R 如何根据另一列中的更改创建二进制变量？

我试图计算每个ID的柱面从西向东和从西向东的变化次数

structure(list(ID = c(30767L, 30767L, 30767L, 30767L, 30767L, 
30767L, 30767L, 30759L, 30759L, 30759L, 30759L, 30759L, 30759L, 
30759L, 30759L, 30759L, 30759L, 30759L, 30759L, 30759L, 30759L, 
30759L, 30759L, 30759L, 30759L, 30759L, 30759L, 30759L, 30759L, 
30759L, 30759L, 30759L, 30759L, 30759L, 30759L, 30759L, 30759L, 
30759L, 30759L, 30759L), shore = c("West", "West", "West", "West", 
"West", "West", "West", "West", "West", "West", "West", "West", 
"East", "West", "East", "East", "West", "West", "West", "West", 
"West", "West", "West", "West", "West", "West", "East", "West", 
"West", "West", "West", "West", "East", "East", "East", "East", 
"East", "East", "East", "East")), row.names = c(NA, -40L), groups = structure(list(
    ID = c(30759L, 30767L), .rows = list(8:40, 1:7)), row.names = c(NA, 
-2L), class = c("tbl_df", "tbl", "data.frame"), .drop = TRUE), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"))

基本上，我首先要做的是确定没有变化为-东西向运动为0，东西向运动为1。。见下面的例子

      ID Shore Direction
1  30759  West         -
2  30759  West         -
3  30759  West         -
4  30759  East         0
5  30759  West         1
6  30759  East         0
7  30759  East         -
8  30759  West         1
9  30759  West         -
10 30759  West         -

---

按

ID

分组，然后使用

case\u when

和

lag

计算变量

library(dplyr)

DF %>%
  group_by(ID) %>%
  mutate(dir = case_when(
    shore == "West" & lag(shore) == "East" ~ 1L,
    shore == "East" & lag(shore) == "West" ~ 0L,
    TRUE ~ NA_integer_)) %>%
  ungroup

---

这里有一种方法，使用

dplyr

：

df %>% 
  dplyr::mutate(prev = lag(shore),
                direction = dplyr::case_when(shore == "West" & prev == "East" ~ 1,
                                             shore == "East" & prev == "West" ~ 0,
                                             TRUE ~ NA_real_))

---

lag（）

函数给出了

shore

列的上一个条目（在本例中）。然后，我添加了一个方向列，当方向从东到西变化时，它是

1

，当方向从西到东变化时，它是

0

，反之亦然。然后，您可以删除

prev

列。

我认为这不是我想要的，请参阅编辑后的文章。请修改它。