R - 统计所有组合

我想计算data.frame中的所有组合.

数据看起来像这样

   9 10 11 12
1  1  1  1  1
2  0  0  0  0
3  0  0  0  0
4  1  1  1  1
5  1  1  1  1
6  0  0  0  0
7  1  0  0  1
8  1  0  0  1
9  1  1  1  1
10 1  1  1  1

我想要的输出很简单

comb     n 
1 1 1 1  5
0 0 0 0  3 
1 0 0 1  2 

你知道有什么简单的功能吗？

谢谢

dt = structure(list(`9` = c(1, 0, 0, 1, 1, 0, 1, 1, 1, 1), `10` = c(1, 
0, 0, 1, 1, 0, 0, 0, 1, 1), `11` = c(1, 0, 0, 1, 1, 0, 0, 0, 
1, 1), `12` = c(1, 0, 0, 1, 1, 0, 1, 1, 1, 1)), .Names = c("9", 
"10", "11", "12"), class = "data.frame", row.names = c(NA, -10L
))

Cath.. 11

基础R解决方案aggregate:

aggregate(seq(nrow(dt))~., data=dt, FUN=length)
#  9 10 11 12 seq(nrow(dt))
#1 0  0  0  0             3
#2 1  0  0  1             2
#3 1  1  1  1             5

编辑

要使colnames更符合您的输出,您可以执行以下操作:

`colnames<-`(aggregate(seq(nrow(dt))~., data=dt, FUN=length), c("c", "o", "m", "b", "n"))
#  c o m b n
#1 0 0 0 0 3
#2 1 0 0 1 2
#3 1 1 1 1 5

或者,更短:

aggregate(cbind(n = 1:nrow(dt))~., dt, length)
#  9 10 11 12 n
#1 0  0  0  0 3
#2 1  0  0  1 2
#3 1  1  1  1 5

---

akrun.. 10

我们可以使用data.table或dplyr.这些非常有效.我们将'data.frame'转换为'data.table'(setDT(dt)),按'dt'(names(dt))的所有列分组,我们将nrow(.N)转换为'Count'

library(data.table)
setDT(dt)[,list(Count=.N) ,names(dt)]

---

或者我们可以使用类似的方法dplyr.

library(dplyr)
names(dt) <- make.names(names(dt))
dt %>%
   group_by_(.dots=names(dt)) %>%
   summarise(count= n())

基准

如果有人想要查看某些指标(以及之前备份我的声明(efficient!)),

set.seed(24)
df1 <- as.data.frame(matrix(sample(0:1, 1e6*6, replace=TRUE), ncol=6))

akrunDT <-  function() {
  as.data.table(df1)[,list(Count=.N) ,names(df1)]
 }

akrunDplyr <- function() {
  df1 %>%
    group_by_(.dots=names(df1)) %>%
    summarise(count= n())
}

cathG <- function() {
 aggregate(cbind(n = 1:nrow(df1))~., df1, length)
  }

docendoD <- function() {
  as.data.frame(table(comb = do.call(paste, df1)))
}

deena <- function() {
   table(apply(df1, 1, paste, collapse = ","))
}

下面是microbenchmark结果

library(microbenchmark)
microbenchmark(akrunDT(), akrunDplyr(), cathG(), docendoD(),  deena(),
  unit='relative', times=20L)
#   Unit: relative
#        expr       min        lq      mean   median        uq        max neval  cld
#     akrunDT()  1.000000  1.000000  1.000000  1.00000  1.000000  1.0000000    20     a   
#  akrunDplyr()  1.512354  1.523357  1.307724  1.45907  1.365928  0.7539773    20     a   
#       cathG() 43.893946 43.592062 37.008677 42.10787 38.556726 17.9834245    20    c 
#    docendoD() 18.778534 19.843255 16.560827 18.85707 17.296812  8.2688541    20    b  
#       deena() 90.391417 89.449547 74.607662 85.16295 77.316143 34.6962954    20    d

---

talat.. 7

您可以仅使用基本R尝试以下方法:

as.data.frame(table(comb = do.call(paste, dt)))
#     comb Freq
#1 0 0 0 0    3
#2 1 0 0 1    2
#3 1 1 1 1    5

---

小智.. 5

也许那样: table(apply(dt, 1, paste, collapse = ","))

1> Cath..：

---

基础R解决方案aggregate:

aggregate(seq(nrow(dt))~., data=dt, FUN=length)
#  9 10 11 12 seq(nrow(dt))
#1 0  0  0  0             3
#2 1  0  0  1             2
#3 1  1  1  1             5

编辑

要使colnames更符合您的输出,您可以执行以下操作:

`colnames<-`(aggregate(seq(nrow(dt))~., data=dt, FUN=length), c("c", "o", "m", "b", "n"))
#  c o m b n
#1 0 0 0 0 3
#2 1 0 0 1 2
#3 1 1 1 1 5

或者,更短:

aggregate(cbind(n = 1:nrow(dt))~., dt, length)
#  9 10 11 12 n
#1 0  0  0  0 3
#2 1  0  0  1 2
#3 1  1  1  1 5

2> akrun..：

---

我们可以使用data.table或dplyr.这些非常有效.我们将'data.frame'转换为'data.table'(setDT(dt)),按'dt'(names(dt))的所有列分组,我们将nrow(.N)转换为'Count'

library(data.table)
setDT(dt)[,list(Count=.N) ,names(dt)]

---

或者我们可以使用类似的方法dplyr.

library(dplyr)
names(dt) <- make.names(names(dt))
dt %>%
   group_by_(.dots=names(dt)) %>%
   summarise(count= n())

基准

如果有人想要查看某些指标(以及之前备份我的声明(efficient!)),

set.seed(24)
df1 <- as.data.frame(matrix(sample(0:1, 1e6*6, replace=TRUE), ncol=6))

akrunDT <-  function() {
  as.data.table(df1)[,list(Count=.N) ,names(df1)]
 }

akrunDplyr <- function() {
  df1 %>%
    group_by_(.dots=names(df1)) %>%
    summarise(count= n())
}

cathG <- function() {
 aggregate(cbind(n = 1:nrow(df1))~., df1, length)
  }

docendoD <- function() {
  as.data.frame(table(comb = do.call(paste, df1)))
}

deena <- function() {
   table(apply(df1, 1, paste, collapse = ","))
}

下面是microbenchmark结果

library(microbenchmark)
microbenchmark(akrunDT(), akrunDplyr(), cathG(), docendoD(),  deena(),
  unit='relative', times=20L)
#   Unit: relative
#        expr       min        lq      mean   median        uq        max neval  cld
#     akrunDT()  1.000000  1.000000  1.000000  1.00000  1.000000  1.0000000    20     a   
#  akrunDplyr()  1.512354  1.523357  1.307724  1.45907  1.365928  0.7539773    20     a   
#       cathG() 43.893946 43.592062 37.008677 42.10787 38.556726 17.9834245    20    c 
#    docendoD() 18.778534 19.843255 16.560827 18.85707 17.296812  8.2688541    20    b  
#       deena() 90.391417 89.449547 74.607662 85.16295 77.316143 34.6962954    20    d

3> talat..：

---

您可以仅使用基本R尝试以下方法:

as.data.frame(table(comb = do.call(paste, dt)))
#     comb Freq
#1 0 0 0 0    3
#2 1 0 0 1    2
#3 1 1 1 1    5

4> 小智..：

---

也许那样: table(apply(dt, 1, paste, collapse = ","))