折叠因素和与dplyr

鉴于突变行：折叠因素和与dplyr

x <- data.frame(Day = c(1,2,3,4,5,6,7,8,9,10), 
       var1 = c(5,4,2,3,4,5,1,2,3,4), 
       var2 = c(3,6,2,3,4,5,7,8,1,2), 
       var3 = c(1,2,3,4,6,2,4,7,8,4), 
       var4 = c(1,3,7,5,3,7,2,3,1,2))

此刻一天变量是数字，但对应于1 =星期一，5 =星期五，6 =星期一，10 =星期五。我想所有的各天坍塌在一起，并通过日平均它们的值了：

z <- data.frame(Day = c("Monday", "Tuesday", "Wednesday", "Thursday","Friday"), 
       var1 = c(5,2.5,2,3,4), 
       var2 = c(4,6.5,5,2,3), 
       var3 = c(1.5,3,5,6,5), 
       var4 = c(4,2.5,5,3,2.5))

来源

2017-10-11 D500

使用modular%%：

days = c("Monday", "Tuesday", "Wednesday", "Thursday", "Friday") 
x %>% group_by(Day = days[(Day - 1) %% 5 + 1]) %>% summarise_all(mean) 

# A tibble: 5 x 5 
#  Day var1 var2 var3 var4 
#  <chr> <dbl> <dbl> <dbl> <dbl> 
#1 Friday 4.0 3.0 5.0 2.5 
#2 Monday 5.0 4.0 1.5 4.0 
#3 Thursday 3.0 2.0 6.0 3.0 
#4 Tuesday 2.5 6.5 3.0 2.5 
#5 Wednesday 2.0 5.0 5.0 5.0

来源

2017-10-11 18:59:04 Psidom

如果数据是有序的，通过复制天，然后用summarise_at创建分组变量得到的mean的“ VAR”列

library(dplyr) 
v1 <- c("Monday", "Tuesday", 
      "Wednesday", "Thursday","Friday") 
x %>% 
    group_by(Day = factor(rep(v1, 2), levels = v1)) %>% 
    summarise_at(vars(matches('var')), mean) 
# A tibble: 5 x 5 
#  Day var1 var2 var3 var4 
#  <chr> <dbl> <dbl> <dbl> <dbl> 
# 1 Monday 5.0 4.0 1.5 4.0 
# 2 Tuesday 2.5 6.5 3.0 2.5 
# 3 Wednesday 2.0 5.0 5.0 5.0 
# 4 Thursday 3.0 2.0 6.0 3.0 
# 5 Friday 4.0 3.0 5.0 2.5

如果数据没有排序，然后创建一个键/值数据集，与原始数据集的加入，由分组后‘天’，得到mean如上

x1 <- data.frame(Day = 1:10, DayC = c("Monday", "Tuesday", 
     "Wednesday", "Thursday","Friday"), stringsAsFactors= FALSE) 

x %>% 
    left_join(., x1) %>% 
    group_by(Day = DayC) %>% 
    summarise_at(vars(matches('var')), mean) %>% 
    arrange(factor(Day, levels = v1))

来源

2017-10-11 18:50:54 akrun

折叠因素和与dplyr

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