计算增长增长，按类型分组(使用R)

Question

我有一个数据集z，我希望按以下类型计算增长增长：

location   size     type     date

ny         5        hello    10/01/2020
ny         7        ai       10/02/2020
ny         8        ai       10/03/2020
ny         6        hello    10/04/2020
ca         15       cool     10/05/2020
ca         10       name     10/06/2020
ca         5        name     10/07/2020
ca         16       cool     10/08/2020

期望输出

location  type    increase   percent_increase     start_date    end_date

ca        cool    1           6.67%               10/05/2020    10/08/2020
ca        name    -5         -50%                 10/6/2020     10/7/2020
ny        hello   1           20%                 10/01/2020    10/4/2020
ny        ai      1           14.28%              10/2/2020     10/3/2020

这就是我要做的：

library(tidyverse)
z %>%
group_by(type, location) %>% 
mutate(percent_increase = (size/lead(size) - 1) * 100)

我没有得到我想要的输出。如能提供任何协助，我们将不胜感激。

Answer 1

要获得所需的结果，需要在mutate行中进行不同的计算：

我还添加了一个filter来删除NA对percent_increase变量的任何结果。

最后，添加‘arrange’按位置按字母顺序排序，以匹配与请求输出相同的顺序。

码

Z %>% group_by(类型，位置) %>%变异(增加=(铅(大小)-大小)，percent_increase =(增大/大小)* 100，start_date =日期，end_date =铅(日期) %>%过滤器(！is.na(Percent_increase)) %>%排列(位置)

输出

# A tibble: 4 x 8
# Groups:   type, location [4]
  location  size type  date       increase percent_increase start_date end_date  
  <chr>    <int> <chr> <chr>         <int>            <dbl> <chr>      <chr>     
1 ca          15 cool  10/05/2020        1             6.67 10/05/2020 10/08/2020
2 ca          10 name  10/06/2020       -5           -50    10/06/2020 10/07/2020
3 ny           5 hello 10/01/2020        1            20    10/01/2020 10/04/2020
4 ny           7 ai    10/02/2020        1            14.3  10/02/2020 10/03/2020

输入

z <- structure(list(location = c("ny", "ny", "ny", "ny", "ca", "ca", 
"ca", "ca"), size = c(5L, 7L, 8L, 6L, 15L, 10L, 5L, 16L), type = c("hello", 
"ai", "ai", "hello", "cool", "name", "name", "cool"), date = c("10/01/2020", 
"10/02/2020", "10/03/2020", "10/04/2020", "10/05/2020", "10/06/2020", 
"10/07/2020", "10/08/2020")), class = "data.frame", row.names = c(NA, 
-8L))

Answer 2

您缺少按日期组织的排列功能

就像这样：

z %>%
group_by(type, location) %>% 
arrange(date) %>%
mutate(percent_increase = (size/lead(size) - 1) * 100)