R 按行计数列，避免熔化/聚集

我正在使用这样的数据帧：

   idno      08:00      08:05      08:10    08:15    08:20    08:25
1     1   Domestic   Domestic   Domestic Domestic Domestic Domestic
2     2    Leisure    Leisure    Leisure  Leisure  Leisure  Leisure
3     3        Eat        Eat        Eat      Eat      Eat      Eat
4     4       Paid       Paid       Paid     Paid     Paid     Paid
5     5      Sleep      Sleep      Sleep    Sleep    Sleep    Sleep
6     6        Eat        Eat        Eat  Missing  Missing  Missing
7     7      Sleep      Sleep      Sleep    Sleep    Sleep    Sleep
8     8       Paid       Paid       Paid     Paid     Paid     Paid
9     9      Sleep      Sleep      Sleep    Sleep    Sleep    Sleep
10   10 Child Care Child Care Child Care   Travel   Travel   Travel

我感兴趣的是这样总结这个数据框架

（需要输出）

然而，我想知道是否有更直接的方法来做到这一点。我的问题是，我正在处理一个非常大的数据库，并使用

melt

，然后

count

然后

spread

可能有点慢

是否有一种直接的方法来计算每行的列（变量的分布），最好使用

data.table

setDT(df)[,.N,by=] # 

类似于每行的

by

列

df = structure(list(idno = 1:10, `08:00` = c("Domestic", "Leisure", 
"Eat", "Paid", "Sleep", "Eat", "Sleep", "Paid", "Sleep", "Child Care"
), `08:05` = c("Domestic", "Leisure", "Eat", "Paid", "Sleep", 
"Eat", "Sleep", "Paid", "Sleep", "Child Care"), `08:10` = c("Domestic", 
"Leisure", "Eat", "Paid", "Sleep", "Eat", "Sleep", "Paid", "Sleep", 
"Child Care"), `08:15` = c("Domestic", "Leisure", "Eat", "Paid", 
"Sleep", "Missing", "Sleep", "Paid", "Sleep", "Travel"), `08:20` =    c("Domestic", 
"Leisure", "Eat", "Paid", "Sleep", "Missing", "Sleep", "Paid", 
"Sleep", "Travel"), `08:25` = c("Domestic", "Leisure", "Eat", 
"Paid", "Sleep", "Missing", "Sleep", "Paid", "Sleep", "Travel"
)), .Names = c("idno", "08:00", "08:05", "08:10", "08:15", "08:20", 
"08:25"), row.names = c(NA, 10L), class = "data.frame")

---

您可以从

qdapTools

中尝试

mtabulate

library(qdapTools)

mtabulate(split(df[-1], seq(nrow(df))))

#   Child Care Domestic Eat Leisure Missing Paid Sleep Travel
#1           0        6   0       0       0    0     0      0
#2           0        0   0       6       0    0     0      0
#3           0        0   6       0       0    0     0      0
#4           0        0   0       0       0    6     0      0
#5           0        0   0       0       0    0     6      0
#6           0        0   3       0       3    0     0      0
#7           0        0   0       0       0    0     6      0
#8           0        0   0       0       0    6     0      0
#9           0        0   0       0       0    0     6      0
#10          3        0   0       0       0    0     0      3

---

库（data.table）；dcast（melt（setDT（df），id=1，idno~value）

这仍然是一个有趣的解决方案，感谢@jaapalternine（基于重复目标）：

xtabs（~idno+value，data.table:：melt（df，id=1））

备选方案二：

library（dplyr）；图书馆（tidyr）；df%%>%聚集（键，值，-1）%%>%group\U by（idno，值）%%>%tally%%>%排列（键=值，值=n，填充=0）

感谢您提供的所有选择。让我保持警觉！非常有趣谢谢

df = structure(list(idno = 1:10, `08:00` = c("Domestic", "Leisure", 
"Eat", "Paid", "Sleep", "Eat", "Sleep", "Paid", "Sleep", "Child Care"
), `08:05` = c("Domestic", "Leisure", "Eat", "Paid", "Sleep", 
"Eat", "Sleep", "Paid", "Sleep", "Child Care"), `08:10` = c("Domestic", 
"Leisure", "Eat", "Paid", "Sleep", "Eat", "Sleep", "Paid", "Sleep", 
"Child Care"), `08:15` = c("Domestic", "Leisure", "Eat", "Paid", 
"Sleep", "Missing", "Sleep", "Paid", "Sleep", "Travel"), `08:20` =    c("Domestic", 
"Leisure", "Eat", "Paid", "Sleep", "Missing", "Sleep", "Paid", 
"Sleep", "Travel"), `08:25` = c("Domestic", "Leisure", "Eat", 
"Paid", "Sleep", "Missing", "Sleep", "Paid", "Sleep", "Travel"
)), .Names = c("idno", "08:00", "08:05", "08:10", "08:15", "08:20", 
"08:25"), row.names = c(NA, 10L), class = "data.frame")

library(qdapTools)

mtabulate(split(df[-1], seq(nrow(df))))

#   Child Care Domestic Eat Leisure Missing Paid Sleep Travel
#1           0        6   0       0       0    0     0      0
#2           0        0   0       6       0    0     0      0
#3           0        0   6       0       0    0     0      0
#4           0        0   0       0       0    6     0      0
#5           0        0   0       0       0    0     6      0
#6           0        0   3       0       3    0     0      0
#7           0        0   0       0       0    0     6      0
#8           0        0   0       0       0    6     0      0
#9           0        0   0       0       0    0     6      0
#10          3        0   0       0       0    0     0      3