6.6 数据拆分

数据拆分通常是按找某一个分类变量分组，分完组就是计算，计算完就把结果按照原来的分组方式合并

## Notice that assignment form is not used since a variable is being added
g <- airquality$Month
l <- split(airquality, g) # 分组
l <- lapply(l, transform, Oz.Z = scale(Ozone)) # 计算：按月对 Ozone 标准化
aq2 <- unsplit(l, g) # 合并
head(aq2)
##   Ozone Solar.R Wind Temp Month Day       Oz.Z
## 1    41     190  7.4   67     5   1  0.7822293
## 2    36     118  8.0   72     5   2  0.5572518
## 3    12     149 12.6   74     5   3 -0.5226399
## 4    18     313 11.5   62     5   4 -0.2526670
## 5    NA      NA 14.3   56     5   5         NA
## 6    28      NA 14.9   66     5   6  0.1972879

tapply 自带分组的功能，按月份 Month 对 Ozone 中心标准化，其返回一个列表

with(airquality, tapply(Ozone, Month, scale))
## $`5`
##              [,1]
##  [1,]  0.78222929
##  [2,]  0.55725184
##  [3,] -0.52263993
##  [4,] -0.25266698
##  [5,]          NA
##  [6,]  0.19728792
##  [7,] -0.02768953
##  [8,] -0.20767149
....

上面的过程等价于

do.call("rbind", lapply(split(airquality, airquality$Month), transform, Oz.Z = scale(Ozone)))
##       Ozone Solar.R Wind Temp Month Day         Oz.Z
## 5.1      41     190  7.4   67     5   1  0.782229293
## 5.2      36     118  8.0   72     5   2  0.557251841
## 5.3      12     149 12.6   74     5   3 -0.522639926
## 5.4      18     313 11.5   62     5   4 -0.252666984
## 5.5      NA      NA 14.3   56     5   5           NA
## 5.6      28      NA 14.9   66     5   6  0.197287919
## 5.7      23     299  8.6   65     5   7 -0.027689532
## 5.8      19      99 13.8   59     5   8 -0.207671494
## 5.9       8      19 20.1   61     5   9 -0.702621887
....

由于上面对 Ozone 正态标准化，所以标准化后的 Oz.z 再按月分组计算方差自然每个月都是 1，而均值都是 0。

with(aq2, tapply(Oz.Z, Month, sd, na.rm = TRUE))
## 5 6 7 8 9 
## 1 1 1 1 1
with(aq2, tapply(Oz.Z, Month, mean, na.rm = TRUE))
##             5             6             7             8             9 
## -4.240273e-17  1.052760e-16  5.841432e-17  5.898060e-17  2.571709e-17

循着这个思路，我们可以用 tapply 实现分组计算，上面函数 sd 和 mean 完全可以用自定义的更加复杂的函数替代

cut 函数可以将连续型变量划分为分类变量

set.seed(2019)
Z <- stats::rnorm(10)
cut(Z, breaks = -6:6)
##  [1] (0,1]   (-1,0]  (-2,-1] (0,1]   (-2,-1] (0,1]   (-1,0]  (0,1]   (-2,-1]
## [10] (-1,0] 
## 12 Levels: (-6,-5] (-5,-4] (-4,-3] (-3,-2] (-2,-1] (-1,0] (0,1] (1,2] ... (5,6]
# labels = FALSE 返回每个数所落的区间位置
cut(Z, breaks = -6:6, labels = FALSE)
##  [1] 7 6 5 7 5 7 6 7 5 6

我们还可以指定参数 dig.lab 设置分组的精度，ordered 将分组变量看作是有序的，breaks 传递单个数时，表示分组数，而不是断点

cut(Z, breaks = 3, dig.lab = 4, ordered = TRUE)
##  [1] (0.06396,0.9186]  (-0.7881,0.06396] (-1.643,-0.7881]  (0.06396,0.9186] 
##  [5] (-1.643,-0.7881]  (0.06396,0.9186]  (-0.7881,0.06396] (0.06396,0.9186] 
##  [9] (-1.643,-0.7881]  (-0.7881,0.06396]
## Levels: (-1.643,-0.7881] < (-0.7881,0.06396] < (0.06396,0.9186]

此时，统计每组的频数，如图 6.2

# 条形图
plot(cut(Z, breaks = -6:6))
# 直方图
hist(Z, breaks = -6:6)
连续型变量分组统计连续型变量分组统计

图 6.2: 连续型变量分组统计

在指定分组数的情况下，我们还想获取分组的断点

labs <- levels(cut(Z, 3))
labs
## [1] "(-1.64,-0.788]" "(-0.788,0.064]" "(0.064,0.919]"

用正则表达式抽取断点

cbind(
  lower = as.numeric(sub("\\((.+),.*", "\\1", labs)),
  upper = as.numeric(sub("[^,]*,([^]]*)\\]", "\\1", labs))
)
##       lower  upper
## [1,] -1.640 -0.788
## [2,] -0.788  0.064
## [3,]  0.064  0.919

更多相关函数可以参考 findInterval 和 embed

tabulate 和 table 有所不同，它表示排列，由 0 和 1 组成的一个长度为 5 数组，其中 1 有 3 个，则排列组合为

combn(5, 3, tabulate, nbins = 5)
##      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,]    1    1    1    1    1    1    0    0    0     0
## [2,]    1    1    1    0    0    0    1    1    1     0
## [3,]    1    0    0    1    1    0    1    1    0     1
## [4,]    0    1    0    1    0    1    1    0    1     1
## [5,]    0    0    1    0    1    1    0    1    1     1