Question

我有一个看起来像这样的数据框（但有更多的变量/列）

set.seed(5)
id<-seq(5)*floor(runif(5,min=1000, max=10000))
vals1<-c("Y","N","N","N","N")
vals2<-c("N","N","N","N","N")
vals3<-c("N","N","N","Y","N")
df<-data.frame(id,vals1,vals2,vals3)

我想在框架中创建一个最终列，以便生成一个具有以下逻辑的最终标志：如果有任何值，那么＆＃39; Y＆＃39;对于任何id，最终标志是＆＃39; Y＆＃39;否则它将是＆＃39; N＆＃39;因此，对于这个数据帧，第1和第4个ID（2801,14236）有一个＆＃39; Y＆＃39;在最后一栏中，其余的有一个＆＃39; n＆＃39;为最后一栏。我尝试了一些方法，如申请，如果......其他无济于事。

Answer 1

通过为每一行指定“N”进行初始化。在下一步中，对于带有“Y”的行（使用apply检查），请指定“Y”

df$final = "N"
df$final[apply(df, 1, function(a) "Y" %in% a)] = "Y"

Answer 2

下面的信件编码解决方案。

set.seed(5)
id <- seq(5) * floor(runif(5, min=1000, max=10000))
vals1 <- c("Y","N","N","N","N")
vals2 <- c("N","N","N","N","N")
vals3 <- c("N","N","N","Y","N")

df <- data.frame(id, vals1, vals2, vals3)

# If you really want to use the letter encoding, my solution works as below
df$Final <- apply(df[,2:4], MARGIN = 1, FUN = function(x) {any(x == 'Y')})

但是，我认为你应该使用布尔值（TRUE / FALSE）。

与apply和any

结合使用效果很好

set.seed(5)
id <- seq(5) * floor(runif(5, min=1000, max=10000))
vals1 <- c("Y","N","N","N","N")
vals2 <- c("N","N","N","N","N")
vals3 <- c("N","N","N","Y","N")

df <- data.frame(id, vals1, vals2, vals3)

# Convert your labels into booleans:
df[,2:4] <- df[,2:4] == 'Y'

# Then summarise across rows
df$Final <- apply(df[,2:4], MARGIN = 1, FUN = function(x) {any(x)})

Answer 3

与@ d.b回答有点相似：

df$final <- apply(df, 1, function(x) c("N","Y")[any(x == "Y")+1])

基于多列在数据框的最后一列中创建值

3 个答案: