[英]Get a value based on the value of another column in R - dplyr
我得到了这个 df:
df <- data.frame(month = c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4),
day = c(1,2,3,4,5,1,2,3,4,5,1,2,3,4,5,1,2,3,4,5),
flow = c(2,5,7,8,5,4,6,7,9,2,NA,1,6,10,2,NA,NA,NA,NA,NA))
我想达到这个结果:
month day flow dayofminflow
1 1 1 2 1
2 1 2 5 1
3 1 3 7 1
4 1 4 8 1
5 1 5 5 1
6 2 1 4 5
7 2 2 6 5
8 2 3 7 5
9 2 4 9 5
10 2 5 2 5
11 3 1 NA 2
12 3 2 1 2
13 3 3 6 2
14 3 4 10 2
15 3 5 2 2
16 4 1 NA NA
17 4 2 NA NA
18 4 3 NA NA
19 4 4 NA NA
20 4 5 NA NA
我正在使用这个解决方案,但当第一个值为 NA 时它返回 NA:
newdf <- df %>% group_by(month) %>% mutate(Val=day[flow==min(flow)][1])
当所有数据均为 NA 时,此解决方案会返回错误:
library(dplyr)
df <- df %>%
group_by(month) %>%
mutate(dayminflowofthemonth = day[which.min(flow)]) %>%
ungroup
您只需更改第一个解决方案中min()
中的默认na.rm = TRUE
即可忽略 NA 吗?
df %>%
group_by(month) %>%
mutate(dayofminflow = day[which(min(flow, na.rm = TRUE) == flow)][1])
# A tibble: 20 x 4
# Groups: month [4]
month day flow dayofminflow
<dbl> <dbl> <dbl> <dbl>
1 1 1 2 1
2 1 2 5 1
3 1 3 7 1
4 1 4 8 1
5 1 5 5 1
6 2 1 4 5
7 2 2 6 5
8 2 3 7 5
9 2 4 9 5
10 2 5 2 5
11 3 1 NA 2
12 3 2 1 2
13 3 3 6 2
14 3 4 10 2
15 3 5 2 2
16 4 1 NA NA
17 4 2 NA NA
18 4 3 NA NA
19 4 4 NA NA
20 4 5 NA NA
尽管您收到警告no non-missing arguments to min; returning Inf
从第4
个月开始no non-missing arguments to min; returning Inf
,因为所有流量值都是 NA。
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