Question

我正在使用purrr和功能编程来步入婴儿的脚步，我可能会淹没在一杯水中。考虑清单

zz<-list(structure(list(year = c(2000, 2001, 2002, 2003, 2000, 2001, 
2002, 2003, 2000, 2001, 2002, 2003), tot_i = c(22393349.081, 
23000574.372, 21682040.898, 21671102.853, 34361300.338, 35297814.942, 
34745691.204, 35878883.117, 11967951.257, 12297240.57, 13063650.306, 
14207780.264), relation = c("EU28-Algeria", "EU28-Algeria", "EU28-Algeria", 
"EU28-Algeria", "World-Algeria", "World-Algeria", "World-Algeria", 
"World-Algeria", "Extra EU28-Algeria", "Extra EU28-Algeria", 
"Extra EU28-Algeria", "Extra EU28-Algeria"), g_rate = c(0.736046372770467, 
0.0271163231905857, -0.0573261107603093, -0.000504474880914325, 
0.614846575418334, 0.0272549232650638, -0.0156418673197543,     0.0326138831530727, 
0.428272657063707, 0.0275142592018328, 0.0623237165799383, 0.0875811837579971
)), row.names = c(NA, -12L), class = c("tbl_df", "tbl", "data.frame"
)), structure(list(year = c(2000, 2001, 2002, 2003, 2000, 2001, 
2002, 2003, 2000, 2001, 2002, 2003), tot_i = c(9233346.648, 7869288.171, 
7271485.687, 6395999.102, 21393949.287, 19851236.26, 19449339.887, 
16055014.309, 12160602.639, 11981948.089, 12177854.2, 9659015.207
), relation = c("EU28-Egypt", "EU28-Egypt", "EU28-Egypt", "EU28-Egypt", 
"World-Egypt", "World-Egypt", "World-Egypt", "World-Egypt", "Extra EU28-Egypt", 
"Extra EU28-Egypt", "Extra EU28-Egypt", "Extra EU28-Egypt"), 
 g_rate = c(0.0970653722744164, -0.147731751985664, -0.0759665259436081, 
 -0.120399959882366, 0.124744629514854, -0.0721097823643728, 
-0.0202454077789513, -0.174521376957825, 0.146712116047648, 
 -0.0146912579338002, 0.0163501051368976, -0.206837670383671
)), row.names = c(NA, -12L), class = c("tbl_df", "tbl", "data.frame"
)))

例如，我可以使用地图做非常简单的事情，例如迭代地取某列的平均值

map(zz, function(x) mean(x$tot_i))

或过滤年份的值

map(zz, function(x) filter(x, year==2000))

但是，一旦想增加一点复杂性，我就会把头撞在墙上。例如

1）我想按关系对zz中的数据进行迭代分组，并通过取tot_i和

的平均值进行汇总

2）给出年份列表

   ll<-list(c(2000, 2001), c(2001, 2003))

我想根据ll中列出的年份过滤zz列表的两个元素。

然后我将有很多其他操作可以对数据进行处理，但是已经了解了1和2将使我从现在的困境中走很长一段路。

欢迎任何建议。

Answer 1

当我们从'll'的相应元素中进行子集设置时，请使用map2遍历list和filter这两个基于'year'元素的行{{ 1}} %in%

.y

如果我们使用匿名函数，则有两个参数而不是1

map2(zz, ll, ~ .x %>% 
               filter(year %in% .y))
#[[1]]
# A tibble: 6 x 4
#   year     tot_i relation           g_rate
#  <dbl>     <dbl> <chr>               <dbl>
#1  2000 22393349. EU28-Algeria       0.736 
#2  2001 23000574. EU28-Algeria       0.0271
#3  2000 34361300. World-Algeria      0.615 
#4  2001 35297815. World-Algeria      0.0273
#5  2000 11967951. Extra EU28-Algeria 0.428 
#6  2001 12297241. Extra EU28-Algeria 0.0275

#[[2]]
# A tibble: 6 x 4
#   year     tot_i relation          g_rate
#  <dbl>     <dbl> <chr>              <dbl>
#1  2001  7869288. EU28-Egypt       -0.148 
#2  2003  6395999. EU28-Egypt       -0.120 
#3  2001 19851236. World-Egypt      -0.0721
#4  2003 16055014. World-Egypt      -0.175 
#5  2001 11981948. Extra EU28-Egypt -0.0147
#6  2003  9659015. Extra EU28-Egypt -0.207

类似于我们从map2(zz, ll, function(x, y) filter(x, year %in% y))使用Map的方式

base R

r，分组和过滤

1 个答案: