment +0.5)~trials$ReInvestRatio +(1|SubjectID)) 在模型中有效，因为这似乎针对正确的、清洁的 ReInvestRatio 。