我为大学任务写了一些代码。赋值基于各种具体样品及其拉伸强度。有20种混凝土混合物（由四种不同的促进剂和五种不同的增塑剂制成）。我们的工作是对这个数据框进行统计分析：
TStrength accelerator plasticiser
1 3.417543 1 1
2 2.887113 1 2
3 3.600988 1 3
4 3.702631 1 4
5 3.686944 1 5
6 3.699785 1 1
7 3.112972 1 2
8 3.918160 1 3
9 3.600538 1 4
10 2.748832 1 5
11 3.404498 1 1
12 3.735437 1 2
13 3.347577 1 3
14 3.101556 1 4
15 3.527621 1 5
16 3.856831 1 1
17 3.492118 1 2
18 3.928343 1 3
19 3.511689 1 4
20 3.371985 1 5
21 3.069794 2 1
22 3.168010 2 2
23 3.316657 2 3
24 3.455162 2 4
25 2.818250 2 5
26 4.054507 2 1
27 3.065984 2 2
28 3.201351 2 3
29 3.417554 2 4
30 3.364320 2 5
31 3.218677 2 1
32 2.647151 2 2
33 3.222705 2 3
34 3.145210 2 4
35 3.636642 2 5
36 3.317620 2 1
37 3.645922 2 2
38 2.556071 2 3
39 3.177663 2 4
40 3.014374 2 5
41 3.838183 3 1
42 4.155951 3 2
43 3.886330 3 3
44 3.723898 3 4
45 4.425442 3 5
46 3.738460 3 1
47 3.217834 3 2
48 3.942241 3 3
49 3.699851 3 4
50 3.797089 3 5
51 3.652456 3 1
52 4.851609 3 2
53 3.359099 3 3
54 4.089559 3 4
55 4.282991 3 5
56 3.803784 3 1
57 3.519551 3 2
58 3.935084 3 3
59 3.890324 3 4
60 4.611936 3 5
61 3.343098 4 1
62 3.713952 4 2
63 3.629883 4 3
64 3.082509 4 4
65 3.346548 4 5
66 3.277845 4 1
67 3.509506 4 2
68 3.490567 4 3
69 3.235009 4 4
70 3.970925 4 5
71 3.504646 4 1
72 3.270798 4 2
73 3.547298 4 3
74 3.278489 4 4
75 3.322743 4 5
76 2.975010 4 1
77 3.384996 4 2
78 3.399486 4 3
79 3.703567 4 4
80 3.214973 4 5
我的第一步是尝试找出20种混凝土类型中每种类型的Tstrength值的均值（每种独特的混凝土样本有四种类型）。我是R的新手，我的代码肯定不漂亮，但这是我写的代码找到的代码：
#Setting the correct directory
setwd("C:/Users/Matthew/Desktop/Work/Engineering")
#Creating the data frame object, Concrete.
#Note that this will only work if the file
#s...-CW.dat is in the current working directory
#Therefore for this code to work, CreateData.r must
#be run on the individual computer with the
#given matriculation number, and the file must be saved
#in the specified directory
Concrete<-read.table(file='s...-CW.dat',header=TRUE)
#Since the samples of concrete are made from 4 different accelerators and
#5 different plasticisers there will be 4*5=20 unique combinations from
#which concrete samples can come from (i.e. 1,1; 1,2; 4,5 etc).
# There are four samples of each combination
#The next section of code is used to find the mean of the four samples,
#for each combination (20 total)
#creating a list with Tstrength from all (1,1) combinations
#Then finding average
combo1 = list(Concrete[1,1],Concrete[6,1],Concrete[11,1],Concrete[16,1])
combo1mean = mean(unlist(combo1))
#Repeating for (1,2)
combo2 = list(Concrete[2,1],Concrete[7,1],Concrete[12,1],Concrete[17,1])
combo2mean = mean(unlist(combo2))
#Repeating for (1,3)
combo3 = list(Concrete[3,1],Concrete[8,1],Concrete[13,1],Concrete[18,1])
combo3mean = mean(unlist(combo3))
#Repeating for (1,4)
combo4 = list(Concrete[4,1],Concrete[9,1],Concrete[14,1],Concrete[19,1])
combo4mean = mean(unlist(combo4))
#Repeating for (1,5)
combo5 = list(Concrete[5,1],Concrete[10,1],Concrete[15,1],Concrete[20,1])
combo5mean = mean(unlist(combo5))
#Repeating for (2,1)
combo6 = list(Concrete[21,1],Concrete[26,1],Concrete[31,1],Concrete[36,1])
combo6mean = mean(unlist(combo6))
#Repeating for (2,2)
combo7 = list(Concrete[22,1],Concrete[27,1],Concrete[32,1],Concrete[37,1])
combo7mean = mean(unlist(combo7))
#Repeating for (2,3)
combo8 = list(Concrete[23,1],Concrete[28,1],Concrete[33,1],Concrete[38,1])
combo8mean = mean(unlist(combo8))
#Repeating for (2,4)
combo9 = list(Concrete[24,1],Concrete[29,1],Concrete[34,1],Concrete[39,1])
combo9mean = mean(unlist(combo9))
#Repeating for (2,5)
combo10 = list(Concrete[25,1],Concrete[30,1],Concrete[35,1],Concrete[40,1])
combo10mean = mean(unlist(combo10))
#Repeating for (3,1)
combo11 = list(Concrete[41,1],Concrete[46,1],Concrete[51,1],Concrete[56,1])
combo11mean = mean(unlist(combo11))
#Repeating for (3,2)
combo12 = list(Concrete[42,1],Concrete[47,1],Concrete[52,1],Concrete[57,1])
combo12mean = mean(unlist(combo12))
#Repeating for (3,3)
combo13 = list(Concrete[43,1],Concrete[48,1],Concrete[53,1],Concrete[58,1])
combo13mean = mean(unlist(combo13))
#Repeating for (3,4)
combo14 = list(Concrete[44,1],Concrete[49,1],Concrete[54,1],Concrete[59,1])
combo14mean = mean(unlist(combo14))
#Repeating for (3,5)
combo15 = list(Concrete[45,1],Concrete[50,1],Concrete[55,1],Concrete[60,1])
combo15mean = mean(unlist(combo15))
#Repeating for (4,1)
combo16 = list(Concrete[61,1],Concrete[66,1],Concrete[71,1],Concrete[76,1])
combo16mean = mean(unlist(combo16))
#Repeating for (4,2)
combo17 = list(Concrete[62,1],Concrete[67,1],Concrete[72,1],Concrete[77,1])
combo17mean = mean(unlist(combo17))
#Repeating for (4,3)
combo18 = list(Concrete[63,1],Concrete[68,1],Concrete[73,1],Concrete[78,1])
combo18mean = mean(unlist(combo18))
#Repeating for (4,4)
combo19 = list(Concrete[64,1],Concrete[69,1],Concrete[74,1],Concrete[79,1])
combo19mean = mean(unlist(combo19))
#Repeating for (4,5)
combo20 = list(Concrete[65,1],Concrete[70,1],Concrete[75,1],Concrete[80,1])
combo20mean = mean(unlist(combo20))
关于代码的一些注释：＆＃34; s ...＆＃34;只是我的预科号码。我有三重检查，我没有在文件名或存储位置的目录中犯了错误。 CreataData.r只是一个提供给我们的脚本，用于生成用于创建“混凝土”的数据。根据我们的预科编号（所以我们不仅仅是盲目地互相抄袭）。
我对代码的问题是，无论何时运行，都会创建对象Concrete，combo1mean，combo2mean和combo3mean也是如此。但是，我无法弄清楚为什么其他物体没有被创造出来。
我在Rgui中运行脚本没有成功。运行脚本后，它告诉我检查Concrete是否已初始化，然后检查combo4mean及以上是否已初始化，但他们从未这样做过。我认为它可能与运行错误的文件有关，或者我没有正确保存数据，但是脚本肯定包含所有代码，我创建了一个新文件以查看是否可行，但不幸的是它没有。另外，我读过W.N.Venables，D.M。的R介绍。史密斯和R核心团队，但没有任何帮助我解决这个问题。
PS我不是这样做的一个简单的家庭作业方式。我真的试图弄清楚出了什么问题，但我似乎无法找到问题所在。如果问题不准确，或者如果我有误解，我也很抱歉，我对R很新，我正在尽我所能去学习它！提前干杯。
编辑：以防万一有人好奇，我设法从一个空的工作区开始，获得完全相同的代码在另一台计算机上工作。我还不太清楚为什么它在第一台计算机上没有工作，但感谢42代码建议。
答案 0 :(得分：2)
添加应绕过与阅读文本文件相关的问题的代码。这可以在任何R安装上成功：
public boolean isDiagonal(int row,int col){
if (row == board.length-1 && board[row-1][col+1] == 'T') {
return true;
} else if (col == board.length -1 && board[row+1][col-1] == 'T') {
return true;
} else if (board[row][col] == 'T' && board[row+1][col+1] == 'T' ||
board[row][col] == 'T' && board[row-1][col-1] == 'T') {
return true;
} else if (board[row][col] == 'T' && board[row-1][col+1] == 'T' ||
board[row][col] == 'T' && board[row+1][col-1] == 'T') {
return true;
}
return false;
}
这可能与您尝试的约1/10（或更少）代码有关（更重要的是没有错误）：
Concrete <- read.table(text="TStrength accelerator plasticiser
1 3.417543 1 1
2 2.887113 1 2
3 3.600988 1 3
4 3.702631 1 4
5 3.686944 1 5
6 3.699785 1 1
7 3.112972 1 2
8 3.918160 1 3
9 3.600538 1 4
10 2.748832 1 5
11 3.404498 1 1
12 3.735437 1 2
13 3.347577 1 3
14 3.101556 1 4
15 3.527621 1 5
16 3.856831 1 1
17 3.492118 1 2
18 3.928343 1 3
19 3.511689 1 4
20 3.371985 1 5
21 3.069794 2 1
22 3.168010 2 2
23 3.316657 2 3
24 3.455162 2 4
25 2.818250 2 5
26 4.054507 2 1
27 3.065984 2 2
28 3.201351 2 3
29 3.417554 2 4
30 3.364320 2 5
31 3.218677 2 1
32 2.647151 2 2
33 3.222705 2 3
34 3.145210 2 4
35 3.636642 2 5
36 3.317620 2 1
37 3.645922 2 2
38 2.556071 2 3
39 3.177663 2 4
40 3.014374 2 5
41 3.838183 3 1
42 4.155951 3 2
43 3.886330 3 3
44 3.723898 3 4
45 4.425442 3 5
46 3.738460 3 1
47 3.217834 3 2
48 3.942241 3 3
49 3.699851 3 4
50 3.797089 3 5
51 3.652456 3 1
52 4.851609 3 2
53 3.359099 3 3
54 4.089559 3 4
55 4.282991 3 5
56 3.803784 3 1
57 3.519551 3 2
58 3.935084 3 3
59 3.890324 3 4
60 4.611936 3 5
61 3.343098 4 1
62 3.713952 4 2
63 3.629883 4 3
64 3.082509 4 4
65 3.346548 4 5
66 3.277845 4 1
67 3.509506 4 2
68 3.490567 4 3
69 3.235009 4 4
70 3.970925 4 5
71 3.504646 4 1
72 3.270798 4 2
73 3.547298 4 3
74 3.278489 4 4
75 3.322743 4 5
76 2.975010 4 1
77 3.384996 4 2
78 3.399486 4 3
79 3.703567 4 4
80 3.214973 4 5", header=TRUE)
重要的是，您忘记在> means.by.type <- with( Concrete, tapply(TStrength,
list( acc=accelerator, plas=plasticiser),
FUN=mean))
> means.by.type
plas
acc 1 2 3 4 5
1 3.594664 3.306910 3.698767 3.479103 3.333845
2 3.415150 3.131767 3.074196 3.298897 3.208397
3 3.758221 3.936236 3.780689 3.850908 4.279364
4 3.275150 3.469813 3.516808 3.324893 3.463797
上提供str或dput，因此无法确定您的问题是数据准备还是编码。