示例数据
dat <- data.table(yr = c(2013,2013,2013,2013,2013,2013,2013,2013,2013,2013,2012,2012,2012,2012,2012,2012,2012,2012,2012,2012,2012),
location = c("Bh","Bh","Bh","Bh","Bh","Go","Go","Go","Go","Go","Bh","Bh","Bh","Bh","Bh","Bh","Go","Go","Go","Go","Go"),
time.period = c("t4","t5","t6","t7","t8","t3","t4","t5","t6","t7","t3","t4","t5","t6","t7","t8","t3","t4","t5","t6","t7"),
period = c(20,21,22,23,24,19,20,21,22,23,19,20,21,22,23,24,19,20,21,22,23),
value = c(runif(21)))
key <- data.table(time.period = c("t1","t2","t3","t4","t5","t6","t7","t8","t9","t10"),
period = c(17,18,19,20,21,22,23,24,25,26))
key为每个time.period相关联的period
在数据表dat中，对于每个location和yr，如果缺少一对time.period和period，我想插入其他行
例如。位置Bh和yr 2013
dat[location == "Bh" & yr == 2013,]
yr location time.period period value
1: 2013 Bh t4 20 0.7167561
2: 2013 Bh t5 21 0.5659722
3: 2013 Bh t6 22 0.8549229
4: 2013 Bh t7 23 0.1046213
5: 2013 Bh t8 24 0.8144670
我想这样做：
yr location time.period period value
1: 2013 Bh t1 17 0
1: 2013 Bh t2 18 0
1: 2013 Bh t3 19 0
1: 2013 Bh t4 20 0.7167561
2: 2013 Bh t5 21 0.5659722
3: 2013 Bh t6 22 0.8549229
4: 2013 Bh t7 23 0.1046213
5: 2013 Bh t8 24 0.8144670
1: 2013 Bh t9 25 0
1: 2013 Bh t10 26 0
我试过了：
dat %>% group_by(location,yr) %>% complete(period = seq(17, max(26), 1L))
A tibble: 40 x 5
Groups: location, yr [4]
location yr period time.period value
<chr> <dbl> <dbl> <chr> <dbl>
1 Bh 2012 17 <NA> NA
2 Bh 2012 18 <NA> NA
3 Bh 2012 19 t3 0.46757583
4 Bh 2012 20 t4 0.07041745
5 Bh 2012 21 t5 0.58707367
6 Bh 2012 22 t6 0.83271673
7 Bh 2012 23 t7 0.76918731
8 Bh 2012 24 t8 0.25368225
9 Bh 2012 25 <NA> NA
10 Bh 2012 26 <NA> NA
# ... with 30 more rows
如您所见，time.period未填写。我如何填写该栏目？
答案 0 :(得分：3)
tidyr::complete可用于查找解决方案。
library(dplyr)
library(tidyr)
dat %>% complete(yr, location, key, fill = list(value = 0)) )
# # A tibble: 40 x 5
# yr location time.period period value
# <dbl> <chr> <chr> <dbl> <dbl>
# 1 2012 Bh t1 17.0 0
# 2 2012 Bh t2 18.0 0
# 3 2012 Bh t3 19.0 0.177
# 4 2012 Bh t4 20.0 0.687
# 5 2012 Bh t5 21.0 0.384
# 6 2012 Bh t6 22.0 0.770
# 7 2012 Bh t7 23.0 0.498
# 8 2012 Bh t8 24.0 0.718
# 9 2012 Bh t9 25.0 0
# 10 2012 Bh t10 26.0 0
# # ... with 30 more rows
数据
dat <- data.table(yr = c(2013,2013,2013,2013,2013,2013,2013,2013,2013,2013,2012,2012,2012,2012,2012,2012,2012,2012,2012,2012,2012),
location = c("Bh","Bh","Bh","Bh","Bh","Go","Go","Go","Go","Go","Bh","Bh","Bh","Bh","Bh","Bh","Go","Go","Go","Go","Go"),
time.period = c("t4","t5","t6","t7","t8","t3","t4","t5","t6","t7","t3","t4","t5","t6","t7","t8","t3","t4","t5","t6","t7"),
period = c(20,21,22,23,24,19,20,21,22,23,19,20,21,22,23,24,19,20,21,22,23),
value = c(runif(21)))
key <- data.table(time.period = c("t1","t2","t3","t4","t5","t6","t7","t8","t9","t10"),
period = c(17,18,19,20,21,22,23,24,25,26))
答案 1 :(得分：2)
由于您使用的是data.table，因此您可以执行以下操作：
dat_new <- dat[,.SD[key, on='time.period'],.(location, yr)]
dat_new[, period := i.period][, i.period := NULL]
dat_new[is.na(value), value := 0]
print(head(dat_new), 10)
location yr time.period period value
1: Bh 2013 t1 17 0.0000000
2: Bh 2013 t2 18 0.0000000
3: Bh 2013 t3 19 0.0000000
4: Bh 2013 t4 20 0.9255600
5: Bh 2013 t5 21 0.3816035
6: Bh 2013 t6 22 0.5202268
7: Bh 2013 t7 23 0.5326466
8: Bh 2013 t8 24 0.5091590
9: Bh 2013 t9 25 0.0000000
10: Bh 2013 t10 26 0.0000000
<强>解释
1.首先，我们将key数据框与dat中的每组.(location, yr)一起加入。
2.这会将列键数据帧添加为i.period。
3.最后，我们将NA设置为0并在设置i.period后删除period := i.period列。
答案 2 :(得分：0)
你需要这样的东西吗？
x <- merge(dat, key, by = "time.period", all.y = T)
x[is.na(x)] <- 0