本文于2026-03-31更新。 如发现问题或者有建议，欢迎提交 Issue
Matt Dowle | DataCamp
这个人之牛逼，可以和Hadley Wickhams肩并肩。他是data.table包的作者，语法非常类sql，非常支持sql使用者。
data.table包是十大最流行的的包，超过ggplot2。
1 intro
我感觉DT分为DT[where, select, group by]三种结构，是非常类似sql的语言。
而且data.table建表自带recycling的功能，非常方便，本身也是一种data.frame。
1.1 recycling
data.table(x = c("A", "B"), y = 1:4)my_first_data_table <-
data.table(x = c("a","b","c","d","e"),
y = c(1,2,3,4,5))DT <- data.table(a = c(1L,2L),b = LETTERS[1:4])
DT[3,]
DT[2:3,]1.2 where
# DT and the data.table package are pre-loaded
# Print the second to last row of DT using .N
DT[2:.N]
# .N 相当于最后一行的意思
# Print the column names of DT
names(DT)
# Print the number or rows and columns of DT
dim(DT)
# Print a new data.table containing rows 2, 2, and 3 of DT
DT[c(2,2,3)]1.3 select
typeof(DT[,.(b)])datatable-faq.pdf 这个地方反馈了一些
data.table在select功能上的一些bug。
D <- 5
DT[, .(D)]
# DT[, D]
DT[,b] # vector
DT[,.("b")] # data.table by only one chr.
DT[,.(b)] # data.table
DT[,"b"] # data.table "b" = .(b).()相当于select函数，其中的数字可以相当于真的就是select 5。
然而，, D]相当于select No.5 col。
# DT and the data.table package are pre-loaded
DT <- data.table(
A = 1:5,
B = letters[1:5],
C = 6:10
)
# Subset rows 1 and 3, and columns B and C
DT[c(1,3), .(B,C)]
# Assign to ans the correct value
ans <- DT[,.(B, val = A*C)]
# Fill in the blanks such that ans2 equals target
target <- data.table(B = c("a", "b", "c", "d", "e",
"a", "b", "c", "d", "e"),
val = as.integer(c(6:10, 1:5)))
ans2 <- DT[, .(B, val = c(C,A))]我觉得
val = c(C,A)这步操作太强了。
1.4 group by
# iris is already available in your workspace
# Convert iris to a data.table: DT
DT <- as.data.table(iris)
# For each Species, print the mean Sepal.Length
DT[, mean(Sepal.Length), by = Species]
# Print mean Sepal.Length, grouping by first letter of Species
DT[, mean(Sepal.Length), by = substr(Species, 1, 1)]# data.table version of iris: DT
DT <- as.data.table(iris)
# Group the specimens by Sepal area (to the nearest 10 cm2) and count how many occur in each group
DT[, .N, by = 10 * round(Sepal.Length * Sepal.Width / 10)]
# Now name the output columns `Area` and `Count`
DT[, .N, by = 10 * round(Sepal.Length * Sepal.Width / 10)][,.(Area = round,Count = N)]# Create the data.table DT
DT <- data.table(A = rep(letters[2:1], each = 4L),
B = rep(1:4, each = 2L),
C = sample(8))
# Create the new data.table, DT2
DT2 <- DT[,.(C = cumsum(C)),by = .(A,B)]
DT2
# Select from DT2 the last two values from C while you group by A
DT2[,.(C = tail(C,2)),by =.(A)]2 intermediate
2.1 chaining
这里好的地方是，不需要像dplyr一样进行ungroup有点麻烦。
# The data.table package has already been loaded
# Build DT
DT <- data.table(A = rep(letters[2:1], each = 4L),
B = rep(1:4, each = 2L),
C = sample(8))
# Combine the two steps in a one-liner
DT[, .(C = cumsum(C)), by = .(A, B)][, .(C = tail(C, 2)), by = A]# The data.table DT is loaded in your workspace
# Perform chained operations on DT
as.data.table(iris)[, .(Sepal.Length = median(Sepal.Length),
Sepal.Width = median(Sepal.Width),
Petal.Length = median(Petal.Length),
Petal.Width = median(Petal.Width)),
by = Species][order(-Species)]order(-Species)倒序。
这里很多函数一定要定义是data.table。
2.2 .SD
这里类似于mutate_all
DT <- data.table(
x = c(2,1,2,1,2,2,1),
y = seq(1,13,2),
z = seq(2,14,2)
)
DT[,.(lapply(.SD,mean)),by = x]# A new data.table DT is available
# Mean of columns
DT[,lapply(.SD,mean),by = x]
# Median of columns
DT[,lapply(.SD,median),by = x]2.3 .SDcols
.SDcols是一个参数，是可以限定对哪些col执行函数。
> DT[, .SD[-1], by = grp, .SDcols = paste0("Q", 1:3)]
grp Q1 Q2 Q3
1: 6 4 1 4
2: 8 1 3 1
3: 8 5 2 5这个的SD[-1]相当于踢掉了每组的第一行。
paste0("Q", 1:3)相当于选用了以"Q"为首字母的前三列。
这点有点难理解，很正常，但是data.table包本身就是base在R的基础函数上开发的，非常快。
> DT[,.(lapply(.SD,sum),.N),by = x]
x V1 N
1: 2 26 4
2: 2 30 4
3: 1 23 3
4: 1 26 3
> DT[, lapply(.SD, cumsum), by = .(by1 = x, by2 = z > 8), .SDcols = c("x", "y")]
by1 by2 x y
1: 2 FALSE 2 1
2: 2 FALSE 4 6
3: 1 FALSE 1 3
4: 1 FALSE 2 10
5: 2 TRUE 2 9
6: 2 TRUE 4 20
7: 1 TRUE 1 13
> 2.4 :=
这是计算机一个符号，类似于x := x + 10，x在循环中每次增加10.
# The data.table DT
DT <- data.table(A = letters[c(1, 1, 1, 2, 2)], B = 1:5)
# Add column by reference: Total
DT[, Total := sum(B), by = A]
# Add 1 to column B
DT[c(2, 4), B := B + 1L]
# Add a new column Total2
DT[2:4, Total2 := sum(B), by = A]
# Remove the Total column
DT[, Total := NULL]
# Select the third column using `[[`
DT[[3]]我感觉这个包到这地方开始比tidyverse复杂了，所以作为backup了。