Question

我有一个数据框：

dt <- read.table(text = "
350 16 
366 11 
376  0
380  0
397  0
398 45  
400 19  
402 0
510 0
525 0
537 0
549 0
569 112
578 99")

我想删除第二列中所有带有零的行，除了非零值之前和之后的行。

结果将是：

dt1 <- read.table(text = "
350 16 
366 11 
376  0
397  0
398 45  
400 19  
402 0
549 0
569 112
578 99")

Answer 1

library(data.table)
setDT(dt)

dt[{n0 <- V2 != 0; n0 | shift(n0) | shift(n0, type = 'lead')}]
#or
dt[(n0 <- V2 != 0) | shift(n0) | shift(n0, type = 'lead')] # thanks @Frank

#      V1  V2
#  1: 350  16
#  2: 366  11
#  3: 376   0
#  4: 397   0
#  5: 398  45
#  6: 400  19
#  7: 402   0
#  8: 549   0
#  9: 569 112
# 10: 578  99

Answer 2

使用dplyr：

dt %>%
  filter(lag(V2, 1) != 0 | lead(V2, 1) != 0 | V2 != 0)

    V1  V2
1  350  16
2  366  11
3  376   0
4  397   0
5  398  45
6  400  19
7  402   0
8  549   0
9  569 112
10 578  99

或者：

dt %>%
  group_by(cond = lag(V2, 1) != 0 | lead(V2, 1) != 0 | V2 != 0) %>%
  filter(cond == TRUE) %>%
  ungroup() %>%
  select(-cond)

# A tibble: 10 x 2
      V1    V2
   <int> <int>
 1   350    16
 2   366    11
 3   376     0
 4   397     0
 5   398    45
 6   400    19
 7   402     0
 8   549     0
 9   569   112
10   578    99

Answer 3

使用base R比较向上和向下位移向量的简单解决方案

dt[ !(c(dt$V2[-1],0) == 0 & c(0,dt$V2[-length(dt$V2)]) == 0 & dt$V2 == 0), ]

Answer 4

与其他回答相比，这并不是什么新鲜事物，但是我发现这个问题很有趣，因此我制定了自己的解决方案-贴吧：

## Function to test if both neighbors of a vector element have the value 0
## Returns a logical vector.
neighbors_zero <- function(x) {
  ## left neighbor is zero?
  rn0 <- c(x[2:length(x)], x[1]) == 0
  ## right neighbor is zero?
  ln0 <- c(x[length(x)], x[1:(length(x)-1)]) == 0
  return(rn0 & ln0)
}

## Test if a value is itsself zero and between other zeros
zero_between_zeros <- dt$V2 == 0 & neighbors_zero(dt$V2)

dt[!zero_between_zeros, ]

在条件下删除值为零的行

4 个答案: