Data Operation
name <- c("Alex", "Rosie", "Greg")
gender <- c("M", "W", "M")
age <- c(18,19,20)
a_score <- c(98, 99, 100)
b_score <- c(1, 2, 3)
df <- data.frame(name, gender, age, a_score, b_score)
创建新变量
transform(df, ...): 对df进行...操作
df <- transform(df, age = age + 1, avg_score = (a_score+b_score)/2)
name gender age a_score b_score avg_score
1 Alex M 19 98 1 49.5
2 Rosie W 20 99 2 50.5
3 Greg M 21 100 3 51.5
变量重编码
# 未指定区间
grades <- c("D", "C", "B", "A")
tmp <- quantile(stu_tot, probs=c(0.4, 0.6, 0.8))
break_pts <- c(-Inf, tmp, Inf)
score_level <- cut(scores, break_pts, labels=grades, ordered_result=TRUE)
# 指定区间
scores <- c(50, 70, 93, 67)
grades <- c("F", "D", "C", "B", "A")
break_pts <- c(-Inf, 60, 70, 80, 90, Inf)
score_level <- cut(scores, break_pts, labels=grades, ordered_result=TRUE)
注意
cut(x, breaks, labels = NULL,
include.lowest = FALSE, right = TRUE, dig.lab = 3,
ordered_result = FALSE, …)
right=T 表示左开右闭
right=F 表示左闭右开
补充
pretty(x, …)
# S3 method for default
pretty(x, n = 5, min.n = n %/% 3, shrink.sml = 0.75,
high.u.bias = 1.5, u5.bias = .5 + 1.5*high.u.bias,
eps.correct = 0, …)
n
integer giving the desired number of intervals. Non-integer values are rounded down.
min.n
nonnegative integer giving the minimal number of intervals. If min.n == 0, pretty(.) may return a single value.
pretty(1:15) # 0 2 4 6 8 10 12 14 16
pretty(1:15, high.u.bias = 2) # 0 5 10 15
pretty(1:15, n = 4) # 0 5 10 15
pretty(1:15 * 2) # 0 5 10 15 20 25 30
pretty(1:20) # 0 5 10 15 20
pretty(1:20, n = 2) # 0 10 20
pretty(1:20, n = 10) # 0 2 4 ... 20
变量的重命名
# 法一
fix(df) # 跳出来一个编辑框
# 法二
names(df)[1] <- "sname"
# 法三
install.packages("plyr")
library(plyr)
df <- rename(df, c(name="sname", age="sage"))
缺失值
df == NA永远不会为T，因为NA不可比- R不会把
Inf或NaN标记为NA
is.na(c(1,2,3))
## [1] FALSE FALSE FALSE
is.na(df)
name gender age a_score b_score
[1,] FALSE FALSE FALSE FALSE FALSE
[2,] FALSE FALSE FALSE FALSE FALSE
[3,] FALSE FALSE FALSE FALSE FALSE
df$test <- NA
df
## name gender age a_score b_score test
## 1 Alex M 18 98 1 NA
## 2 Rosie W 19 99 2 NA
## 3 Greg M 20 100 3 NA
df[is.na(df)]
## [1] NA NA NA
df[is.na(df)] <- T
df
## name gender age a_score b_score test
## 1 Alex M 18 98 1 TRUE
## 2 Rosie W 19 99 2 TRUE
## 3 Greg M 20 100 3 TRUE
日期
# 获得当前日期
Sys.Date()
## [1] "2021-10-24"
date()
## [1] "Sun Oct 24 20:34:34 2021"
# 字符串转换为日期
as.Date(c("1970-1-5", "2017-9-12"))
as.Date("1/5/1970", format="%m/%d/%Y")
# 默认format=yyyy-mm-dd
# 指定输出形式
format(Sys.Date(), format="%B %d %Y")
## [1] "十月 24 2021"
# 算术运算
sdate <- as.Date("2021-02-13")
edate <- as.Date("2022-02-13")
days <- edate - sdate
days
## Time difference of 365 days
# 或
difftime(edate, sdate, units="weeks")
## Time difference of 52.14286 weeks
# 逻辑运算
sdate > edate
## [1] FALSE
# 日期转换为字符串
sdate <- as.character(Sys.Date())
sdate
## [1] "2021-10-24"
类型转换
is.numeric() as.numeric()
排序
一些绘图需要排序预处理，如打乱的时序数据
# order() 默认是升序，加一个减号变为降序
new_df <- df[order(df$a_score, -df$b_score),]
# 优先按a_score升序，若a_score相同则按b_score降序
合并
1. 添加列
id1 <- c(2, 3, 4, 5, 7)
heights <- c(62, 65, 71, 71, 67)
df1 <- data.frame(id = id1, heights)
id2 <- c(1, 2, 6:10)
weights <- c(147, 113, 168, 135, 142, 159, 160)
df2 <- data.frame(id = id2, weights)
df1
## id heights
## 1 2 62
## 2 3 65
## 3 4 71
## 4 5 71
## 5 7 67
df2
## id weights
## 1 1 147
## 2 2 113
## 3 6 168
## 4 7 135
## 5 8 142
## 6 9 159
## 7 10 160
merge(df1, df2, all = FALSE) # colnames(df1) ∩ colnames(df2) = id
## id heights weights
## 1 2 62 113
## 2 7 67 135
merge(df2, df1, all = FALSE)
## id weights heights
## 1 2 113 62
## 2 7 135 67
merge(df1, df2, by = "id", all = TRUE)
## id heights weights
## 1 1 NA 147
## 2 2 62 113
## 3 3 65 NA
## 4 4 71 NA
## 5 5 71 NA
## 6 6 NA 168
## 7 7 67 135
## 8 8 NA 142
## 9 9 NA 159
## 10 10 NA 160
merge(df1, df2, by = "id", all.x = TRUE)
## id heights weights
## 1 2 62 113
## 2 3 65 NA
## 3 4 71 NA
## 4 5 71 NA
## 5 7 67 135
merge(df1, df2, by = "id", all.y = TRUE)
## id heights weights
## 1 1 NA 147
## 2 2 62 113
## 3 6 NA 168
## 4 7 67 135
## 5 8 NA 142
## 6 9 NA 159
## 7 10 NA 160
# 按行名进行合并
z <- matrix(c(0,0,1,1,0,0,1,1,0,0,0,0,1,0,1,1,0,1,1,1,1,0,0,0,"RND1","WDR", "PLAC8","TYBSA","GRA","TAF"), nrow=6, dimnames=list(c("ILMN_1651838","ILMN_1652371","ILMN_1652464","ILMN_1652952","ILMN_1653026","ILMN_1653103"),c("A","B","C","D","symbol")))
t<-matrix(c("GO:0002009", 8, 342, 1, 0.07, 0.679, 0, 0, 1, 0,
"GO:0030334", 6, 343, 1, 0.07, 0.065, 0, 0, 1, 0,
"GO:0015674", 7, 350, 1, 0.07, 0.065, 1, 0, 0, 0), nrow=10, dimnames= list(c("GO.ID","LEVEL","Annotated","Significant","Expected","resultFisher","ILMN_1652464","ILMN_1651838","ILMN_1711311","ILMN_1653026")))
> z
A B C D symbol
ILMN_1651838 "0" "1" "1" "1" "RND1"
ILMN_1652371 "0" "1" "0" "1" "WDR"
ILMN_1652464 "1" "0" "1" "1" "PLAC8"
ILMN_1652952 "1" "0" "1" "0" "TYBSA"
ILMN_1653026 "0" "0" "0" "0" "GRA"
ILMN_1653103 "0" "0" "1" "0" "TAF"
> t
[,1] [,2] [,3]
GO.ID "GO:0002009" "GO:0030334" "GO:0015674"
LEVEL "8" "6" "7"
Annotated "342" "343" "350"
Significant "1" "1" "1"
Expected "0.07" "0.07" "0.07"
resultFisher "0.679" "0.065" "0.065"
ILMN_1652464 "0" "0" "1"
ILMN_1651838 "0" "0" "0"
ILMN_1711311 "1" "1" "0"
ILMN_1653026 "0" "0" "0"
cbind(t, z[, "symbol"][match(rownames(t), rownames(z))])
[,1] [,2] [,3] [,4]
GO.ID "GO:0002009" "GO:0030334" "GO:0015674" NA
LEVEL "8" "6" "7" NA
Annotated "342" "343" "350" NA
Significant "1" "1" "1" NA
Expected "0.07" "0.07" "0.07" NA
resultFisher "0.679" "0.065" "0.065" NA
ILMN_1652464 "0" "0" "1" "PLAC8"
ILMN_1651838 "0" "0" "0" "RND1"
ILMN_1711311 "1" "1" "0" NA
ILMN_1653026 "0" "0" "0" "GRA"
df1 <- data.frame(name, age, gender)
df2 <- data.frame(name, a_score, b_score)
df <- merge(df1, df2, by="name")
name age gender a_score b_score
1 Alex 18 M 98 1
2 Greg 20 M 100 3
3 Rosie 19 W 99 2
# dff <- merge(df1, df2, by=c("name", "age")) 也有这样的语法
# 直接合并，不指定索引
df <- cbind(df1, df2)
name age gender name a_score b_score
1 Alex 18 M Alex 98 1
2 Rosie 19 W Rosie 99 2
3 Greg 20 M Greg 100 3
2. 添加行
两个df必须拥有相同的变量。不过顺序不必相同
df1 <- data.frame(name, age, gender)
df2 <- data.frame(name, age, gender)
df <- rbind(df1, df2)
子集操作
访问见 Grammar
丢弃变量
# 丢弃name与gender列
# 法一
tmp <- names(df) %in% c("name", "gender")
new_df <- df[!tmp]
# 法二
df$name <- df$gender <- NULL
随机抽样
# 从第一行到最后一行中，随机抽2个，无放回
mysample <- df[sample(1:nrow(df), 2, replace=F), ]
SQL语句
install.packages('sqldf')
library(sqldf)
new_df <- sqldf('select * from df where name=\'Alex\'', row.names=T) # 保留行名
常用函数
1. 数学函数P86 统计函数P88
2. scale()
- 归一化 默认均值为0 方差为1
- 存在非数值列会报错
new_df <- scale(df)
# 若要化为均值为M，方差为SD
new_df <- scale(df) * SD + M
# 若想要只针对某个列（col_name）进行标准化
new_df <- transform(df, col_name=scale(col_name)*SD+M)
3. 概率函数
set.seed(1234) # 指定种子，用于复现，但只是一次性的，每次随机都要重新指定
runif(5) # 生成5个服从均匀分布U(0,1)的伪随机数
runif(5, min=0, max=100) # U[0, 100]
rnorm(10) # 生成标准正态分布的随机数
rnorm(10,mean=1,sd=4) # 生成...随机数 sd是标准差
mvrnorm(100, mu=mean, Sigma=sigma) # 生成100个服从N(mean, sigma)
# mean是平均值向量，sigma是协方差矩阵
dnorm(0) # 表示标准正态分布密度 就是咱们记得表达式，像凸字形
## [1] 0.3989423
pnorm(0) # 标准正态分布函数 形似 sigmoid
## [1] 0.5
sample(x, size, replace = FALSE, prob = NULL)：从 x 中选 size 个
- size: 个数
- replace: 若为T，则有放回的抽
4. 字符处理函数P93
5. 其他实用函数
option(digits=2) # 后续都保留两位小数
x <- 1:10
length(x)
## [1] 10
m <- matrix(1:20, nrow=4)
length(m)
## [1] 20
df <- data.frame(name=c("A", "B", "C"), score=1:3)
length(df)
## [1] 2
arr <- array(1:20, dim=c(2,2,5))
length(arr)
## [1] 20
cut(x, 5) # 分割为一个有着5个水平的因子
## [1] (0.991,2.8] (0.991,2.8] (2.8,4.6] (2.8,4.6] (4.6,6.4]
## [6] (4.6,6.4] (6.4,8.2] (6.4,8.2] (8.2,10] (8.2,10]
## 5 Levels: (0.991,2.8] (2.8,4.6] (4.6,6.4] ... (8.2,10]
cut(x, breaks=0:3)
## [1] (0,1] (1,2] (2,3] <NA> <NA> <NA> <NA> <NA> <NA> <NA>
## Levels: (0,1] (1,2] (2,3]
labels = c("差", "中", "优")
cut(x, breaks=0:3, labels)
## [1] 差 中 优 <NA> <NA> <NA> <NA> <NA> <NA> <NA>
## Levels: 差 中 优
cut(x, breaks=0:3, labels, ordered_result=TRUE)
## [1] 差 中 优 <NA> <NA> <NA> <NA> <NA> <NA> <NA>
## Levels: 差 < 中 < 优
pretty(x, 5) # 创建美观的分割点
## [1] 0 2 4 6 8 10
quantile(x, c(.8, .6, .4, .2)) # 返回一根向量
## 80% 60% 40% 20%
## 8.2 6.4 4.6 2.8
x <- list(1:2, 'aaa', matrix(1:20,nrow=4))
"["(x, 2) # 返回x的第二个元素，仍是列表
## [[1]]
## [1] "aaa"
cat("我后面会有一个空格", "我前面有一个空格", "不会自动\n换行")
## 我后面会有一个空格 我前面有一个空格 不会自动
## 换行
cat(..., file='xxx', append=FALSE)
6. 批量操作函数
apply(x, MARGIN, FUN)
输入一个df或matrix，输出vector或list或array
MARGIN=1表示行，MARGIN=2表示列
若MARGIN=1，则表示对x的每一行进行FUN操作
即 MARGIN 为多少，就保留这些维度，即若 MARGIN = c(1, 2)，那么最后的 shape 是 (len(第一维度), len(第二维度))
apply(X, # Array, matrix or data frame
MARGIN, # 1: columns, 2: rows, c(1, 2): rows and columns
FUN, # Function to be applied
...) # Additional arguments to FUN
m <- matrix(1:20, nrow=4)
apply(m, 1, mean) # 对每一行求平均值
apply(m, 1, mean, trim=0.2) # 对每一行的20%~80%的部分求平均
x <- array(1:24, c(3,4,2))
apply(x, c(1, 2), sum) # 在第三维度值求和
[,1] [,2] [,3] [,4]
[1,] 14 20 26 32
[2,] 16 22 28 34
[3,] 18 24 30 36
# Example
# 给出有2列的矩阵，求斜边长
apply(m, 1, function(x) sqrt(sum(x^2)))
sapply(x, FUN, options)
返回一个长度与 X 一致的向量(通过length()函数可以反向推出处理的单元)，每个元素为 FUN 计算出的结果，且分别对应到X中的每个元素。
sapply(X, # Vector, list or expression object
FUN, # Function to be applied
..., # Additional arguments to be passed to FUN
simplify = TRUE, # If FALSE returns a list. If "array" returns an array if possible
USE.NAMES = TRUE) # If TRUE and if X is a character vector, uses the names of X
# 输入列表
l <- list(a=1:10, b=matrix(1:20, nrow=4))
sapply(l, sum)
## a b
## 55 210
# 输入向量
movies <- c("SPYDERMAN","BATMAN","VERTIGO","CHINATOWN")
movies_lower <-sapply(movies, tolower)
movies_lower
## SPYDERMAN BATMAN VERTIGO CHINATOWN
## "spyderman" "batman" "vertigo" "chinatown"
tolower(movies)
## [1] "spyderman" "batman" "vertigo" "chinatown"
# 输入df
df
## age score
## 1 18 99
## 2 19 98
## 3 20 97
sapply(df, sum)
## age score
## 57 294
# 输入矩阵
m <- matrix(1:20, nrow=4)
sapply(m, sum)
## [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
# 事与愿违是因为
length(m)
## [1] 20
sum(m)
## [1] 210
mean(m)
## [1] 10.5
# FUN使用参数
sapply(x, "[", 2) # 若x是一个列表，则取列表中每一个元素的第2个元素 见课本P99
# 可以基于这个实现向量切片
below_ave <- function(x) {
ave <- mean(x)
return(x[x > ave])
}
dt_s <- sapply(df, below_ave)
## age score
## 20.0 51.5
dt_l <- lapply(df, below_ave)
## $age
## [1] 20
## $score
## [1] 51.5
identical(dt_s, dt_l)
## [1] TRUE
# 实例，假设df$name 由Alex Sean类似格式组成，现要求从中抽取出姓和名
name <- strsplit((df$name), " ")
fname <- sapply(name, "[", 2)
lname <- sapply(name, "[", 1)
lapply(x, FUN, options)
返回一个长度与 X 一致的列表，每个元素为 FUN 计算出的结果，且分别对应到X中的每个元素。
与apply()区别在于输出的是list，且无需MARGIN
lapply(X, # List or vector
FUN, # Function to be applied
...) # Additional arguments to be passed to FUN
# 输入列表
l <- list(a=1:10, b=matrix(1:20, nrow=4))
lapply(l, sum)
## $a
## [1] 55
## $b
## [1] 210
# 输入向量
movies <- c("SPYDERMAN","BATMAN","VERTIGO","CHINATOWN")
movies_lower <-lapply(movies, tolower)
movies_lower
## [[1]]
## [1] "spyderman"
## [[2]]
## [1] "batman"
## [[3]]
## [1] "vertigo"
## [[4]]
## [1] "chinatown"
unlist(movies_lower)
## [1] "spyderman" "batman" "vertigo" "chinatown"
# 输入df
df
## age score
## 1 18 99
## 2 19 98
## 3 20 97
lapply(df, sum)
## $age
## [1] 57
## $score
## [1] 294
# 输入矩阵
m <- matrix(1:20, nrow=4)
lapply(m, sum)
## [[1]]
## [1] 1
## [[2]]
## [1] 2
## ...
# 事与愿违是因为
length(m)
## [1] 20
tapply(X, INDEX, FUN): 对一维数据按 label 分组后分别 apply fun
tapply(X, # Object you can split (matrix, data frame, ...)
INDEX, # List of factors of the same length
FUN, # Function to be applied to factors (or NULL)
..., # Additional arguments to be passed to FUN
default = NA, # If simplify = TRUE, is the array initialization value
simplify = TRUE) # If set to FALSE returns a list object
set.seed(2)
data_set <- data.frame(price = round(rnorm(25, sd = 10, mean = 30)),
type = sample(1:4, size = 25, replace = TRUE),
store = sample(paste("Store", 1:4),
size = 25, replace = TRUE))
head(data_set)
price type store
1 21 2 Store 2
2 32 3 Store 3
3 46 4 Store 4
4 19 3 Store 4
5 29 1 Store 4
6 31 3 Store 4
price <- data_set$price
store <- data_set$store
type <- factor(data_set$type,
labels = c("toy", "food", "electronics", "drinks"))
# Mean price by product type
mean_prices <- tapply(price, type, mean)
mean_prices
toy food electronics drinks
39.50000 30.33333 32.20000 29.33333
class(mean_prices)
## "array"
# 等价于
by(data_set, data_set$type, function(x) mean(x$price))
data_set$type: 1
[1] 39.5
------------------------------------------
data_set$type: 2
[1] 30.33333
------------------------------------------
data_set$type: 3
[1] 32.2
------------------------------------------
data_set$type: 4
[1] 29.33333
# 由于array不能用 $ 操作，所以可考虑返回list
# Mean price by product type
mean_prices_list <- tapply(price, type, mean, simplify = FALSE)
mean_prices_list
$toy
[1] 39.5
$food
[1] 30.33333
$electronics
[1] 32.2
$drinks
[1] 29.33333
# Mean price by product type and store
tapply(price, list(type, store), mean)
Store 1 Store 2 Store 3 Store 4
toy 46 31.00000 49 36.66667
food 26 30.33333 39 NA
electronics 50 29.00000 32 25.00000
drinks 22 40.00000 20 36.00000
# Mean price by product type and store, changing default argument
tapply(price, list(type, store), mean, default = 0)
Store 1 Store 2 Store 3 Store 4
toy 46 31.00000 49 36.66667
food 26 30.33333 39 0.00000
electronics 50 29.00000 32 25.00000
drinks 22 40.00000 20 36.00000
mapply()
mapply(FUN, …, MoreArgs = NULL, SIMPLIFY = TRUE,
USE.NAMES = TRUE)
mapply(function(x,y){x^y},x=c(2,3),y=c(3,4))
## [1] 8 81
# the values in y are recycled.
# i.e. for both the values in x the same value (4) of y is used.
mapply(function(x,y){x^y},x=c(2,3),y=c(4))
## [1] 16 81
mapply(function(x,y){x^y},c(a=2,b=3),c(A=3,B=4))
## a b
## 8 81
mapply(function(x,y){x^y},c(a=2,b=3),c(A=3,B=4),USE.NAMES=FALSE)
## [1] 8 81
# 指定fun中的额外参数
mapply(function(x,y,z,k){(x+k)^(y+z)},c(a=2,b=3),c(A=3,B=4),MoreArgs=list(1,2))
## a b
## 256 3125
vapply()
vapply(.x, .f, fun_value, ..., use_names = TRUE)
The vapply function is very similar compared to the sapply function, but when using vapply you need to specify the output type explicitly.
test <- list(a = c(1, 3, 5), b = c(2,4,6), c = c(9,8,7), d=c("A", "B", "C"))
sapply(test, max)
a b c d
"5" "6" "9" "C"
vapply(test, max, numeric(1))
## Error in vapply(test, max, numeric(1)) : values must be type 'double',
## but FUN(X[[4]]) result is type 'character'
rapply()
rapply stands for recursive apply, and as the name suggests it is used to apply a function to all elements of a list recursively.
x <- list(1,2,3,4)
rapply(x,function(x){x^2},class=c("numeric"))
## [1] 1 4 9 16
x <- list(1,2,3,4,"a")
rapply(x,function(x){x^2},class=c("numeric"))
## [1] 1 4 9 16
7. 聚合函数
aggregate(x, by, FUN)
FUN只能是单返回值函数
整合数据，相当于group by by_list and apply fun on each group
by中的变量必须在一个list中，即使只有一个元素；可以在列表中自定义各组的别名
# Data frame
aggregate(x, # R object
by, # List of variables (grouping elements)
FUN, # Function to be applied for summary statistics
..., # Additional arguments to be passed to FUN
simplify = TRUE, # Whether to simplify results as much as possible or not
drop = TRUE) # Whether to drop unused combinations of grouping values or not.
# Formula
aggregate(formula, # Input formula
data, # List or data frame where the variables are stored
FUN, # Function to be applied for summary statistics
..., # Additional arguments to be passed to FUN
subset, # Observations to be used (optional)
na.action = na.omit) # How to deal with NA values
aggregate(mtcars, by=list(mtcars$cyl, mtcars$gear), FUN=mean, na.rm=TRUE)
# group by(cyl, gear) and apply mean on each group
aggregate(mtcars$disp, by=list(mtcars$cyl, mtcars$gear), FUN=mean, na.rm=TRUE)
Group.1 Group.2 x
1 4 3 120.1000
2 6 3 241.5000
3 8 3 357.6167
4 4 4 102.6250
5 6 4 163.8000
6 4 5 107.7000
7 6 5 145.0000
8 8 5 326.0000
# 等价于
aggregate(disp~cyl+gear, data=mtcars, FUN=mean, na.rm=TRUE)
cyl gear disp
1 4 3 120.1000
2 6 3 241.5000
3 8 3 357.6167
4 4 4 102.6250
5 6 4 163.8000
6 4 5 107.7000
7 6 5 145.0000
8 8 5 326.0000
# 统计频率
aggregate(chickwts$feed, by = list(chickwts$feed), FUN = length)
aggregate(feed ~ feed, data = chickwts, FUN = length) # Equivalent
states <- data.frame(state.region, state.x77)
## state.region Population Income Illiteracy Life.Exp Murder HS.Grad Frost Area
## Alabama South 3615 3624 2.1 69.05 15.1 41.3 20 50708
## Alaska West 365 6315 1.5 69.31 11.3 66.7 152 566432
## Arizona West 2212 4530 1.8 70.55 7.8 58.1 15 113417
aggregate(states$Illiteracy, by=list(state.region), FUN=mean)
# Group.1 x
1 Northeast 1.000000
2 South 1.737500
3 North Central 0.700000
4 West 1.023077
by(data, INDICES, FUN)
indices是一个因子或因子组成的列表（定义分组），fun是任意函数
即将data按照indices因子水平进行分组，然后对每组应用fun函数
df
## name gender age a_score b_score
## 1 Alex M 18 98 1
## 2 Rosie W 19 99 2
## 3 Greg M 20 100 3
by(df, df$gender, function(x) mean(x[,4]))
## df$gender: M
## [1] 99
## ------------------------------------------------
## df$gender: W
## [1] 99
table()
类似count on group-by
默认会忽略NA，需要NA的话，添加参数useNa='ifany'
table(Arthritis$Improved) # 统计频数
## None Some Marked
## 42 14 28
table(Arthritis$Improved, Arthritis$Treatment)
## Placebo Treated
## None 29 13
## Some 7 7
## Marked 7 21
8. 转置t()
df <- data.frame(name, gender, age, a_score, b_score)
row.names(df) <- df$name # 或rownames(df) <- df$name
df$name <- NULL
t(df)
Alex Rosie Greg
gender "M" "W" "M"
age "18" "19" "20"
a_score " 98" " 99" "100"
b_score "1" "2" "3"