获得 3 个最常见的组元素，连接关系，并忽略不常见的值
obtaining 3 most common elements of groups, concatenating ties, and ignoring less common values
我正在尝试使用函数获取每组数据帧中最常见的 3 个数字，但忽略不太常见的值（每组），并允许使用唯一数字（如果存在）。接受的答案将具有最低 system.time
#my current function
library(plyr)
get.3modes.andcounts<- function(origtable,groupby,columnname) {
data <- ddply (origtable, groupby, .fun = function(xx){
c(m1 = paste(names(sort(table(xx[,columnname]),decreasing=TRUE)[1])),
m2 = paste(names(sort(table(xx[,columnname]),decreasing=TRUE)[2])),
m3 = paste(names(sort(table(xx[,columnname]),decreasing=TRUE)[3])),
counts=length2(xx[[columnname]], na.rm=TRUE) #http://www.cookbook-r.com/Manipulating_data/Summarizing_data/
) } )
return(data)
}
length2 <- function (x, na.rm=FALSE) {
if (na.rm) sum(!is.na(x))
else length(x)
}
# example df
col2<-c(4, 4, 4, 4, 5, 3, 3, 3, 2, 2, # group1 "5" is the less common
2, 2, 2, 4, 4, 3, 3, 2, 2, 2, # group2 "3" and "4" are equally less common, and there is 2 more frequent
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, # group3 "4" is unique
4, 4, 4, 4, 5, 5, 5, 5, 2, 2, # group4 "2" is the less common, other ties more frequent
4, 4, 4, 4, 4, 5, 5, 5, 5, 5) # group5 "4" and "5" are equally common and no value is less common (similar to unique)
col1<-paste(c(rep("group1",10),rep("group2",10),rep("group3",10),rep("group4",10),rep("group5",10)), sep=", ")
df<-data.frame(col1=col1,col2=col2)
get.3modes.andcounts(df,"col1","col2")
#CURRENT result
col1 m1 m2 m3 counts
1 group1 4 3 2 10 # ok
2 group2 2 3 4 10 # no, 3 and 4 are the less common
3 group3 4 NA NA 10 # ok
4 group4 4 5 2 10 # no, 2 is less common
5 group5 4 5 NA 10 # ok
# desired
col1 m1 m2 m3 counts
1 group1 4 3 2 10
2 group2 2 NA NA 10
3 group3 4 NA NA 10
4 group4 4 5 NA 10
5 group5 4 5 NA 10
编辑：实际样本有多个关系，不希望有超过 3 列。超过 3 个数字（在 3 列中）仅在出现平局时才被接受。这就是为什么，我决定要求另一种类型的输出。
编辑：group7。只有三个最常见的通缉令。例外，包括第三个最常见的关系（就像在其他组中一样）。
# EXAMPLE 2
# new proposal
col2<-c(4, 4, 4, 4, 5, 3, 3, 3, 2, 2, 6, 6, # group1 2 and 6 tied in the 3rd position, 5 less common
2, 2, 2, 4, 4, 3, 3, 2, 2, 2, 6, 6, # group2 4, 3 and 6 tied in the less common, excluded.
4, 4, 4, 7, 7, 7, 5, 5, 5, 4, 4, 6, # group3 4, 7 and 5 more common, 3 most common present, exclude everything else
4, 4, 4, 4, 5, 5, 5, 5, 2, 2, 6, 6, # group4 2 and 6 less common, excluded (4 AND 5 tied)
4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, # group5 6 less common, excluded, (4 and 5 tied)
4, 4, 4, 3, 3, 3, 2, 2, 2, 1, 1, 1, # group6 all tied
14,14,14,16,16,16,16,34,34,42,42,42,80,80,84,92, #group7 16, 14, 42 are the three most freq.
20,52,40,40,40,20,20,60,60,50) #group 8 20,40 tied, 60 next.
col1<-paste(c(rep("group1",12),rep("group2",12),rep("group3",12),rep("group4",12),rep("group5",12),
rep("group6",12),rep("group7",16),rep("group8", 10)), sep=", ")
df<-data.frame(col1=col1,col2=col2)
#desired output
col1 m1 m2 m3 counts
1 group1 4 3 2,6 12 # 2 and 6 tied in the 3rd position, 5 less common
2 group2 2 NA NA 12 # 4, 3 and 6 tied in the less common, excluded.
3 group3 4 7,5 NA 12 # three most common numbers present, exclude everything else
4 group4 4,5 NA NA 12 # 2 and 6 less common excluded (4 AND 5 tied)
5 group5 4,5 NA NA 12 # 6 less common, excluded, (4 and 5 tied)
6 group6 4,3,2,1 NA NA 12 # all tied
7 group7 16 14,42 NA 16 # three most frequent present, discard others
8 group8 20,40 60 NA 10 # three most frequent present
使用 dplyr 和 tidyr（plyr 的更新版本）：
library(dplyr)
library(tidyr)
df %>%
group_by(col1, col2) %>%
summarise(n = n()) %>%
mutate(m = min_rank(desc(n)), count = sum(n)) %>%
filter(m <= 3 & (m != max(m) | m == 1)) %>%
group_by(col1, m, count) %>%
summarize(a = paste(col2, collapse = ',')) %>%
spread(m, a, sep = '') %>%
ungroup
# # A tibble: 7 × 5
# col1 count m1 m2 m3
# * <fctr> <int> <chr> <chr> <chr>
# 1 group1 12 4 3 2,6
# 2 group2 12 2 <NA> <NA>
# 3 group3 12 4 5,7 <NA>
# 4 group4 12 4,5 <NA> <NA>
# 5 group5 12 4,5 <NA> <NA>
# 6 group6 12 1,2,3,4 <NA> <NA>
# 7 group7 16 16 14,42 <NA>
如果你在函数中需要它：
get.3modes.andcounts <- function(origtable, groupby, columnname) {
origtable %>%
group_by_(groupby, columnname) %>%
summarise(n = n()) %>%
mutate(r = min_rank(desc(n)), count = sum(n)) %>%
filter(r <= 3 & (r != max(r) | r == 1)) %>%
group_by_(groupby, 'r', 'count') %>%
summarize_(a = paste0('paste(',columnname, ', collapse = ",")')) %>%
spread(r, a, sep = '') %>%
ungroup
}
get.3modes.andcounts(df, 'col1', 'col2')
# # A tibble: 7 × 5
# col1 count m1 m2 m3
# * <fctr> <int> <chr> <chr> <chr>
# 1 group1 12 4 3 2,6
# 2 group2 12 2 <NA> <NA>
# 3 group3 12 4 5,7 <NA>
# 4 group4 12 4,5 <NA> <NA>
# 5 group5 12 4,5 <NA> <NA>
# 6 group6 12 1,2,3,4 <NA> <NA>
# 7 group7 16 16 14,42 <NA>
System.time
system.time(get.3modes.andcounts(df, 'col1', 'col2'))
# user system elapsed
# 0.012 0.000 0.011
benchmark(get.3modes.andcounts(df, 'col1', 'col2'), replications = 10, columns = c("test", "replications", "elapsed"))
# test replications elapsed
# 1 get.3modes.andcounts(df, "col1", "col2") 10 0.08
benchmark(get.3modes.andcounts(df, 'col1', 'col2'), replications = 100, columns = c("test", "replications", "elapsed"))
# test replications elapsed
# 1 get.3modes.andcounts(df, "col1", "col2") 100 0.684
benchmark(get.3modes.andcounts(df, 'col1', 'col2'), replications = 1000, columns = c("test", "replications", "elapsed"))
# test replications elapsed
# 1 get.3modes.andcounts(df, "col1", "col2") 1000 6.796
数据：
col2 <- c(4, 4, 4, 4, 5, 3, 3, 3, 2, 2, 6, 6, # group1 2 and 6 tied in the 3rd position, 5 less common
2, 2, 2, 4, 4, 3, 3, 2, 2, 2, 6, 6, # group2 4, 3 and 6 tied in the less common, excluded.
4, 4, 4, 7, 7, 7, 5, 5, 5, 4, 4, 6, # group3 4, 7 and 5 more common, 3 most common present, exclude everything else
4, 4, 4, 4, 5, 5, 5, 5, 2, 2, 6, 6, # group4 2 and 6 less common, excluded (4 AND 5 tied)
4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, # group5 6 less common, excluded, (4 and 5 tied)
4, 4, 4, 3, 3, 3, 2, 2, 2, 1, 1, 1, # group6 all tied
14,14,14,16,16,16,16,34,34,42,42,42,80,80,84,92) #group7 16, 14, 42 are the three most freq.
col1 <- paste(c(rep("group1", 12), rep("group2", 12), rep("group3", 12), rep("group4", 12), rep("group5", 12),
rep("group6", 12), rep("group7", 16)), sep = ", ")
df <- data.frame(col1=col1,col2=col2)
您可以更改 n > 0，它会起作用。你的问题要求 3，但我的回答会更通用，接受任何正整数。
使用基数 R:
myfun <- function( data, n = 3, col1, col2 )
{
## n: numeric: total number of most common elements per group
stopifnot( n > 0 )
a1 <- lapply( split( data, data[[col1]] ), function( x ) { # split data by col1
# browser()
val <- factor( x[[col2]] ) # factor of data values
z1 <- tabulate( val ) # frequency table of levels of val
z2 <- sort( z1[ z1 > 0 ], decreasing = TRUE ) # sorted frequency table with >0
lenx <- length( unique( z2 ) ) # length of unique of z2
if ( lenx == 1 ) { # lenx == 1
return( c( paste( ( levels(val)[ which( z1 %in% z2 ) ] ), collapse = ','), rep(NA_character_, n - 1 ), sum( z1 ) ) )
} else if ( lenx > 1 ) { # lenx > 1
# remove the minimum, and and extract values by using levels of val with indices from the match of z1 and z2
z2 <- setdiff( z2, min( z2 ) )
z2 <- sapply( z2, function( y ) paste( levels(val)[ which( z1 %in% y ) ], collapse = ',') )
# count the length of z2 and get indices of length >= n
z2_ind <- which( cumsum( lengths(unlist( lapply(z2, strsplit, split = "," ),
recursive = F ) ) ) >= n )
if( length( z2_ind ) > 0 ) {
z2 <- z2[ seq_len( z2_ind[1] ) ]
}
# adjust length by assigning NA
if( length(z2) != n ) { z2[ (length(z2)+1):n ] <- NA_character_ }
return( c( z2, sum( z1 ) ) )
} else { # lenx < 1
return( as.list( rep(NA_character_, n ), NA_character_ ) )
}})
a1 <- do.call('rbind', a1) # row bind values of a1
a1 <- data.frame( group = rownames( a1 ), a1, stringsAsFactors = FALSE )
colnames( a1 ) <- c( 'group', paste( 'm', 1:n, sep = '' ), 'count' )
rownames( a1 ) <- NULL # remove row names
return( a1 )
}
输出：
# example1:
myfun(df, 3, 'col1', 'col2')
# group m1 m2 m3 count
# 1 group1 4 3 2 10
# 2 group2 2 NA NA 10
# 3 group3 4 NA NA 10
# 4 group4 4, 5 NA NA 10
# 5 group5 4, 5 NA NA 10
# example 2
myfun(df3, 3, 'col1', 'col2')
# group m1 m2 m3 count
# 1 group1 4 3 2, 6 12
# 2 group2 2 NA NA 12
# 3 group3 4 5, 7 NA 12
# 4 group4 4, 5 NA NA 12
# 5 group5 4, 5 NA NA 12
# 6 group6 4, 3, 2, 1 NA NA 12
# 7 group7 16 14, 42 NA 16
通过将字母分配给 example 1 data df.
的第 3 列来创建 字符数据而不是数字数据
set.seed(1L)
df$col3 <- sample( letters, 50, TRUE )
myfun(df, 3, 'col1', 'col3')
# group m1 m2 m3 count
# 1 group1 x <NA> <NA> 10
# 2 group2 j,u <NA> <NA> 10
# 3 group3 a,d,f,g,i,j,k,q,w,y <NA> <NA> 10
# 4 group4 m <NA> <NA> 10
# 5 group5 u <NA> <NA> 10
不需要额外的包裹。试试这个：
count <- function(df) {
count_n <- function(vec, n) {
fac <- factor(table(vec), levels = sort(unique(table(vec)), decreasing = T))
top3 <- na.omit(names(sort(fac)[1:3]))
min <- names(fac[fac == min(levels(fac))])
if(length(levels(fac))==1){min <- 'NA'}
top3 <- setdiff(top3,min)
nums <- na.omit(names(fac[fac == levels(fac)[n]]))
ifelse(length(intersect(nums, top3))>0, paste(nums, collapse = ','),'NA')
} ##Get the number of rank n.
group <- unique(as.character(df$col1))
m1 <- aggregate(df, list(df$col1), count_n, 1)$col2
m2 <- aggregate(df, list(df$col1), count_n, 2)$col2
m3 <- aggregate(df, list(df$col1), count_n, 3)$col2
count <- aggregate(df, list(df$col1), length)$col2
res <- data.frame(col1 = group, m1, m2, m3, count)
res
}
我正在尝试使用函数获取每组数据帧中最常见的 3 个数字，但忽略不太常见的值（每组），并允许使用唯一数字（如果存在）。接受的答案将具有最低 system.time
#my current function
library(plyr)
get.3modes.andcounts<- function(origtable,groupby,columnname) {
data <- ddply (origtable, groupby, .fun = function(xx){
c(m1 = paste(names(sort(table(xx[,columnname]),decreasing=TRUE)[1])),
m2 = paste(names(sort(table(xx[,columnname]),decreasing=TRUE)[2])),
m3 = paste(names(sort(table(xx[,columnname]),decreasing=TRUE)[3])),
counts=length2(xx[[columnname]], na.rm=TRUE) #http://www.cookbook-r.com/Manipulating_data/Summarizing_data/
) } )
return(data)
}
length2 <- function (x, na.rm=FALSE) {
if (na.rm) sum(!is.na(x))
else length(x)
}
# example df
col2<-c(4, 4, 4, 4, 5, 3, 3, 3, 2, 2, # group1 "5" is the less common
2, 2, 2, 4, 4, 3, 3, 2, 2, 2, # group2 "3" and "4" are equally less common, and there is 2 more frequent
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, # group3 "4" is unique
4, 4, 4, 4, 5, 5, 5, 5, 2, 2, # group4 "2" is the less common, other ties more frequent
4, 4, 4, 4, 4, 5, 5, 5, 5, 5) # group5 "4" and "5" are equally common and no value is less common (similar to unique)
col1<-paste(c(rep("group1",10),rep("group2",10),rep("group3",10),rep("group4",10),rep("group5",10)), sep=", ")
df<-data.frame(col1=col1,col2=col2)
get.3modes.andcounts(df,"col1","col2")
#CURRENT result
col1 m1 m2 m3 counts
1 group1 4 3 2 10 # ok
2 group2 2 3 4 10 # no, 3 and 4 are the less common
3 group3 4 NA NA 10 # ok
4 group4 4 5 2 10 # no, 2 is less common
5 group5 4 5 NA 10 # ok
# desired
col1 m1 m2 m3 counts
1 group1 4 3 2 10
2 group2 2 NA NA 10
3 group3 4 NA NA 10
4 group4 4 5 NA 10
5 group5 4 5 NA 10
编辑：实际样本有多个关系，不希望有超过 3 列。超过 3 个数字（在 3 列中）仅在出现平局时才被接受。这就是为什么，我决定要求另一种类型的输出。
编辑：group7。只有三个最常见的通缉令。例外，包括第三个最常见的关系（就像在其他组中一样）。
# EXAMPLE 2
# new proposal
col2<-c(4, 4, 4, 4, 5, 3, 3, 3, 2, 2, 6, 6, # group1 2 and 6 tied in the 3rd position, 5 less common
2, 2, 2, 4, 4, 3, 3, 2, 2, 2, 6, 6, # group2 4, 3 and 6 tied in the less common, excluded.
4, 4, 4, 7, 7, 7, 5, 5, 5, 4, 4, 6, # group3 4, 7 and 5 more common, 3 most common present, exclude everything else
4, 4, 4, 4, 5, 5, 5, 5, 2, 2, 6, 6, # group4 2 and 6 less common, excluded (4 AND 5 tied)
4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, # group5 6 less common, excluded, (4 and 5 tied)
4, 4, 4, 3, 3, 3, 2, 2, 2, 1, 1, 1, # group6 all tied
14,14,14,16,16,16,16,34,34,42,42,42,80,80,84,92, #group7 16, 14, 42 are the three most freq.
20,52,40,40,40,20,20,60,60,50) #group 8 20,40 tied, 60 next.
col1<-paste(c(rep("group1",12),rep("group2",12),rep("group3",12),rep("group4",12),rep("group5",12),
rep("group6",12),rep("group7",16),rep("group8", 10)), sep=", ")
df<-data.frame(col1=col1,col2=col2)
#desired output
col1 m1 m2 m3 counts
1 group1 4 3 2,6 12 # 2 and 6 tied in the 3rd position, 5 less common
2 group2 2 NA NA 12 # 4, 3 and 6 tied in the less common, excluded.
3 group3 4 7,5 NA 12 # three most common numbers present, exclude everything else
4 group4 4,5 NA NA 12 # 2 and 6 less common excluded (4 AND 5 tied)
5 group5 4,5 NA NA 12 # 6 less common, excluded, (4 and 5 tied)
6 group6 4,3,2,1 NA NA 12 # all tied
7 group7 16 14,42 NA 16 # three most frequent present, discard others
8 group8 20,40 60 NA 10 # three most frequent present
使用 dplyr 和 tidyr（plyr 的更新版本）：
library(dplyr)
library(tidyr)
df %>%
group_by(col1, col2) %>%
summarise(n = n()) %>%
mutate(m = min_rank(desc(n)), count = sum(n)) %>%
filter(m <= 3 & (m != max(m) | m == 1)) %>%
group_by(col1, m, count) %>%
summarize(a = paste(col2, collapse = ',')) %>%
spread(m, a, sep = '') %>%
ungroup
# # A tibble: 7 × 5
# col1 count m1 m2 m3
# * <fctr> <int> <chr> <chr> <chr>
# 1 group1 12 4 3 2,6
# 2 group2 12 2 <NA> <NA>
# 3 group3 12 4 5,7 <NA>
# 4 group4 12 4,5 <NA> <NA>
# 5 group5 12 4,5 <NA> <NA>
# 6 group6 12 1,2,3,4 <NA> <NA>
# 7 group7 16 16 14,42 <NA>
如果你在函数中需要它：
get.3modes.andcounts <- function(origtable, groupby, columnname) {
origtable %>%
group_by_(groupby, columnname) %>%
summarise(n = n()) %>%
mutate(r = min_rank(desc(n)), count = sum(n)) %>%
filter(r <= 3 & (r != max(r) | r == 1)) %>%
group_by_(groupby, 'r', 'count') %>%
summarize_(a = paste0('paste(',columnname, ', collapse = ",")')) %>%
spread(r, a, sep = '') %>%
ungroup
}
get.3modes.andcounts(df, 'col1', 'col2')
# # A tibble: 7 × 5
# col1 count m1 m2 m3
# * <fctr> <int> <chr> <chr> <chr>
# 1 group1 12 4 3 2,6
# 2 group2 12 2 <NA> <NA>
# 3 group3 12 4 5,7 <NA>
# 4 group4 12 4,5 <NA> <NA>
# 5 group5 12 4,5 <NA> <NA>
# 6 group6 12 1,2,3,4 <NA> <NA>
# 7 group7 16 16 14,42 <NA>
System.time
system.time(get.3modes.andcounts(df, 'col1', 'col2'))
# user system elapsed
# 0.012 0.000 0.011
benchmark(get.3modes.andcounts(df, 'col1', 'col2'), replications = 10, columns = c("test", "replications", "elapsed"))
# test replications elapsed
# 1 get.3modes.andcounts(df, "col1", "col2") 10 0.08
benchmark(get.3modes.andcounts(df, 'col1', 'col2'), replications = 100, columns = c("test", "replications", "elapsed"))
# test replications elapsed
# 1 get.3modes.andcounts(df, "col1", "col2") 100 0.684
benchmark(get.3modes.andcounts(df, 'col1', 'col2'), replications = 1000, columns = c("test", "replications", "elapsed"))
# test replications elapsed
# 1 get.3modes.andcounts(df, "col1", "col2") 1000 6.796
数据：
col2 <- c(4, 4, 4, 4, 5, 3, 3, 3, 2, 2, 6, 6, # group1 2 and 6 tied in the 3rd position, 5 less common
2, 2, 2, 4, 4, 3, 3, 2, 2, 2, 6, 6, # group2 4, 3 and 6 tied in the less common, excluded.
4, 4, 4, 7, 7, 7, 5, 5, 5, 4, 4, 6, # group3 4, 7 and 5 more common, 3 most common present, exclude everything else
4, 4, 4, 4, 5, 5, 5, 5, 2, 2, 6, 6, # group4 2 and 6 less common, excluded (4 AND 5 tied)
4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, # group5 6 less common, excluded, (4 and 5 tied)
4, 4, 4, 3, 3, 3, 2, 2, 2, 1, 1, 1, # group6 all tied
14,14,14,16,16,16,16,34,34,42,42,42,80,80,84,92) #group7 16, 14, 42 are the three most freq.
col1 <- paste(c(rep("group1", 12), rep("group2", 12), rep("group3", 12), rep("group4", 12), rep("group5", 12),
rep("group6", 12), rep("group7", 16)), sep = ", ")
df <- data.frame(col1=col1,col2=col2)
您可以更改 n > 0，它会起作用。你的问题要求 3，但我的回答会更通用，接受任何正整数。
使用基数 R:
myfun <- function( data, n = 3, col1, col2 )
{
## n: numeric: total number of most common elements per group
stopifnot( n > 0 )
a1 <- lapply( split( data, data[[col1]] ), function( x ) { # split data by col1
# browser()
val <- factor( x[[col2]] ) # factor of data values
z1 <- tabulate( val ) # frequency table of levels of val
z2 <- sort( z1[ z1 > 0 ], decreasing = TRUE ) # sorted frequency table with >0
lenx <- length( unique( z2 ) ) # length of unique of z2
if ( lenx == 1 ) { # lenx == 1
return( c( paste( ( levels(val)[ which( z1 %in% z2 ) ] ), collapse = ','), rep(NA_character_, n - 1 ), sum( z1 ) ) )
} else if ( lenx > 1 ) { # lenx > 1
# remove the minimum, and and extract values by using levels of val with indices from the match of z1 and z2
z2 <- setdiff( z2, min( z2 ) )
z2 <- sapply( z2, function( y ) paste( levels(val)[ which( z1 %in% y ) ], collapse = ',') )
# count the length of z2 and get indices of length >= n
z2_ind <- which( cumsum( lengths(unlist( lapply(z2, strsplit, split = "," ),
recursive = F ) ) ) >= n )
if( length( z2_ind ) > 0 ) {
z2 <- z2[ seq_len( z2_ind[1] ) ]
}
# adjust length by assigning NA
if( length(z2) != n ) { z2[ (length(z2)+1):n ] <- NA_character_ }
return( c( z2, sum( z1 ) ) )
} else { # lenx < 1
return( as.list( rep(NA_character_, n ), NA_character_ ) )
}})
a1 <- do.call('rbind', a1) # row bind values of a1
a1 <- data.frame( group = rownames( a1 ), a1, stringsAsFactors = FALSE )
colnames( a1 ) <- c( 'group', paste( 'm', 1:n, sep = '' ), 'count' )
rownames( a1 ) <- NULL # remove row names
return( a1 )
}
输出：
# example1:
myfun(df, 3, 'col1', 'col2')
# group m1 m2 m3 count
# 1 group1 4 3 2 10
# 2 group2 2 NA NA 10
# 3 group3 4 NA NA 10
# 4 group4 4, 5 NA NA 10
# 5 group5 4, 5 NA NA 10
# example 2
myfun(df3, 3, 'col1', 'col2')
# group m1 m2 m3 count
# 1 group1 4 3 2, 6 12
# 2 group2 2 NA NA 12
# 3 group3 4 5, 7 NA 12
# 4 group4 4, 5 NA NA 12
# 5 group5 4, 5 NA NA 12
# 6 group6 4, 3, 2, 1 NA NA 12
# 7 group7 16 14, 42 NA 16
通过将字母分配给 example 1 data df.
set.seed(1L)
df$col3 <- sample( letters, 50, TRUE )
myfun(df, 3, 'col1', 'col3')
# group m1 m2 m3 count
# 1 group1 x <NA> <NA> 10
# 2 group2 j,u <NA> <NA> 10
# 3 group3 a,d,f,g,i,j,k,q,w,y <NA> <NA> 10
# 4 group4 m <NA> <NA> 10
# 5 group5 u <NA> <NA> 10
不需要额外的包裹。试试这个：
count <- function(df) {
count_n <- function(vec, n) {
fac <- factor(table(vec), levels = sort(unique(table(vec)), decreasing = T))
top3 <- na.omit(names(sort(fac)[1:3]))
min <- names(fac[fac == min(levels(fac))])
if(length(levels(fac))==1){min <- 'NA'}
top3 <- setdiff(top3,min)
nums <- na.omit(names(fac[fac == levels(fac)[n]]))
ifelse(length(intersect(nums, top3))>0, paste(nums, collapse = ','),'NA')
} ##Get the number of rank n.
group <- unique(as.character(df$col1))
m1 <- aggregate(df, list(df$col1), count_n, 1)$col2
m2 <- aggregate(df, list(df$col1), count_n, 2)$col2
m3 <- aggregate(df, list(df$col1), count_n, 3)$col2
count <- aggregate(df, list(df$col1), length)$col2
res <- data.frame(col1 = group, m1, m2, m3, count)
res
}