忽略要应用于数据帧列表 R 的 Ifelse 语句 R 中的 NA

Question

我有一个函数可以计算数据帧列表中多列值的 z 分数。下面是我的数据帧的 sn-p

df <- list(Al2O3 = structure(list(Determination_No = c(1, 2, 3, 4, 
5, 6, 7, 8, 9, 10), `2` = c(2.04, 2.07, 2.05, 2.07, 2.1, 2.08, 
NA, NA, NA, NA), `3` = c(2.08, 2.1, 2.08, 2.13, 2.1, 2.08, NA, 
NA, NA, NA), `4` = c(2.08, 2.08, 2.09, 2.06, 2.08, 2.07, 2.07, 
2.06, 2.08, 2.08), `5` = c(2.11, 2.09, 2.1, 2.08, 2.09, 2.09, 
NA, NA, NA, NA), `7` = c(2.06, 2.05, 2.04, 2.05, 2.04, 2.03, 
NA, NA, NA, NA), `8` = c(2.078, 2.065, 2.057, 2.063, 2.067, 2.066, 
NA, NA, NA, NA), `10` = c(2.191776681, 2.153987428, 2.153987428, 
2.097303548, 2.116198175, 2.116198175, NA, NA, NA, NA), `12` = c(2.24, 
2.08, 2.12, 2.15, 2.15, 2.15, NA, NA, NA, NA), `36` = c(2.07, 
2.082, 2.048, 2.046, 2.086, 2.069, NA, NA, NA, NA)), class = "data.frame", row.names = c(NA, 
-10L)), As = structure(list(Determination_No = c(1, 2, 3, 4, 
5, 6, 7, 8, 9, 10), `2` = c(0.002, 0.001, 0.001, 0.001, 0.002, 
0.001, NA, NA, NA, NA), `3` = c(0.003, 0.002, 0.002, 0.002, 0.001, 
0.002, NA, NA, NA, NA), `4` = c(0.001, 0.002, 0.001, 0.002, 0.002, 
0.002, 0.001, 0.002, 0.002, 0.003), `5` = c(0.002, 0.001, 0.001, 
0.001, 0.001, 0.002, NA, NA, NA, NA), `7` = c(NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_), `8` = c(NA, 0.001, NA, NA, NA, NA, NA, NA, NA, NA), 
    `10` = c(NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
    NA_real_, NA_real_, NA_real_, NA_real_, NA_real_), `12` = c(NA_real_, 
    NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
    NA_real_, NA_real_, NA_real_), `36` = c(0.0053, 0.0053, 0.0053, 
    0.00454, 0.0053, 0.0053, NA, NA, NA, NA)), class = "data.frame", row.names = c(NA, 
-10L)), Ba = structure(list(Determination_No = c(1, 2, 3, 4, 
5, 6, 7, 8, 9, 10), `2` = c(NA_real_, NA_real_, NA_real_, NA_real_, 
NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_), 
    `3` = c(NA, NA, NA, NA, 0.001, NA, NA, NA, NA, NA), `4` = c(0.004, 
    0.003, 0.003, 0.004, 0.003, 0.002, 0.004, 0.002, 0.005, NA
    ), `5` = c(NA, NA, NA, NA, NA, 0.003, NA, NA, NA, NA), `7` = c(NA_real_, 
    NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
    NA_real_, NA_real_, NA_real_), `8` = c(0.002, 0.003, NA, 
    NA, NA, 0.002, NA, NA, NA, NA), `10` = c(NA_real_, NA_real_, 
    NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
    NA_real_, NA_real_), `12` = c(NA_real_, NA_real_, NA_real_, 
    NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, NA_real_, 
    NA_real_), `36` = c(0.00089566, 0.00089566, 0.00089566, 0.00089566, 
    0.00089566, 0.00089566, NA, NA, NA, NA)), class = "data.frame", row.names = c(NA, 
-10L))

我的目的是让函数根据 Z 得分值计算更多统计数据。我的挑战是我的数据框中有很多 NA。当我去应用我的 if 语句时，由于存在 NA 值，它不起作用。我的功能如下

ZMax <- 3.5
FinalStats <- function(x,...){ 
  unlistdata <- unlist(x[-1])
  GrandMean <- mean(unlistdata,na.rm = T)
  GrandSD <- sd(unlistdata,na.rm=T)
  ZScore <- abs(((x[-1])-GrandMean)/GrandSD)

  if(ZScore > ZMax){
    LabMean <- mapply(mean, x[-1], na.rm = T) #Calculate Mean by columns
    SD.All <- unlist(x[-1])
    ConsensusValue <- mean(LabMean)
    Uncertainty <- sd(SD.All, na.rm = T)
  }else{ 
    LabMedian <- mapply(median, x[-1], na.rm = T) #Calculate Median by columns
    LabMedian[is.infinite(LabMedian)] <- NA #convert any Inf values to NA
    SD.All <- unlist(x[-1])
    ConsensusValue <- LabMedian
    Uncertainty <- sd(SD.All, na.rm = T)
  }

  FinalValues <- cbind(ConsensusValue,Uncertainty) #combined the desired Info
  
  return(FinalValues)
}

df.stats <- lapply(df,FinalStats)

如何让 if 语句忽略 NA 值？

我已尝试通过以下方式在 base R 中使用 ifelse

FinalStats <- function(x,...){ 
  unlistdata <- unlist(x[-1])
  GrandMean <- mean(unlistdata,na.rm = T)
  GrandSD <- sd(unlistdata,na.rm=T)
  ZScore <- abs(((x[-1])-GrandMean)/GrandSD)
    
  ConsensusValue   <- ifelse((is.na(ZScore > ZMax)),
                        mean(mapply(mean, x[-1], na.rm = T)),
                       median(mapply(median,x[-1],na.rm=T))) 
  
  return(ConsensusValue)
} 

不幸的是，我尝试使用 ifelse 语句部分适用于第一个数据帧，并且在我的示例中仅在其他两个数据帧上返回 NA。

我正在寻找的结果是单个值，它可以是每列平均值的平均值，也可以是每列中位数的中位数。根据 Z 分数，我得到的是具有平均值（看起来正确）或一系列 NA 的数据帧列表

我尝试在 ifelse 语句之外计算平均值和中值，然后使用 ifelse 语句来选择我想要的值，但我得到的是值的数据框，而不是单个值。但是，如果我返回在 ifelse 之外计算的平均值或中位数，那么我会得到正确的结果。

FinalStats <- function(x,...){ 
  unlistdata <- unlist(x[-1])
  GrandMean <- mean(unlistdata,na.rm = T)
  GrandSD <- sd(unlistdata,na.rm=T)
  ZScore <- abs(((x[-1])-GrandMean)/GrandSD)
  
  LabMean <- mean(mapply(mean, x[-1], na.rm = T),na.rm=T) #Calculate Mean by columns
  LabMedian <- median(mapply(median, x[-1], na.rm = T),na.rm = T) #Calculate Median by columns
  LabMedian[is.infinite(LabMedian)] <- NA #convert any Inf values to NA
    

  ConsensusValue <- ifelse(!is.na(ZScore > ZMax),
                       LabMean,
                      LabMedian)
  return(ConsensusValue)
}   
CatergoreisStats <- lapply(df,FinalStats) 

我已经意识到我的 if 语句正在检查每个数据帧中的每个值，然后分配平均值或中值。我想要做的是检查每个数据帧是否有任何值超过我的 Zmax 然后分配平均值或中值。

Answer 1

我认为您应该使用if(any(...)) 作为条件，因为您想检查ZScore 中的任何值是否大于ZMax。 NA 的值可以在 any 中使用 na.rm = TRUE 忽略。

ZMax <- 3.5

FinalStats <- function(x,...){ 
  unlistdata <- unlist(x[-1])
  GrandMean <- mean(unlistdata,na.rm = T)
  GrandSD <- sd(unlistdata,na.rm=T)
  ZScore <- abs(((x[-1])-GrandMean)/GrandSD)
  if(any(ZScore > ZMax, na.rm = TRUE)){
    LabMean <- mapply(mean, x[-1], na.rm = T) #Calculate Mean by columns
    SD.All <- unlist(x[-1])
    ConsensusValue <- mean(LabMean)
    Uncertainty <- sd(SD.All, na.rm = T)
  }else{ 
    LabMedian <- mapply(median, x[-1], na.rm = T) #Calculate Median by columns
    LabMedian[is.infinite(LabMedian)] <- NA #convert any Inf values to NA
    SD.All <- unlist(x[-1])
    ConsensusValue <- LabMedian
    Uncertainty <- sd(SD.All, na.rm = T)
  }
  
  FinalValues <- cbind(ConsensusValue,Uncertainty) #combined the desired Info
  
  return(FinalValues)
}

这会返回 -

CatergoreisStats <- lapply(df,FinalStats)
CatergoreisStats
#$Al2O3
#     ConsensusValue Uncertainty
#[1,]       2.088453  0.03880474

#$As
#   ConsensusValue Uncertainty
#2          0.0010 0.001475832
#3          0.0020 0.001475832
#4          0.0020 0.001475832
#5          0.0010 0.001475832
#7              NA 0.001475832
#8          0.0010 0.001475832
#10             NA 0.001475832
#12             NA 0.001475832
#36         0.0053 0.001475832

#$Ba
#   ConsensusValue Uncertainty
#2              NA 0.001303559
#3      0.00100000 0.001303559
#4      0.00300000 0.001303559
#5      0.00300000 0.001303559
#7              NA 0.001303559
#8      0.00200000 0.001303559
#10             NA 0.001303559
#12             NA 0.001303559
#36     0.00089566 0.001303559