我有一个员工名单,我需要知道他们最常在哪个部门.将员工ID与部门名称制表是微不足道的,但从频率表中返回部门名称而不是名册数量是很棘手的.下面是一个简单示例(列名=部门,行名=员工ID).
DF <- matrix(sample(1:9,9),ncol=3,nrow=3)
DF <- as.data.frame.matrix(DF)
> DF
V1 V2 V3
1 2 7 9
2 8 3 6
3 1 5 4
现在我该怎么办？
> DF2
RE
1 V3
2 V1
3 V2
the*_*ail 81
使用您的数据的一个选项(供将来参考,用于set.seed()使用可sample重复的示例):
DF <- data.frame(V1=c(2,8,1),V2=c(7,3,5),V3=c(9,6,4))
colnames(DF)[apply(DF,1,which.max)]
[1] "V3" "V1" "V2"
比使用更快的解决方案apply可能是max.col:
colnames(DF)[max.col(DF,ties.method="first")]
#[1] "V3" "V1" "V2"
......哪里ties.method可以是"random" "first"或"last"
如果碰巧有两列等于最大值,这当然会引起问题.我不确定你想在那个实例中做什么,因为你会有一些行的结果.例如:
DF <- data.frame(V1=c(2,8,1),V2=c(7,3,5),V3=c(7,6,4))
apply(DF,1,function(x) which(x==max(x)))
[[1]]
V2 V3
2 3
[[2]]
V1
1
[[3]]
V2
2
Aru*_*run 14
如果您对data.table解决方案感兴趣,请参阅此处.这有点棘手,因为你更喜欢获得第一个最大值的id.如果你想要最后的最大值,那就容易多了.然而,它并不复杂,而且速度快!
在这里,我生成了您的尺寸数据(26746*18).
set.seed(45)
DF <- data.frame(matrix(sample(10, 26746*18, TRUE), ncol=18))
data.table 回答:require(data.table)
DT <- data.table(value=unlist(DF, use.names=FALSE),
colid = 1:nrow(DF), rowid = rep(names(DF), each=nrow(DF)))
setkey(DT, colid, value)
t1 <- DT[J(unique(colid), DT[J(unique(colid)), value, mult="last"]), rowid, mult="first"]
# data.table solution
system.time({
DT <- data.table(value=unlist(DF, use.names=FALSE),
colid = 1:nrow(DF), rowid = rep(names(DF), each=nrow(DF)))
setkey(DT, colid, value)
t1 <- DT[J(unique(colid), DT[J(unique(colid)), value, mult="last"]), rowid, mult="first"]
})
# user system elapsed
# 0.174 0.029 0.227
# apply solution from @thelatemail
system.time(t2 <- colnames(DF)[apply(DF,1,which.max)])
# user system elapsed
# 2.322 0.036 2.602
identical(t1, t2)
# [1] TRUE
它在这些尺寸的数据上快了大约11倍,并且data.table也很好地扩展.
DT <- data.table(value=unlist(DF, use.names=FALSE),
colid = 1:nrow(DF), rowid = rep(names(DF), each=nrow(DF)))
setkey(DT, colid, value)
t1 <- DT[J(unique(colid)), rowid, mult="last"]
Val*_*tin 10
基于以上建议,以下data.table解决方案对我来说非常快:
library(data.table)
set.seed(45)
DT <- data.table(matrix(sample(10, 10^7, TRUE), ncol=10))
system.time(
DT[, col_max := colnames(.SD)[max.col(.SD, ties.method = "first")]]
)
#> user system elapsed
#> 0.15 0.06 0.21
DT[]
#> V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 col_max
#> 1: 7 4 1 2 3 7 6 6 6 1 V1
#> 2: 4 6 9 10 6 2 7 7 1 3 V4
#> 3: 3 4 9 8 9 9 8 8 6 7 V3
#> 4: 4 8 8 9 7 5 9 2 7 1 V4
#> 5: 4 3 9 10 2 7 9 6 6 9 V4
#> ---
#> 999996: 4 6 10 5 4 7 3 8 2 8 V3
#> 999997: 8 7 6 6 3 10 2 3 10 1 V6
#> 999998: 2 3 2 7 4 7 5 2 7 3 V4
#> 999999: 8 10 3 2 3 4 5 1 1 4 V2
#> 1000000: 10 4 2 6 6 2 8 4 7 4 V1
并且还具有以下优点:.SD通过在.SDcols以下方面提及它们,可以始终指定要考虑的列:
DT[, MAX2 := colnames(.SD)[max.col(.SD, ties.method="first")], .SDcols = c("V9", "V10")]
如果我们需要最小值的列名,如@lwshang所建议,只需要使用-.SD:
DT[, col_min := colnames(.SD)[max.col(-.SD, ties.method = "first")]]
一种解决方案是将日期从宽到长整形重整，将所有部门放在一列中，然后按雇主ID（在本例中为行号）进行分组，然后用最大值。也有两种方法可以处理这种关系。
library(tidyverse)
# sample data frame with a tie
df <- data_frame(V1=c(2,8,1),V2=c(7,3,5),V3=c(9,6,5))
# If you aren't worried about ties:
df %>%
rownames_to_column('id') %>% # creates an ID number
gather(dept, cnt, V1:V3) %>%
group_by(id) %>%
slice(which.max(cnt))
# A tibble: 3 x 3
# Groups: id [3]
id dept cnt
<chr> <chr> <dbl>
1 1 V3 9.
2 2 V1 8.
3 3 V2 5.
# If you're worried about keeping ties:
df %>%
rownames_to_column('id') %>%
gather(dept, cnt, V1:V3) %>%
group_by(id) %>%
filter(cnt == max(cnt)) %>% # top_n(cnt, n = 1) also works
arrange(id)
# A tibble: 4 x 3
# Groups: id [3]
id dept cnt
<chr> <chr> <dbl>
1 1 V3 9.
2 2 V1 8.
3 3 V2 5.
4 3 V3 5.
# If you're worried about ties, but only want a certain department, you could use rank() and choose 'first' or 'last'
df %>%
rownames_to_column('id') %>%
gather(dept, cnt, V1:V3) %>%
group_by(id) %>%
mutate(dept_rank = rank(-cnt, ties.method = "first")) %>% # or 'last'
filter(dept_rank == 1) %>%
select(-dept_rank)
# A tibble: 3 x 3
# Groups: id [3]
id dept cnt
<chr> <chr> <dbl>
1 2 V1 8.
2 3 V2 5.
3 1 V3 9.
# if you wanted to keep the original wide data frame
df %>%
rownames_to_column('id') %>%
left_join(
df %>%
rownames_to_column('id') %>%
gather(max_dept, max_cnt, V1:V3) %>%
group_by(id) %>%
slice(which.max(max_cnt)),
by = 'id'
)
# A tibble: 3 x 6
id V1 V2 V3 max_dept max_cnt
<chr> <dbl> <dbl> <dbl> <chr> <dbl>
1 1 2. 7. 9. V3 9.
2 2 8. 3. 6. V1 8.
3 3 1. 5. 5. V2 5.
dplyr解决方案：主意：
代码：
DF = data.frame(V1=c(2,8,1),V2=c(7,3,5),V3=c(9,6,4))
DF %>%
rownames_to_column() %>%
gather(column, value, -rowname) %>%
group_by(rowname) %>%
filter(rank(-value) == 1)
结果：
# A tibble: 3 x 3
# Groups: rowname [3]
rowname column value
<chr> <chr> <dbl>
1 2 V1 8
2 3 V2 5
3 1 V3 9
这种方法可以很容易地扩展到获得顶部n列。示例n=2：
DF %>%
rownames_to_column() %>%
gather(column, value, -rowname) %>%
group_by(rowname) %>%
mutate(rk = rank(-value)) %>%
filter(rk <= 2) %>%
arrange(rowname, rk)
结果：
# A tibble: 6 x 4
# Groups: rowname [3]
rowname column value rk
<chr> <chr> <dbl> <dbl>
1 1 V3 9 1
2 1 V2 7 2
3 2 V1 8 1
4 2 V3 6 2
5 3 V2 5 1
6 3 V3 4 2
一种选择dplyr 1.0.0可能是：
DF %>%
rowwise() %>%
mutate(row_max = names(.)[which.max(c_across(everything()))])
V1 V2 V3 row_max
<dbl> <dbl> <dbl> <chr>
1 2 7 9 V3
2 8 3 6 V1
3 1 5 4 V2
在某些情况下，使用pmap()(requires purrr)可能更安全：
DF %>%
mutate(row_max = pmap(across(everything()), ~ names(c(...)[which.max(c(...))])))
样本数据：
DF <- structure(list(V1 = c(2, 8, 1), V2 = c(7, 3, 5), V3 = c(9, 6,
4)), class = "data.frame", row.names = c(NA, -3L))