Question

我有一个数据集，想要计算3个不同的条件

coins和copay列中包含“0”的行数 - 将任一列中的N / A计为0
coins.copay列= 1但排除满足条件1的值的行数
coins.copay列= 0但排除满足条件1的值的行数

以下是来自更大数据集的示例：

Plan   Year   Coins   Copay   Type   Coins.Copay
 A     2018     0      NA      HMO        1
 B     2018    10      NA      HMO        1
 C     2017    NA       0      SNP        0
 D     2015    20      20      SNP        0
 E     2016    20       0      HMO        1
 F     2018    10      10      HMO        0
 G     2016    NA      NA      HMO        0
 H     2014    NA      NA      HMO        0
 I     2012    NA      10      PPO        0
 J     2011     0       0      HMO        0
 K     2014     5      10      SNP        0
 L     2013    10      NA      HMO        1

因此，我希望得到以下数量（基于上述条件）：

5（计划A，C，G，H，J满足条件）
3（方案B，E，L满足条件;方案A不满足条件1）
4（计划D，F，I，K满足条件;计划C，G，H，J不计入满足条件1）

Answer 1

这可以使用布尔逻辑非常有效地完成：

zeros_or_na <- (is.na(df$Coins) | !df$Coins) & (is.na(df$Copay) | !df$Copay)
sum(zeros_or_na)                     # [1] 5
sum(df$Coins.Copay & !zeros_or_na)   # [1] 3
sum(!df$Coins.Copay & !zeros_or_na)  # [1] 4

Answer 2

一个选项可能是：

方法：

1.在内部条件下使用&因此决定会很快。

2.筛选条件2&3，因此过滤器仅应用一次。

3.使用过滤后的数据计算了一次条件2&3，从总行数中减去sum个来计算   条件1

Excluded_conditino_one = which((!is.na(df$Coins) & 
                  df$Coins) | (!is.na(df$Copay) & df$Copay))


coins.copay_1 = sum(df[Excluded_conditino_one,"Coins.Copay"]==1) #3
coins.copay_0 = sum(df[Excluded_conditino_one,"Coins.Copay"]==0) #4


Condition_One = length(df$Plan) - (coins.copay_1+coins.copay_0)  #5


#Test
paste(Condition_One, coins.copay_1, coins.copay_0)
[1] "5 3 4"

Workbench性能分析：

CBarun <- function(df){
  zeros_or_na <- (is.na(df$Coins) | !df$Coins) & (is.na(df$Copay) | !df$Copay)
  Condition_One = sum(zeros_or_na)                     # [1] 5
  coins.copay_1 = sum(df$Coins.Copay & !zeros_or_na)   # [1] 3
  coins.copay_0 = sum(!df$Coins.Copay & !zeros_or_na)  # [1] 4
}


Masoud <- function(df){
  Condition_One = length(which(rowSums(cbind(df$Coins, df$Copay), na.rm=T)==0)) #5

  coins.copay_1 = length(which(rowSums(cbind(df$Coins, df$Copay), na.rm=T)!=0 &
  df$Coins.Copay!=0)) #3

  coins.copay_0 = length(which(rowSums(cbind(df$Coins, df$Copay), na.rm=T)!=0 &
  df$Coins.Copay==0))   
}

MKR <- function(df){

  Excluded_conditino_one = which((!is.na(df$Coins) &
          df$Coins) | (!is.na(df$Copay) & df$Copay))

  coins.copay_1 = sum(df[Excluded_conditino_one,"Coins.Copay"]==1) #3
  coins.copay_0 = sum(df[Excluded_conditino_one,"Coins.Copay"]==0) #4


  Condition_One = length(df$Plan) - (coins.copay_1+coins.copay_0)  #5
}


library(microbenchmark)

microbenchmark(CBarun(df),
               #AyushNigam(),
               Masoud(df),
               MKR(df),
               times = 10
)

# Unit: microseconds
#       expr     min      lq     mean   median      uq     max neval
# CBarun(df)  60.790  61.185  71.7644  62.7645  67.896 137.370    10
# Masoud(df) 185.923 186.317 209.9624 201.3180 222.633 273.949    10
# MKR(df)    101.054 102.633 122.1330 106.1850 121.975 227.370    10

数据

df <- read.table(text = "Plan Year Coins Copay Type Coins.Copay A 2018 0 NA HMO 1 B 2018 10 NA HMO 1 C 2017 NA 0 SNP 0 D 2015 20 20 SNP 0 E 2016 20 0 HMO 1 F 2018 10 10 HMO 0 G 2016 NA NA HMO 0 H 2014 NA NA HMO 0 I 2012 NA 10 PPO 0 J 2011 0 0 HMO 0 K 2014 5 10 SNP 0 L 2013 10 NA HMO 1", header = TRUE, stringsAsFactors = FALSE)

Answer 3

另一种选择是将length与rowSums：

一起使用

length(which(rowSums(cbind(df$Coins, df$Copay), na.rm=T)==0)) #5

length(which(rowSums(cbind(df$Coins, df$Copay), na.rm=T)!=0 & df$Coins.Copay!=0)) #3

length(which(rowSums(cbind(df$Coins, df$Copay), na.rm=T)!=0 & df$Coins.Copay==0)) #4

基于3个条件的计数

3 个答案: