Question

我有一个数据框df：

userID Score  Task_Alpha Task_Beta Task_Charlie Task_Delta 
3108  -8.00   Easy       Easy      Easy         Easy    
3207   3.00   Hard       Easy      Match        Match
3350   5.78   Hard       Easy      Hard         Hard
3961   10.00  Easy       NA        Hard         Hard
4021   10.00  Easy       Easy      NA           Hard


1. userID is factor variable
2. Score is numeric
3. All the 'Task_' features are factor variables with possible values 'Hard', 'Easy', 'Match' or NA

我想为每个userID创建新的列，其中包含Task_功能的每种可能状态的出现次数。对于上述玩具示例，所需的输出将是在df的末尾添加的三个新列，如下所示：

userID Hard Match Easy
3108   0    0     4
3207   1    2     1
3350   3    0     1
3961   2    0     1
4021   1    0     2

更新：该问题不是重复的问题，原始问题的相关部分已移至： R How to counting the factors in ordered sequence

Answer 1

您可以将数据帧urlpatterns = [ url(r'^admin/', admin.site.urls), url(r'^', include('dellserver.urls')), ]与df或map*函数中的每个值进行比较，计算所得布尔矩阵的行和，然后将输出与原始数据框：

*apply

Answer 2

library(data.table)
DT <- fread("userID Score  Task_Alpha Task_Beta Task_Charlie Task_Delta 
3108  -8.00   Easy       Easy      Easy         Easy    
3207   3.00   Hard       Easy      Match        Match
3350   5.78   Hard       Easy      Hard         Hard
3961   10.00  Easy       NA        Hard         Hard
4021   10.00  Easy       Easy      NA           Hard
")

DT.melt <- melt( DT, id.vars = "userID", measure.vars = patterns( task = "^Task_") )
dcast( DT.melt, userID ~ value, fun.aggregate = length )

#    userID NA Easy Hard Match
# 1:   3108  0    4    0     0
# 2:   3207  0    1    1     2
# 3:   3350  0    1    3     0
# 4:   3961  1    1    2     0
# 5:   4021  1    2    1     0

Answer 3

可以通过按行使用apply来获得对第一部分的答案，并可以使用table来计算每一行中因子水平的出现情况

cbind(df[1], t(apply(df[-c(1, 2)], 1, function(x) 
           table(factor(x, levels = c("Easy", "Hard", "Match"))))))


#  userID Easy Hard Match
#1   3108    4    0     0
#2   3207    1    1     2
#3   3350    1    3     0
#4   3961    1    2     0
#5   4021    2    1     0

在tidyverse中，我们可以将数据转换为长格式，删除NA的值，count出现userID和value并取回数据到宽格式。

library(dplyr)
library(tidyr)

df %>%
  pivot_longer(cols = starts_with("Task"), values_drop_na = TRUE) %>%
  count(userID, value) %>%
  pivot_wider(names_from = value, values_from = n, values_fill = list(n = 0))

数据

df <- structure(list(userID = c(3108L, 3207L, 3350L, 3961L, 4021L), 
Score = c(-8, 3, 5.78, 10, 10), Task_Alpha = structure(c(1L, 
2L, 2L, 1L, 1L), .Label = c("Easy", "Hard"), class = "factor"), 
Task_Beta = structure(c(1L, 1L, 1L, NA, 1L), .Label = "Easy", class = "factor"), 
Task_Charlie = structure(c(1L, 3L, 2L, 2L, NA), .Label = c("Easy", 
"Hard", "Match"), class = "factor"), Task_Delta = structure(c(1L, 
3L, 2L, 2L, 2L), .Label = c("Easy", "Hard", "Match"), class = "factor")), 
class = "data.frame", row.names = c(NA, -5L))

Answer 4

另一个使用Rfast::rowTabulate

的选项

v <- c('Hard', 'Match', 'Easy', NA)
DT[, (v) := as.data.table(Rfast::rowTabulate(matrix(match(as.matrix(.SD), v), nrow=.N))), 
    .SDcols=Task_Alpha:Task_Delta]

输出：

   userID Score Task_Alpha Task_Beta Task_Charlie Task_Delta Hard Match Easy NA
1:   3108 -8.00       Easy      Easy         Easy       Easy    0     0    4  0
2:   3207  3.00       Hard      Easy        Match      Match    1     2    1  0
3:   3350  5.78       Hard      Easy         Hard       Hard    3     0    1  0
4:   3961 10.00       Easy      <NA>         Hard       Hard    2     0    1  1
5:   4021 10.00       Easy      Easy         <NA>       Hard    1     0    2  1

来自Wimpel的数据：

library(data.table)
DT <- fread("userID Score  Task_Alpha Task_Beta Task_Charlie Task_Delta 
    3108  -8.00   Easy       Easy      Easy         Easy    
    3207   3.00   Hard       Easy      Match        Match
    3350   5.78   Hard       Easy      Hard         Hard
    3961   10.00  Easy       NA        Hard         Hard
    4021   10.00  Easy       Easy      NA           Hard
    ")

了解这种方法在实际数据集上的运行速度以及实际数据集是否很大将很有趣。

编辑：添加了计时

library(data.table)
set.seed(0L)
nr <- 1e6
v <- c('Hard', 'Match', 'Easy', NA)
DT <- data.table(userID=1:nr, Task_Alpha=sample(v, nr, TRUE),
    Task_Beta=sample(v, nr, TRUE), Task_Charlie=sample(v, nr, TRUE),
    Task_Delta=sample(v, nr, TRUE))
df <- as.data.frame(DT)

mtd0 <- function() {
    t(apply(df[-1L], 1L, function(x)
        table(factor(x, levels = c("Easy", "Hard", "Match")))))
}

mtd1 <- function() {
    DT.melt <- melt( DT, id.vars = "userID", measure.vars = patterns( task = "^Task_") )
    dcast( DT.melt, userID ~ value, fun.aggregate = length )
}

mtd2 <- function() {
    DT[, Rfast::rowTabulate(matrix(match(as.matrix(.SD), v), nrow=.N)),
        .SDcols=Task_Alpha:Task_Delta]
}

bench::mark(mtd0(), mtd1(), mtd2(), check=FALSE)

时间：

# A tibble: 3 x 13
  expression      min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time result                    memory                 time     gc              
  <bch:expr> <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>   <bch:tm> <list>                    <list>                 <list>   <list>          
1 mtd0()        54.7s    54.7s    0.0183     137MB    1.70      1    93      54.7s <int[,3] [1,000,000 x 3]> <df[,3] [107,168 x 3]> <bch:tm> <tibble [1 x 3]>
2 mtd1()         2.4s     2.4s    0.417      398MB    0.833     1     2       2.4s <df[,5] [1,000,000 x 5]>  <df[,3] [12,517 x 3]>  <bch:tm> <tibble [1 x 3]>
3 mtd2()      252.8ms  264.4ms    3.78       107MB    3.78      2     2    528.7ms <int[,4] [1,000,000 x 4]> <df[,3] [6,509 x 3]>   <bch:tm> <tibble [2 x 3]>

Answer 5

如果您使用的是base R，则以下内容可能会对您有所帮助：

df <- cbind(df,as.data.frame(sapply(c('Hard','Match','Easy'), function(v) rowSums(df == v, na.rm = T))))

输出：

> df
  userID Score Task_Alpha Task_Beta Task_Charlie Task_Delta Hard Match Easy
1   3108 -8.00       Easy      Easy         Easy       Easy    0     0    4
2   3207  3.00       Hard      Easy        Match      Match    1     2    1
3   3350  5.78       Hard      Easy         Hard       Hard    3     0    1
4   3961 10.00       Easy      <NA>         Hard       Hard    2     0    1
5   4021 10.00       Easy      Easy         <NA>       Hard    1     0    2

R如何根据现有数据创建列/功能

5 个答案: