使用 R 中的特定行限制计算故障率

Question

我有一个这样的数据框

ID <- c("ID300","ID301","ID302","ID303","ID304","ID305","ID306","ID307","ID308","ID309")
Measurement <- c("Length","Length","Length","Length","Length","Length","Length","Length","Length","Length")
PASSFAIL <- c("FAIL","PASS","FAIL","FAIL#Pts","PASS","PASS","PASS","PASS","PASS","FAIL")

df1 <- data.frame(ID,Measurement,PASSFAIL)

第 1 部分 我正在尝试为每个 ID 创建一个失败率列。我试图计算的方式是使用 5 个 ID 的窗口。例如

Fail Rate = (Number of Fails)/(Number of Fails + Number of Pass)

ID300 <- (Fails of Row1 to Row5)/(Total from Row1 to Row5) = (3/5) = 0.6

注意：在 df1 中，PASSFAIL 列中有 FAIL 的任何内容都被视为失败。

如果窗口大小小于 5，它也应该返回 NA，因此我想要的输出看起来像这样

      ID Measurement PASSFAIL  FR
1  ID300      Length     FAIL 0.6
2  ID301      Length     PASS 0.4
3  ID302      Length     FAIL 0.4
4  ID303      Length FAIL#Pts 0.2
5  ID304      Length     PASS 0.0
6  ID305      Length     PASS 0.2
7  ID306      Length     PASS  NA
8  ID307      Length     PASS  NA
9  ID308      Length     PASS  NA
10 ID309      Length     FAIL  NA

第 2 部分 完成此操作后，我需要考虑到相同的窗口 5，重新计算添加的每个新 ID 的失败率。例如，我想要的输出是

      ID Measurement PASSFAIL  FR
1  ID296      Length     PASS 0.4
2  ID297      Length     FAIL 0.6
3  ID298      Length     PASS 0.6
4  ID299      Length     FAIL 0.6
5  ID300      Length     FAIL 0.8
6  ID301      Length     FAIL 0.6
7  ID302      Length     PASS  NA
8  ID303      Length     FAIL  NA
9  ID304      Length FAIL#Pts  NA
10 ID305      Length     PASS  NA

我目前正在通过执行类似这样的操作来计算失败率，它会为整个数据帧计算失败率。考虑到窗口大小为 5，我不知道如何使用循环顺序计算每个 ID。

setDT(df1)
# aggregate
df1 <- df1[, .( FR = (sum(PASSFAIL != "PASS")/.N))]

请提供一些意见。

Answer 1

我迷失在你的第 2 部分，但这里是第 1 部分，使用stats::filter 和grepl 调用来搜索包含"FAIL" 的所有值：

df1$FR <- NA
vals <- na.omit(filter(grepl("FAIL",df1$PASSFAIL), rep(1,5)/5, sides=1))
df1$FR[seq(1,length(vals))] <- vals

df1
#      ID Measurement PASSFAIL  FR
#1  ID300      Length     FAIL 0.6
#2  ID301      Length     PASS 0.4
#3  ID302      Length     FAIL 0.4
#4  ID303      Length FAIL#Pts 0.2
#5  ID304      Length     PASS 0.0
#6  ID305      Length     PASS 0.2
#7  ID306      Length     PASS  NA
#8  ID307      Length     PASS  NA
#9  ID308      Length     PASS  NA
#10 ID309      Length     FAIL  NA

或者：

rev(filter(grepl("FAIL",rev(df1$PASSFAIL)), rep(1,5)/5, sides=1))

如果你想变得花哨。

Answer 2

您可能想尝试 sapply 函数，为了良好的顺序，我也将声明 df1 没有因素。

df1 <- data.frame(ID,Measurement,PASSFAIL,stringsAsFactors = FALSE)
df1$FR <- sapply(df1$ID,FUN = function(x) {
  if(which(df1$ID == x) > nrow(df1)-4){
    return(NA_real_)
  }else{
    start_ID <- which(df1$ID == x)
    end_ID <- start_ID + 4
    return(sum(grepl("FAIL",df1[start_ID:end_ID,"PASSFAIL"]))/5)
  }
})