我有一个数据帧(my_data),即使可能有联系,我也想仅计算3个最大值的总和。我对R很陌生,并且使用过dplyr
。
A tibble: 15 x 3
city month number
<chr> <chr> <dbl>
1 Lund jan 12
2 Lund feb 12
3 Lund mar 18
4 Lund apr 28
5 Lund may 28
6 Stockholm jan 15
7 Stockholm feb 15
8 Stockholm mar 30
9 Stockholm apr 30
10 Stockholm may 10
11 Uppsala jan 22
12 Uppsala feb 30
13 Uppsala mar 40
14 Uppsala apr 60
15 Uppsala may 30
这是我尝试过的代码:
# For each city, count the top 3 of variable number
my_data %>% group_by(city) %>% top_n(3, number) %>% summarise(top_nr = sum(number))
期望的(期望的)输出为:
# A tibble: 3 x 2
city top_nr
<chr> <dbl>
1 Lund 86
2 Stockholm 75
3 Uppsala 130
但实际的R输出为:
# A tibble: 3 x 2
city top_nr
<chr> <dbl>
1 Lund 86
2 Stockholm 90
3 Uppsala 160
似乎存在联系,所有捆绑值都包含在求和中。我只希望计算3个唯一的具有最高值的实例。
任何帮助将不胜感激! :)
答案 0 :(得分:4)
我们可以执行distinct
来删除重复的元素。 top_n
的工作方式是:如果值重复,它将保留那么多重复行
my_data %>%
distinct(city, number, .keep_all = TRUE) %>%
group_by(city) %>%
top_n(3, number) %>%
summarise(top_nr = sum(number))
基于OP的新输出,在top_n
输出(不是arrange
d)之后,获得降序排列的“数字”并获得前3个的sum
'数字'
my_data %>%
group_by(city) %>%
top_n(3, number) %>%
arrange(city, desc(number)) %>%
summarise(number = sum(head(number, 3)))
# A tibble: 3 x 2
# city number
# <chr> <int>
#1 Lund 74
#2 Stockholm 75
#3 Uppsala 130
my_data <- structure(list(city = c("Lund", "Lund", "Lund", "Lund", "Lund",
"Stockholm", "Stockholm", "Stockholm", "Stockholm", "Stockholm",
"Uppsala", "Uppsala", "Uppsala", "Uppsala", "Uppsala"), month = c("jan",
"feb", "mar", "apr", "may", "jan", "feb", "mar", "apr", "may",
"jan", "feb", "mar", "apr", "may"), number = c(12L, 12L, 18L,
28L, 28L, 15L, 15L, 30L, 30L, 10L, 22L, 30L, 40L, 60L, 30L)),
class = "data.frame", row.names = c("1",
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13",
"14", "15"))
答案 1 :(得分:2)
此tidyverse
(实际上是dplyr
)解决方案几乎等于akrun's,但是filter
的数据帧而不是获取top_n
。
library(tidyverse)
my_data %>%
group_by(city) %>%
arrange(desc(number), .by_group = TRUE) %>%
filter(row_number() %in% 1:3) %>%
summarise(top_nr = sum(number))
## A tibble: 3 x 2
# city top_nr
# <chr> <int>
#1 Lund 74
#2 Stockholm 75
#3 Uppsala 130
答案 2 :(得分:2)
如果没有top_n()
,生活可能会更简单:
dat %>%
group_by(city) %>%
summarize(
top_nr = sum(tail(sort(number), 3))
)
答案 3 :(得分:1)
感谢@akrun,但是当我运行您建议的代码时,我得到:Lund 58,它是对28、18和12的总结。我想要的是一种对Lund 28 + 28 + 18 = 74进行总结的方法。 (我发现我在上面的最初描述中犯了一个错误,对此感到抱歉)。 这是预期的(期望的)结果应为:
# A tibble: 3 x 2
city top_nr
<chr> <dbl>
1 Lund 74
2 Stockholm 75
3 Uppsala 130