回馈姓名

Question

我的df看起来像这样：

我希望获得至少连续2次出现1的所有姓名。

在这种情况下，我只想回到B C D F。

有没有办法做到这一点？

Answer 1

我们可以循环遍历行，使用rle（查找是否存在连续元素）并创建逻辑索引以对＆＃39; Names＆＃39;

进行子集化

df1$Name[apply(df1[-1], 1, function(x) {
                 rl <- rle(x==1)
            any(rl$lengths[rl$values]>=2)})]
#[1] "B" "C" "D" "F"

更快的方法可能是paste每行中的元素，然后使用正则表达式查找以查找1是否后跟1

df1$Name[grepl("(?<=1)1", do.call(paste0, df1[-1]), perl = TRUE)]
#[1] "B" "C" "D" "F"

数据

df1 <- structure(list(Name = c("A", "B", "C", "D", "E", "F"), `2000` = c(1L, 
0L, 1L, 1L, 0L, 0L), `2001` = c(0L, 0L, 1L, 1L, 1L, 1L), `2002` = c(0L, 
1L, 0L, 1L, 0L, 0L), `2003` = c(1L, 1L, 0L, 1L, 1L, 1L), `2004` = c(0L, 
0L, 1L, 1L, 0L, 1L), `2005` = c(1L, 1L, 1L, 0L, 1L, 1L)), .Names = c("Name", 
"2000", "2001", "2002", "2003", "2004", "2005"), class = "data.frame", 
 row.names = c(NA, -6L))

Answer 2

使用lag，这只适用于连续2，如果您需要增加限额，可以查看akrun's answer

s=df$Name[apply(df[,c(2:6)],1,function(x) any((lag(x)==x)&x==1))]
s[!is.na(s)]
[1] "B" "C" "D" "F"

Answer 3

使用zoo::rollapply的选项：

library(zoo)

#One can increase the n for 3, 4 etc consecutive values
# For consecutive 2 values
n =2
df1[apply(df1[-1], 1, function(x)
  length(which(rollapply(x, width=n, function(x)all(x==rep(1,n)), align = "left")))>0),1]

#[1] "B" "C" "D" "F"