R语言实战 - 高级数据管理

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2.4 字符处理函数

> x <- c("ab", "cde", "fghij")
> length(x)
[1] 3
> nchar(x[3])
[1] 5
> 
> 
> x <- "abcdef"
> substr(x, 2, 4)
[1] "bcd"
> substr(x, 2, 4) <- "22222"
> x
[1] "a222ef"
> 
> 
> grep("A", c("b", "A", "c"), fixed=TRUE)
[1] 2
> 
> 
> sub("\\s", ".", "Hello There")
[1] "Hello.There"
> 
> 
> y <- strsplit("abc", "")
> y
[[1]]
[1] "a" "b" "c"

> unlist(y)[2]
[1] "b"
> sapply(y, "[", 2)
[1] "b"
> 
> 
> paste("x", 1:3, sep="")
[1] "x1" "x2" "x3"
> paste("x", 1:3, sep="M")
[1] "xM1" "xM2" "xM3"
> paste("Today is", date())
[1] "Today is Sun Sep 10 20:39:26 2017"
> 
> 
> toupper("abc")
[1] "ABC"
> tolower("ABC")
[1] "abc"
> 

2.5 其他实用函数

> x <- c(2, 5, 6, 9)
> length(x)
[1] 4
> 
> 
> indices <- seq(1, 10, 2)
> indices
[1] 1 3 5 7 9
> 
> 
> y <- rep(1:3, 2)
> y
[1] 1 2 3 1 2 3
> 
> 
> z <- cut(y, 3) 
> z
[1] (0.998,1.67] (1.67,2.33]  (2.33,3]     (0.998,1.67] (1.67,2.33] 
[6] (2.33,3]    
Levels: (0.998,1.67] (1.67,2.33] (2.33,3]
> 
> 
> u <- pretty(y, 3)
> u
[1] 1.0 1.5 2.0 2.5 3.0
> 
> 
> firstname <- c("Jane")
> cat("Hello", firstname, "\n")
Hello Jane 
> 
> 
> name <- "Bob"
> cat("Hello", name, "\b.\n", "Isn\‘t R", "\t", "GREAT?\n")
Hello Bob.
 Isn‘t R         GREAT?
> 

2.6 将函数应用于矩阵和数据框

> a <- 5
> sqrt(a)
[1] 2.24
> b <- c(1.243, 5.654, 2.99)
> round(b)
[1] 1 6 3
> c <- matrix(runif(12), nrow=3)
> c
       [,1]  [,2]  [,3]  [,4]
[1,] 0.9636 0.216 0.289 0.913
[2,] 0.2068 0.240 0.804 0.353
[3,] 0.0862 0.197 0.378 0.931
> log(c)
        [,1]  [,2]   [,3]    [,4]
[1,] -0.0371 -1.53 -1.241 -0.0912
[2,] -1.5762 -1.43 -0.218 -1.0402
[3,] -2.4511 -1.62 -0.972 -0.0710
> mean(c)
[1] 0.465
> 
> mydata <- matrix(rnorm(30), nrow=6)
> mydata
       [,1]   [,2]   [,3]   [,4]   [,5]
[1,]  0.459  1.203  1.234  0.591 -0.281
[2,] -1.261  0.769 -1.891 -0.435  0.812
[3,] -0.527  0.238 -0.223 -0.251 -0.208
[4,] -0.557 -1.415  0.768 -0.926  1.451
[5,] -0.374  2.934  0.388  1.087  0.841
[6,] -0.604  0.935  0.609 -1.944 -0.866
> apply(mydata, 1, mean)
[1]  0.641 -0.401 -0.194 -0.136  0.975 -0.374
> apply(mydata, 2, mean)
[1] -0.478  0.777  0.148 -0.313  0.292
> apply(mydata, 2, mean, trim=0.2)
[1] -0.516  0.786  0.386 -0.255  0.291
> 

3. 数据处理难题的一套解决方案

> options(digits=2)
> 
> Student <- c("John Davis", "Angela Williams", "Bullwinkle Moose",
+              "David Jones", "Janice Markhammer", "Cheryl Cushing",
+              "Reuven Ytzrhak", "Greg Knox", "Joel England",
+              "Mary Rayburn")
> Math <- c(502, 600, 412, 358, 495, 512, 410, 625, 573, 522)
> Science <- c(95, 99, 80, 82, 75, 85, 80, 95, 89, 86)
> English <- c(25, 22, 18, 15, 20, 28, 15, 30, 27, 18)

> roster <- data.frame(Student, Math, Science, English, 
+                      stringsAsFactors=FALSE)
> roster
             Student Math Science English
1         John Davis  502      95      25
2    Angela Williams  600      99      22
3   Bullwinkle Moose  412      80      18
4        David Jones  358      82      15
5  Janice Markhammer  495      75      20
6     Cheryl Cushing  512      85      28
7     Reuven Ytzrhak  410      80      15
8          Greg Knox  625      95      30
9       Joel England  573      89      27
10      Mary Rayburn  522      86      18
> z <- scale(roster[, 2:4])
> z
        Math Science English
 [1,]  0.013   1.078   0.587
 [2,]  1.143   1.591   0.037
 [3,] -1.026  -0.847  -0.697
 [4,] -1.649  -0.590  -1.247
 [5,] -0.068  -1.489  -0.330
 [6,]  0.128  -0.205   1.137
 [7,] -1.049  -0.847  -1.247
 [8,]  1.432   1.078   1.504
 [9,]  0.832   0.308   0.954
[10,]  0.243  -0.077  -0.697
attr(,"scaled:center")
   Math Science English 
    501      87      22 
attr(,"scaled:scale")
   Math Science English 
   86.7     7.8     5.5 
> score <- apply(z, 1, mean)
> score
 [1]  0.56  0.92 -0.86 -1.16 -0.63  0.35 -1.05  1.34  0.70 -0.18
> roster <- cbind(roster, score)
> roster
             Student Math Science English score
1         John Davis  502      95      25  0.56
2    Angela Williams  600      99      22  0.92
3   Bullwinkle Moose  412      80      18 -0.86
4        David Jones  358      82      15 -1.16
5  Janice Markhammer  495      75      20 -0.63
6     Cheryl Cushing  512      85      28  0.35
7     Reuven Ytzrhak  410      80      15 -1.05
8          Greg Knox  625      95      30  1.34
9       Joel England  573      89      27  0.70
10      Mary Rayburn  522      86      18 -0.18
> y <- quantile(roster$score, c(.8, .6, .4, .2))
> y
  80%   60%   40%   20% 
 0.74  0.44 -0.36 -0.89 


> roster$grade[score >= y[1]] <- "A"
> roster$grade[score < y[1] & score >= y[2]] <- "B"
> roster$grade[score < y[2] & score >= y[3]] <- "C"
> roster$grade[score < y[3] & score >= y[4]] <- "D"
> roster$grade[score < y[4]] <- "F"
> roster
             Student Math Science English score grade
1         John Davis  502      95      25  0.56     B
2    Angela Williams  600      99      22  0.92     A
3   Bullwinkle Moose  412      80      18 -0.86     D
4        David Jones  358      82      15 -1.16     F
5  Janice Markhammer  495      75      20 -0.63     D
6     Cheryl Cushing  512      85      28  0.35     C
7     Reuven Ytzrhak  410      80      15 -1.05     F
8          Greg Knox  625      95      30  1.34     A
9       Joel England  573      89      27  0.70     B
10      Mary Rayburn  522      86      18 -0.18     C
> name <- strsplit((roster$Student), " ")
> name
[[1]]
[1] "John"  "Davis"

[[2]]
[1] "Angela"   "Williams"

[[3]]
[1] "Bullwinkle" "Moose"     

[[4]]
[1] "David" "Jones"

[[5]]
[1] "Janice"     "Markhammer"

[[6]]
[1] "Cheryl"  "Cushing"

[[7]]
[1] "Reuven"  "Ytzrhak"

[[8]]
[1] "Greg" "Knox"

[[9]]
[1] "Joel"    "England"

[[10]]
[1] "Mary"    "Rayburn"

> Firstname <- sapply(name, "[", 1)
> Firstname
 [1] "John"       "Angela"     "Bullwinkle" "David"      "Janice"    
 [6] "Cheryl"     "Reuven"     "Greg"       "Joel"       "Mary"      
> Lastname <- sapply(name, "[", 2)
> Lastname
 [1] "Davis"      "Williams"   "Moose"      "Jones"      "Markhammer"
 [6] "Cushing"    "Ytzrhak"    "Knox"       "England"    "Rayburn"   
> roster <- cbind(Firstname, Lastname, roster[ , -1])
> roster
    Firstname   Lastname Math Science English score grade
1        John      Davis  502      95      25  0.56     B
2      Angela   Williams  600      99      22  0.92     A
3  Bullwinkle      Moose  412      80      18 -0.86     D
4       David      Jones  358      82      15 -1.16     F
5      Janice Markhammer  495      75      20 -0.63     D
6      Cheryl    Cushing  512      85      28  0.35     C
7      Reuven    Ytzrhak  410      80      15 -1.05     F
8        Greg       Knox  625      95      30  1.34     A
9        Joel    England  573      89      27  0.70     B
10       Mary    Rayburn  522      86      18 -0.18     C
> roster[order(Lastname, Firstname), ]
    Firstname   Lastname Math Science English score grade
6      Cheryl    Cushing  512      85      28  0.35     C
1        John      Davis  502      95      25  0.56     B
9        Joel    England  573      89      27  0.70     B
4       David      Jones  358      82      15 -1.16     F
8        Greg       Knox  625      95      30  1.34     A
5      Janice Markhammer  495      75      20 -0.63     D
3  Bullwinkle      Moose  412      80      18 -0.86     D
10       Mary    Rayburn  522      86      18 -0.18     C
2      Angela   Williams  600      99      22  0.92     A
7      Reuven    Ytzrhak  410      80      15 -1.05     F
> 

 quantile()      http://blog.csdn.net/u012543538/article/details/17025789

scale()           http://blog.sina.com.cn/s/blog_b623d3f40102v2zg.html

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