Tde*_*eus 3 r percentage dplyr
我想计算第一年2015和最后一年之间的百分比变化,2017作为每个city.
这是我的可重现示例,其中最后一列perct_change_2015_2017是所需的输出。我如何在 R 中为一大堆城市做到这一点?最好在 dplyr 中。
使用正确的百分比变化数字进行编辑
example <- structure(list(city = c("Amsterdam", "Amsterdam", "Amsterdam",
"Rotterdam", "Rotterdam", "Rotterdam"), year = c(2015L, 2016L,
2017L, 2015L, 2016L, 2017L), value = c(30L, 35L, 46L, 23L, 19L,
17L), perct_change_2015_2017 = c(0.5333333333, 0.5333333333,
0.5333333333, -0.2608695652, -0.2608695652, -0.2608695652)), .Names = c("city",
"year", "value", "perct_change_2015_2017"), row.names = c(NA,
-6L), class = c("tbl_df", "tbl", "data.frame"), spec = structure(list(
cols = structure(list(city = structure(list(), class = c("collector_character",
"collector")), year = structure(list(), class = c("collector_integer",
"collector")), value = structure(list(), class = c("collector_integer",
"collector")), perct_change_2015_2017 = structure(list(), class = c("collector_double",
"collector"))), .Names = c("city", "year", "value", "perct_change_2015_2017"
)), default = structure(list(), class = c("collector_guess",
"collector"))), .Names = c("cols", "default"), class = "col_spec"))
example
A tibble: 6 x 4
city year value perct_change_2015_2017
<chr> <int> <int> <dbl>
1 Amsterdam 2015 30 0.533
2 Amsterdam 2016 35 0.533
3 Amsterdam 2017 46 0.533
4 Rotterdam 2015 23 -0.260
5 Rotterdam 2016 19 -0.260
6 Rotterdam 2017 17 -0.260
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无论存在多少年,此方法将始终使用2015和2017。我更喜欢使用 www 的解决方案first和last一般,但如果你有更多的年份并想要这些特定的年份,这里是如何做到的。
example %>% group_by(city) %>%
mutate(perct_change_2015_2017 =
(value[year == 2017] - value[year == 2015]) / value[year == 2015]
)
# # A tibble: 6 x 4
# # Groups: city [2]
# city year value perct_change_2015_2017
# <chr> <int> <int> <dbl>
# 1 Amsterdam 2015 30 0.5333333
# 2 Amsterdam 2016 35 0.5333333
# 3 Amsterdam 2017 46 0.5333333
# 4 Rotterdam 2015 23 -0.2608696
# 5 Rotterdam 2016 19 -0.2608696
# 6 Rotterdam 2017 17 -0.2608696
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