Question

我有以下data.frame：

x <- data.frame(product = c(1,1,1,2,2,2), 
        value_day = c(0,0,150.23,110.98,18.65,0), 
        my_date = c("2019-01-01","2019-01-02","2019-01-03","2019-01-01","2019-01-02","2019-01-03"))

我的分析如下：

library(tidyverse)

x %>% 
  group_by(product) %>% 
  mutate(total = sum(value_day)) %>% 
  arrange(my_date) %>% 
  slice(c(1, n())) %>% 
  spread(key = my_date, value = total) %>% 
  filter(value_day > 0)

# A tibble: 2 x 4
# Groups:   product [2]
  product value_day `2019-01-01` `2019-01-03`
    <dbl>     <dbl>        <dbl>        <dbl>
1       1      150.          NA          150.
2       2      111.         130.          NA

但是，我想把日期设为例。像这样：

      product      total        first_sell       last_sell
1           1     150.23        2019-01-03      2019-01-03
2           2     129.63        2019-01-01      2019-01-02

Answer 1

您可以执行以下操作：

library(dplyr)
#> 
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#> 
#>     filter, lag
#> The following objects are masked from 'package:base':
#> 
#>     intersect, setdiff, setequal, union

x <- tibble(product = c(1, 1, 1, 2, 2, 2),
            value_day = c(0, 0, 150.23, 110.98, 18.65, 0),
            my_date = c("2019-01-01", "2019-01-02", "2019-01-03", "2019-01-01", "2019-01-02", "2019-01-03"))

x %>%
  group_by(product) %>%
  summarise(total = sum(value_day),
            first_sell = first(my_date),
            last_sell = last(my_date))
#> # A tibble: 2 x 4
#>   product total first_sell last_sell 
#>     <dbl> <dbl> <chr>      <chr>     
#> 1       1  150. 2019-01-01 2019-01-03
#> 2       2  130. 2019-01-01 2019-01-03

^{由reprex package（v0.3.0）于2019-05-19创建}

Answer 2

library(dplyr)
group_by(x,product) %>% mutate(total = sum(value_day),
 first_sell = first(my_date),
 last_sell = last(my_date)) %>%
 select(-value_day, -my_date) %>%
 distinct()

##   A tibble: 2 x 4
##   Groups:   product [2]
##   product total first_sell last_sell 
##     <dbl> <dbl> <fct>      <fct>     
## 1       1  150. 2019-01-01 2019-01-03
## 2       2  130. 2019-01-01 2019-01-03

如何插入案例之类的列名

2 个答案: