Question

我正在尝试创建一个连续的相等值，包括出现次数。但是，即使行保持连续，我也希望一旦引入新的ID后就可以重置计数。

我的数据示例如下：

dataset <- data.frame(ID = 
c("a","a","a","a","a","a","a","b","b","b","b","b","b","b")
dataset$YesNO <- c(1,1,0,0,0,1,1,1,1,1,0,0,0,0)

所以我想用以下结果创建一个新列：

c(1,2,1,2,3,1,2,1,2,3,1,2,3,4)

我使用了在该论坛上找到的以下代码：

dataset$Counter <- sequence(rle(as.character(dataset$YesNo))$lengths)

但是，这不会重置新ID号的计数。取而代之的是连续计数，结果输出为：

c(1,2,1,2,3,1,2,3,4,5,1,2,3,4)

我缺少根据ID进行重置的步骤。

谢谢！

Answer 1

您可以这样做：

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          <v-card-text>Test</v-card-text>
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          <v-card-text>Test</v-card-text>
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</div>

输出：

dataset$Counter <- with(dataset,
                        ave(YesNO, ID, FUN = function(x) sequence(rle(as.character(x))$lengths)))

Answer 2

使用rleid（来自data.table包）获取分组变量，然后使用ave在该分组的通用值内应用seq_along：

library(data.table)
transform(dataset, Counter = ave(YesNO, rleid(ID, YesNO), FUN = seq_along))

给予：

   ID YesNO Counter
1   a     1       1
2   a     1       2
3   a     0       1
4   a     0       2
5   a     0       3
6   a     1       1
7   a     1       2
8   b     1       1
9   b     1       2
10  b     1       3
11  b     0       1
12  b     0       2
13  b     0       3
14  b     0       4

Answer 3

还有一种dplyr可能性：

dataset %>%
 group_by(ID, grp = {grp = rle(YesNO); rep(seq_along(grp$lengths), grp$lengths)}) %>%
 mutate(Counter = seq_along(grp)) %>%
 ungroup() %>%
 select(-grp)

   ID    YesNO Counter
   <fct> <dbl>   <int>
 1 a        1.       1
 2 a        1.       2
 3 a        0.       1
 4 a        0.       2
 5 a        0.       3
 6 a        1.       1
 7 a        1.       2
 8 b        1.       1
 9 b        1.       2
10 b        1.       3
11 b        0.       1
12 b        0.       2
13 b        0.       3
14 b        0.       4

R：在单个列中按组统计值的连续出现

3 个答案: