[英]Create column with dplyr based on value and also frequency of another column, in R
[英]Create a block column based on id and the value of another column in R
给定以下前两列(id和time_diff),我想生成“ block”列
test
id time_diff block
1 a NA 1
2 a 1 1
3 a 1 1
4 a 1 1
5 a 3 1
6 a 3 1
7 b NA 2
8 b 11 3
9 b 1 3
10 b 1 3
11 b 1 3
12 b 12 4
13 b 1 4
14 c NA 5
15 c 4 5
16 c 7 5
数据已经按ID和时间排序。 在给定相同ID的情况下,time_diff是基于前一时间与该行的时间值之差计算得出的。 我想创建一个块ID,它是一个自动递增的值,并在遇到新ID或具有相同ID的time_diff> 10时增加。
如何在R中实现这一目标?
将您的数据作为数据框导入,如下所示:
df = read.table(text='
id time_diff block
1 a NA 1
2 a 1 1
3 a 1 1
4 a 1 1
5 a 3 1
6 a 3 1
7 b NA 2
8 b 11 3
9 b 1 3
10 b 1 3
11 b 1 3
12 b 12 4
13 b 1 4
14 c NA 5
15 c 4 5
16 c 7 5')
您可以像这样进行单线获取满足您两个条件的事件:
> new_col = as.vector(cumsum(
na.exclude(
c(F,diff(as.numeric(as.factor(df$id)))) | # change of id OR
df$time_diff > 10 # time_diff greater than 10
)
))
> new_col
[1] 0 0 0 0 0 1 2 2 2 2 3 3 4 4 4
最后,使用cbind
将此新列追加到数据cbind
:
> cbind(df, block = c(0,new_col))
id time_diff block block
1 a NA 1 0
2 a 1 1 0
3 a 1 1 0
4 a 1 1 0
5 a 3 1 0
6 a 3 1 0
7 b NA 2 1
8 b 11 3 2
9 b 1 3 2
10 b 1 3 2
11 b 1 3 2
12 b 12 4 3
13 b 1 4 3
14 c NA 5 4
15 c 4 5 4
16 c 7 5 4
您会发现所需的block
变量与我的block
变量之间存在偏移量:纠正它很容易,可以在几个不同的步骤中完成,我留给您:)
@Jealie方法的另一个变体是:
with(test, cumsum(c(TRUE,id[-1]!=id[-nrow(test)])|time_diff>10))
#[1] 1 1 1 1 1 1 2 3 3 3 3 4 4 5 5 5
向Jealie和akrun学习之后,我想到了这个想法。
mydf %>%
mutate(group = cumsum(time_diff > 10 |!duplicated(id)))
# id time_diff block group
#1 a NA 1 1
#2 a 1 1 1
#3 a 1 1 1
#4 a 1 1 1
#5 a 3 1 1
#6 a 3 1 1
#7 b NA 2 2
#8 b 11 3 3
#9 b 1 3 3
#10 b 1 3 3
#11 b 1 3 3
#12 b 12 4 4
#13 b 1 4 4
#14 c NA 5 5
#15 c 4 5 5
#16 c 7 5 5
这是使用dplyr
的方法:
require(dplyr)
set.seed(999)
test <- data.frame(
id = rep(letters[1:4], each = 3),
time_diff = sample(4:15)
)
test %>%
mutate(
b = as.integer(id) - lag(as.integer(id)),
more10 = time_diff > 10,
increment = pmax(b, more10, na.rm = TRUE),
increment = ifelse(row_number() == 1, 1, increment),
block = cumsum(increment)
) %>%
select(id, time_diff, block)
尝试:
> df
id time_diff
1 a NA
2 a 1
3 a 1
4 a 1
5 a 3
6 a 3
7 b NA
8 b 11
9 b 1
10 b 1
11 b 1
12 b 12
13 b 1
14 c NA
15 c 4
16 c 7
block= c(1)
for(i in 2:nrow(df))
block[i] = ifelse(df$time_diff[i]>10 || df$id[i]!=df$id[i-1],
block[i-1]+1,
block[i-1])
df$block = block
df
id time_diff block
1 a NA 1
2 a 1 1
3 a 1 1
4 a 1 1
5 a 3 1
6 a 3 1
7 b NA 2
8 b 11 3
9 b 1 3
10 b 1 3
11 b 1 3
12 b 12 4
13 b 1 4
14 c NA 5
15 c 4 5
16 c 7 5
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