Question

我想在table我的数据后使用melt做一个简单的dplyr。

我的数据看起来像这样

   cluster   21:30   21:45
4        c   alone   alone
6        b       %       %
12       e partner partner
14       b partner partner
20       b   alone   alone
22       c partner partner

table我可以简单地

table(dta$cluster)
   a b c d e 
   2 8 5 1 4

如何使用melt和summarise获得相同的结果？

 library(dplyr)
 library(reshape2)

 dta %>% 
 melt(id.vars = 'cluster')  %>% 
 group_by(cluster) %>% 
 summarise( n() )

我需要的是table在融化数据之后的群集。

所以要正确计算这个data.frame

dta %>% melt(id.vars = 'cluster')

预期的输出是这个

cluster variable value n_cluster 1 a 21:30 . 2 2 a 21:30 nuclear 2 3 a 21:45 . 2 4 a 21:45 nuclear 2 5 b 21:30 % 8 6 b 21:30 partner 8 7 b 21:30 alone 8 8 b 21:30 partner 8 9 b 21:30 partner 8 10 b 21:30 nuclear 8 11 b 21:30 partner 8 12 b 21:30 partner 8 13 b 21:45 % 8 14 b 21:45 partner 8 15 b 21:45 alone 8 16 b 21:45 partner 8 17 b 21:45 partner 8 18 b 21:45 nuclear 8 19 b 21:45 partner 8 20 b 21:45 partner 8 21 c 21:30 alone 5 22 c 21:30 partner 5 23 c 21:30 % 5 24 c 21:30 partner 5 25 c 21:30 partner 5 26 c 21:45 alone 5 27 c 21:45 partner 5 28 c 21:45 % 5 29 c 21:45 partner 5 30 c 21:45 partner 5 31 d 21:30 partner 1 32 d 21:45 alone 1 33 e 21:30 partner 4 34 e 21:30 nuclear 4 35 e 21:30 nuclear 4 36 e 21:30 nuclear 4 37 e 21:45 partner 4 38 e 21:45 nuclear 4 39 e 21:45 nuclear 4 40 e 21:45 nuclear 4

有什么想法吗？

dta = structure(list(cluster = structure(c(3L, 2L, 5L, 2L, 2L, 3L, 5L, 3L, 1L, 3L, 1L, 2L, 5L, 3L, 2L, 2L, 2L, 2L, 4L, 5L), .Label = c("a", "b", "c", "d", "e"), class = "factor"), `21:30` = structure(c(2L, 7L, 5L, 5L, 2L, 5L, 4L, 7L, 1L, 5L, 4L, 5L, 4L, 5L, 5L, 4L, 5L, 5L, 5L, 4L), .Label = c(".", "alone", "children", "nuclear", "partner", "*", "%"), class = "factor"), `21:45` = structure(c(2L, 7L, 5L, 5L, 2L, 5L, 4L, 7L, 1L, 5L, 4L, 5L, 4L, 5L, 5L, 4L, 5L, 5L, 2L, 4L), .Label = c(".", "alone", "children", "nuclear", "partner", "*", "%"), class = "factor")), .Names = c("cluster", "21:30", "21:45"), row.names = c("4", "6", "12", "14", "20", "22", "23", "28", "30", "32", "36", "38", "40", "42", "44", "48", "50", "56", "57", "60"), class = "data.frame")

Answer 1

我似乎无法为此找到一个好的骗局，但一个简单的dplyr习语就是使用count

count(dta, cluster)
# Source: local data frame [5 x 2]
# 
#   cluster n
# 1       a 2
# 2       b 8
# 3       c 5
# 4       d 1
# 5       e 4

根据您新的所需输出，您可以将此结果连接到您的融化数据集

dta %>% 
  melt(id.vars = 'cluster')  %>% 
  left_join(., count(dta, cluster)) %>%
  arrange(cluster)
#    cluster variable   value n
# 1        a    21:30       . 2
# 2        a    21:30 nuclear 2
# 3        a    21:45       . 2
# 4        a    21:45 nuclear 2
# 5        b    21:30       % 8
# 6        b    21:30 partner 8
# 7        b    21:30   alone 8
#...

Answer 2

在计算重复观测变量的分布时，应考虑观测数量。

在这个例子中

n_episode = 2

然后代码变得简单

dta %>% 
  melt(id.vars = 'cluster')  %>% 
  group_by(cluster) %>% 
  mutate( n_cluster = n() / n_episode) %>% 
  arrange(cluster)

可以使用此结果（n_episode）来计算不同大小的组的平均值。

使用dplyr计算熔化数据上值的出现次数

2 个答案: