c***z
发帖数: 6348 | 1 It took me a whole day. Either I am retarded or SQL is. Why it is so
complicated just to loop over all columns!
I am posting here to help people facing similar problems. If you have a
better solution, please do share! Thanks!
Below is the SQL code:
DECLARE @ColName VARCHAR(150)
DECLARE @sqlStatement VARCHAR(MAX)
SET @sqlStatement = 'initial part of you query, for example select, insert
update etc. '
DECLARE myCursor CURSOR STATIC FOR
SELECT name FROM syscolumns WHERE id = (SELECT id FRO... 阅读全帖 |
|
O*****y
发帖数: 222 | 2 a <- read.csv("A.csv", header=TRUE)
b <- read.csv("B.csv", header=TRUE)
new1 <- cbind(a, b[, setdiff(colnames(b), colnames(a))])
new2 <- b[, setdiff(colnames(b), colnames(a))] |
|
l**********l
发帖数: 361 | 3 Title: Non-invasive genetic identification of the red fox Vulpes vulpes in
the Shiretoko National Park, eastern Hokkaido, Japan
Source: Mammal study [1343-4152] Oishi yr:2010 vol:35 iss:3 pg:201 -207
http://apps.isiknowledge.com/full_record.do?product=UA&colname=
Title: Sex Determination and Individual Identification of American Minks (
Neovison vison) on Hokkaido, Northern Japan, by Fecal DNA Analysis
Source: Zoological science [0289-0003] Shimatani yr:2010 vol:27 iss:3 pg:243
-247
http://apps... 阅读全帖 |
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l*******d
发帖数: 101 | 4 跟DaShagen写的大同小异。把找出来的值加进去。
> n = 4
> threshhold = 5
> X = matrix (1:16, byrow=TRUE, ncol=n, dimnames=list(1:4, c("A", "B", "C",
"D")))
> X
A B C D
1 1 2 3 4
2 5 6 7 8
3 9 10 11 12
4 13 14 15 16
> ind = which (X>threshhold, arr.ind=TRUE)
> findNames = data.frame (ind, rowName=rownames(X)[ind[,1]], colName=
colnames(X)[ind[,2]], value=as.vector(X)[(ind[,2]-1)*n+ind[,1]], row.names=
NULL)
> findNames
row col rowName colName value
1 3 1 3 A 9
2 4 1 4 |
|
r****5
发帖数: 618 | 5 f0008非常感谢，我用了
t(apply(subhw,2,FUN=function(x) c(Missing=sum(is.na(x)),Mean=mean(x),Median
=median(x), sdev=sd(x), Min=min(x), Max=max(x))))
如果col没有排序要求，你的更简单。原来summary里可以再加其他function， 象sd等
，我还以为仅仅能用原有的6个呢。
还有个问题，就是这个显示print 后再屏幕上有row names，就是原来的col names。但
是write。table后就没有了。这样我要再加一个col才行。我在上面的function里加了
mycols=colnames(x)。但是执行后没有显示出来
summary<-t(apply(subhw,2,FUN=function(x) c(mycols=colnames(x),Missing=sum(is
.na(x)),Mean=mean(x),Median=median(x), sdev=sd(x), Min=min(x), Max=max(x))))
print... 阅读全帖 |
|
i*****w
发帖数: 75 | 6 建议一种方法:
1) UNPIVOT TableA, You will get:
ID1, COL1, CONTENT1
ID2, COL2, CONTENT2
ID3, COL3, CONTENT3
2) Join TableA and TableB by LIKE.
EXAMPLE:
-- Prepare Source Table
DECLARE @tblA Table (ID int identity, col1 varchar(100), col2 varchar(100),
col3 varchar(100))
INSERT INTO @tblA (col1, col2, col3)
SELECT 'this is a test', 'I am not sure', 'Give it a try.'
UNION ALL
SELECT 'who cares', 'No one knows', 'Why not'
UNION ALL
SELECT 'it is impossible', 'please let me know', 'be honest'
-- ... 阅读全帖 |
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l**********1
发帖数: 5204 | 7 big aunt or uncle
BTW, including LX some post then by Matrix machine learning
LZ pls refer one E-Book
by NAOMI ALTMAN
its title is 'R lecture'
its pp9
>We will use the grep command to find the columns from the same biological
>replicate. Note that hexbin >resets the number of plots per gure
>back to one. If you want to use a loop to plot several pairs, you need to
>use par(ask=T) to page through
>the plots. Here we do only the rst 6 plots. The grayscale of the plot
>indicates the number of ... 阅读全帖 |
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q**j
发帖数: 10612 | 8 That is very good. I am not aware of the assign tool.
Another issue is:
Result = matrix(0,nrow=2,ncol=3)
for (j in 1:3)
{
colnames(Result)[j] = paste("item",j,sep="")
}
can you see why it always gives error message? but
colnames(Result) = c("Good","Better","Best")
will work as expected. I am quite puzzled here. |
|
o******6
发帖数: 538 | 9 ☆─────────────────────────────────────☆
qqzj (小车车) 于 (Thu Feb 26 15:48:04 2009) 提到:
好奇心强，同一个工作用R和python各搞了一次。数据文件是709 by 88的.csv文件。如
下是R code: （code 写的不好，希望拍砖帮我改进。）
************************************************************************
Data = read.csv("D:/VP_BAP.csv")
NoRow = dim(Data)[1]
NoCol = dim(Data)[2]
RowName = Data[,1]
ColName = colnames(Data)
Output = c("","")
for (i in 1:NoRow){
for (j in 1:NoCol){
if (Data[i,j]==c("N")){
Temp = c(as.character(RowName[i] |
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a***r
发帖数: 420 | 10 有个n*n的matrix（result），有col和row name，我想保留这个matrix中大于5的值，
并且保留这些值对应的colname和rowname，因为这个是这些值的来源信息。
不知道有没有专门的语句，发挥主观能动性写了一个：
threshold<-matrix(,ncol=4,nrow=n*n)
for (i in 1:(n-1)){
for (j in (i+1):n) {
if (result[i,j]>5)
threshold[((i-1)*n+j),1] <-rownames(result)[i]
threshold[((i-1)*n+j),2] <-colnames(result)[j]
threshold[((i-1)*n+j),3] <-as.numeric(result[i,j])
}
}
threshold <- threshold[complete.cases(threshold),]
总觉得有点笨哪，有没有语句直接能实现的呢？
谢谢大家！ |
|
S********a
发帖数: 359 | 11 我怎么找出我的perl是多少呢，我目前的理解是不止一个方法读取XLS，哪个最方便使
用呢？
或者我用
library(xlsReadWrite)
mydata<-read.xls("c:\L\NJ\Study\copy.xls", colNames=TRUE)
出下面错误信息
Error in .Call("ReadXls", file, colNames, sheet, type, from, rowNames, :
Incorrect number of arguments (11), expecting 10 for ReadXls
怎么改呢？ |
|
s*****a
发帖数: 2735 | 12 PROC IML;
X={2 3,
4 5};
CREATE test2 FROM x[colname={'a' 'b'}];
APPEND FROM x;
Quit;
上边这个可以生成 test2；
PROC IML;
a=1;
b=2;
x={a b};
CREATE test2 FROM x[colname={'a' 'b'}];
APPEND FROM x;
Quit;
上面这个就不行了
只给出
a b
A B
请提示。 |
|
n*******y
发帖数: 437 | 13 hello大家好 我R用了一阵也不是太熟 请问，当数据点少的时候density plot出
现波纹怎么解决？ 谢谢！
code：
# show fringes
test = as.data.frame( sample(0:5,10000,replace=T) )
colnames(test) = c('mon')
ggplot(test, aes(x=mon)) + geom_density(binwidth=1, size=1)
# normal
test = as.data.frame( sample(0:20,10000,replace=T) )
colnames(test) = c('mon')
ggplot(test, aes(x=mon)) + geom_density(binwidth=1, size=1) |
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c***z
发帖数: 6348 | 14 老板又有新花样，这次要cumulative的percentages
patient_percentiles_cum <- patient_percentiles_fin[, c(1,102)]
colnames(patient_percentiles_cum)[2] <- "top.0"
for (k in 1:100) {
# k <- 1
temp <- patient_percentiles_fin[, c(102:(102-k))]
top <- apply(temp,
1,
FUN = sum)
top <- data.frame(top)
patient_percentiles_cum <- cbind(patient_percentiles_cum,
top)
colnames(patient_percentiles_cum)[2+k] <- paste("top",
... 阅读全帖 |
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h******3
发帖数: 351 | 15 来自主题: JobHunting版 - 贡献两个题 Palo Alto小公司, 不错的start up.
一上来半个小时, 三个题, 一张纸. 我开始没重视(字迹廖草), 最后就载在上面.
code, code, code强, 才有机会啊.
1. 给你一个整数数组 int[] arr, size较小. 请写出一个sort method.
comments: 我写的是selection sort. 题目里边数据量没有具体数据, 也无人可以问,
发题目的是recruiter. 如果你面世, 写那个啊? quick sort
2. step by step, explain what does server do when execute
select max(colname) from t1;
3. 一张表, 写with statement |
|
b******o
发帖数: 545 | 16 Error in .Call("WriteXls", x, file, colNames, sheet, from - 1, rowNames, :
Incorrect number of arguments (7), expecting 6 for WriteXls
================================================================
what's that means? |
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c*****d
发帖数: 6045 | 17 你是问如何格式化输出吗
set linesize 120
set pagesize 500
format [colname] a20 |
|
c***c
发帖数: 6234 | 18 我写java。使用了大量的Stored Procedure。很多是open cursor，return java一个
list。
同组另一个人只会一般query，也只用一般statement。现在测试发现，每个session 有
30-40个 held unclosed cursor。
我们都是确保在使用cursor fetch data 后，都close resultset 和statement的
会不会cursor从一般query来的？
难道不能使用SP的 out parameter 是cursor？我这么用了10年了，一直没问题。很疑惑
code类似
stmPersonMaster.registerOutParameter(5, OracleTypes.CURSOR);
stmPersonMaster.execute();
rs = (ResultSet) stmPersonMaster.getObject(5);
while(rs.next()){
rs.getString(colName)
}
rs.close
stmPersonMaster.close
谢谢 |
|
B********r
发帖数: 397 | 19
table,
是啊,还有一个大问题,就是用CQL3怎么动态的加column, 比如我定义了一个table只有
两个colum,那就不能直接
set keySpace[colFamily][colName] = value
而必须先add column 然后再 insert? 这样不是多了一个round trip么.
还有如果我要 select top 10 * from table order by id
貌似只能把所有的都返回,我在client side自己sort? |
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l**********1
发帖数: 5204 | 20 Continue:
第四乐章 Finale
找有关的PhD dissertation 里边的 R source code program
while U can debug it or even rewrite it for another task,
then you already masted NGS coding skills.
比如
http://www.dspace.cam.ac.uk/handle/1810/218542
DSpace at Cambridge
title: Genome-wide analyses using bead-based microarrays
Authors: Dunning, Mark J
Issue Date: 4-Sep-2008
Files in This Item:
File Description Size Format
dunning_thesis_.pdf 10.47 MB Adobe PDF
its Appendix B
R source Code f... 阅读全帖 |
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|
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s*****n
发帖数: 2174 | 23 你定义矩阵的时候, 默认的dimnames是NULL, 也就是不存在.
下面你对一个不存在的东西取[j]指标, 显然会报错. 但是你
做colnames(Result) = c("Good","Better","Best")的时候,
是等于正常赋值, 这个是可以的. 问题出在取[j]指标上.
这个就好像下面这个例子
> a[3] <- 1
Error: object "a" not found
在没有的定义a的情况下, 你就取a[3]是不行的.
> a <- c(NA, NA, 1)
你对a整体赋值, 这是可以的.
解决办法:
1. (推荐)使用data.frame. data frame 是R里面专门定义的一种特殊类型的矩阵, 比
矩阵的信息要稍微丰富一些. 能用data frame, 就不要用matrix.
Result <- data.frame(matrix(0,nrow=2,ncol=3))
for (j in 1:3){
names(Result)[j] = paste("item",j,sep="")
}
2. 如果非要用matrix, 那么就事先把dimnames的位置 |
|
b********y
发帖数: 63 | 24 Not sure how SAS is more convenient on those issues, but they should not be a
problem for R at all.
同一个事情先用SAS干了，然后用R干了。有些感想，有些问题。跑来跟大家交流一下啊。
1. R里面改个变量名这么这么难？
需要一个reshape pacakge
data.frame = rename(data.frame, c(oldname = "newname"))
不用这个package就更麻烦了。
you can just change the name: colnames(x)[1] = newName1.
2. 控制时间，日期怎么这么难？
以前“告别棒球场”说了，最好不要用日期，全部用字符。临到用的时候才换。就算这
样还是问题多多。比如 “1990-12-04”，怎么样轻松的把它变成 “1990-12-01”（月
初），“1990-12-31”（月末）。往前，往后移动几个月？我写了一个function来干这
个。但是非常慢（yes, I know it.)而且不能往以前移动。我 |
|
q**j
发帖数: 10612 | 25 多谢帮助！
1. 如果一个data frame 有 （key1, key2, var1)，另外一个有(key1, key2, var2)。
我希望找出在第一个data frame里面，但是不在第二个里面的所有行。请问是否必须要
像SAS那样建立in1, in2然后经过merge()以后挑选？有没有简单的办法？（var1 var2
可能是missing）。
2.
> data
x y z
1 apple 1996 1
2 orange 1997 2
3 orange 1996 3
4 apple 1998 4
5 apple 1997 5
> new
[,1] [,2] [,3]
[1,] NA NA NA
[2,] NA NA NA
> rownames(new) = unique(x)
> colnames(new) = unique(y)
> new
1996 1997 1998
apple NA NA NA
orange NA NA NA
> Row.Index = match(data$x,rownames(new)) |
|
f***a
发帖数: 329 | 26 Anyone has any idea? I have been facing such problems. Thanks! :)
#data set
dat <- matrix(1:10,,2)
colnames(dat) <- letters[1:2]
F <- function(a,b){a+b}
outer(1:2,4:6,F) #wokring
apply(dat,1,F) #not working. how to fix?
x <- 1:5
FF <- function(a,b){ sum(a*x+b) }
outer(1:2,4:6,FF)#not working. why? how to fix? |
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s*****n
发帖数: 2174 | 27 第一个问题, 可以用apply实现. 但你要确定F的参数只有一个, 这个参数在实际操作中就是矩阵的每一行. 比如下面这个例子
dat <- matrix(1:10,,2)
colnames(dat) <- letters[1:2]
F <- function(t){t[1] + t[2]}
apply(dat,1,F)
注意定义的是 F(t), 而不是 F(t[1], t[2]).
seq(along=a) 在这里不是1, 而是6. 如果你觉得是1, 说明你没有理解outer的工作原理. outer(a, b, FUN) 的执行原理, 是先把a, b 做成相应的矩阵(本质还是向量)A, B, 然后一起传给FUN. FUN要对A和B的对应元素进行操作. 这里
a 是 (1,2), b是(4,5,6). 传递给FUN的分别是(1,2,1,2,1,2)和(4,4,5,5,6,6)
你定义的FF, 是把A和B看作向量来和x进行操作的, R会认为需要把一个长度为6的向量和一个长度为5的向量相乘, 然后再和一个长度为6的向量相加, 所以报错.
下面的例子可能能对你的理解有帮助:
x <- 1:5 |
|
f**********e
发帖数: 48 | 28 #R code, simple implementation, your data frame is df
cl<-colnames(df)
rw<-rownames(df)
cl_len<-length(cl)
rw_len<-length(rw)
new_rname<-character(cl_len*rw_len)
new_data<-numeric(cl_len*rw_len)
index<-0
for (i in 1:cl_len){
for (j in 1:rw_len){
index<-index+1
new_rname[index]<-paste(rw[j],cl[i])
new_data[index]<-df[i,j]
}
}
data.frame(new_rname,new_data)
|
|
y****2
发帖数: 34 | 29 data <- matrix(c(1 ,1 ,2, 2, 1, 3, 4, 2,1, 5, 6, 3,2, 7, 8, 3,2, 9, 10, 4),
ncol=4, byrow=T)
colnames(data) <- c("id", "x1", "x2", "e")
### step1:
data[,2:3] <- data[,2:3]*data[,4]
data1 <- data[,1:3]
### step2:
data2 <- aggregate(data1[,2:3], list(id=data1[,1]), sum)
### step3:
data3 <- split(data2[,2:3], f=list(data2[,1]))
data3 <- lapply(data3, as.vector, "numeric")
mprod <- function(x){x %*% t(x)}
data4 <- lapply(data3, mprod)
### step4:
data5 <- 0
for(i in 1:length(data4)){
data5 <- data5 + |
|
f******9
发帖数: 267 | 30 谢谢了，但是有个问题因为我的colnames有数字，这样的话，虽然两个表格相同的
column在一起了，但是column的顺序
全打乱了，都按照从小到大排列了。 |
|
|
s*****n
发帖数: 2174 | 32 给你写个example, 抛砖引玉吧
假设你的矩阵叫做 A, 目标data frame叫做 result.
result <- data.frame(
RowName = rep(dimnames(A)[[1]], dim(A)[2]),
ColName = rep(dimnames(A)[[2]], each = dim(A)[1]),
Value = as.vector(A))
result <- result[result$Value > 5, ]
在R里面, 要尽量避免循环啦. |
|
a***r
发帖数: 420 | 33 抛砖引玉
#original matrix: A
#new matrix: X
#number of obs:n
#number of var:nvar
> A
var1 var2 var3 var4 var5
1 1 2 3 4 5
2 6 7 8 9 10
3 11 12 13 14 15
4 16 17 18 19 20
>n=4
>nvar=5
>value <-as.vector(t(A))
>varname <-rep(colnames(A),n)
>ID <-sort(rep((1:n),nvar))
>X <-data.frame(cbind(ID,varname,value))
ID var value
1 1 var1 1
2 1 var2 2
3 1 var3 3
4 1 var4 4
5 1 var5 5
6 2 var1 6
7 2 var2 7
8 2 var3 |
|
p***r
发帖数: 920 | 34 data<-data.frame(matrix(101:106,2,3))
> data
X1 X2 X3
1 101 103 105
2 102 104 106
> newdata<-
data.frame(id=rep(rownames(data),dim(data)[2]),varname=rep(colnames(data
),dim(data)[1]),value=c(t(as.matrix(data))))
> newdata
id varname value
1 1 X1 101
2 2 X2 103
3 1 X3 105
4 2 X1 102
5 1 X2 104
6 2 X3 106 |
|
f******9
发帖数: 267 | 35 How to find the numbers which are more than 0.5 in a matrix (and with their
rownames and colnames)?
for example, in this matrix:
[,1] [,2] [,3] [,4] [,5]
[1,] 1.0000000000 0.0320075758 0.1183712121 0.0293040293 0.0083333333
[2,] 0.0320075758 1.0000000000 0.0571428571 0.1668650794 0.0004662005
[3,] 0.1183712121 0.0571428571 1.0000000000 0.0053418803 0.0028846154
[4,] 0.0293040293 0.1668650794 0.0053418803 1.0000000000 0.0426767677
[5,] 0.008333 |
|
S******y
发帖数: 1123 | 36 最近在研究R - biglm.
R document 里说 - "biglm creates a linear model object that uses only p^2
memory for p variables. It can be updated with more data using update. This
allows linear regression on data sets larger than memory."
读了下面的 source code， 还是没搞懂update具体是怎么实现的。。。
> biglm::biglm
function (formula, data, weights = NULL, sandwich = FALSE)
{
tt <- terms(formula)
if (!is.null(weights)) {
if (!inherits(weights, "formula"))
stop("`weights' must be a formula")
w <- ... 阅读全帖 |
|
z**********i
发帖数: 88 | 37 Using SAS to subset data, I can get N=48, but using R, I can not, I don't
know anything about. Would anybody check my R code and see where is the
problem? Below is the SAS code and R code. Thanks.
data x;
set ALL_12222010;
where tau>93 & abeta142<192 & PTAU181P>23;run; **n=48;
#R code;
sizing <- function(variable.name,dx.group){
cut.sample <- function(variable.name,dx.group){
attach(variable.name)
variable.name$enrich=as.numeric(tau>93 & abeta142<192 & PTAU181P>23);
... 阅读全帖 |
|
J****n
发帖数: 54 | 38 > aaa <- read.xls(rfile)
Error in .Call("ReadXls", file, colNames, sheet, type, from, rowNames, :
Incorrect number of arguments (11), expecting 10 for ReadXls
大家遇到过这个错么？怎么改？(Windows OS)
Thanks. |
|
q**j
发帖数: 10612 | 39 好像是rownames和column names不一样造成的。这个也太不合理了吧。
> colnames(S)=1:12
> rownames(S)=1:12
> all.equal(S, t(S))
[1] TRUE
请问有没有简单的办法让R不这么变态。明明是个矩阵么。
V
-0
-0
-0
-0 |
|
b*****n
发帖数: 685 | 40 colnames()=NULL
rownames()=NULL |
|
t****a
发帖数: 1212 | 41 You can use fisher exactly test to tell if the overlapping is random. Before
doing this you need to now that the size of the background (the DaShagen's
questions: How many genes are there?). In the following code I just assumed
the background size is 20000, both A and B are sampled from the background.
# R code here.
bg <- 20000; a <- 3000; b <- 4000; ab <- 500
your.data <- matrix(c(ab, a-ab, b-ab, bg-a-b+ab), 2, 2)
colnames(your.data) <- rownames(your.data) <- c('In Group', 'Not In Group')
name... 阅读全帖 |
|
a********a
发帖数: 346 | 42 挺简单的一个R code， 下边的error是为什么呢？
> data=matrix(1:4,2,2)
> data
[,1] [,2]
[1,] 1 3
[2,] 2 4
> colnames(data)=c("x","y")
> data
x y
[1,] 1 3
[2,] 2 4
> data$x
Error in data$x : $ operator is invalid for atomic vectors |
|
b********y
发帖数: 63 | 43 > data = as.data.frame(matrix(1:4,2,2))
> colnames(data)=c("x","y")
> data$x |
|
d*********k
发帖数: 1239 | 44 用了dataframe啊，然后 dim() 给我的是3*4矩阵，我可以改colnames和rownames，这
个没有问题
但是怎么给最开始的一列（暂且叫0列，也就是rownames）加个名字呢？就是rownames
（）表示的是年份1990,1991,1992，我想在最左上角加个名字
不知道我解释清楚我想干嘛了没有~~ 谢谢啦 |
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t****a
发帖数: 1212 | 45 it is possible.
assuming that your matrix is m
them you can define names(dimnames(m)) = c(rownames.title, colnames.title) |
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t****a
发帖数: 1212 | 46 it is possible.
assuming that your matrix is m
them you can define names(dimnames(m)) = c(rownames.title, colnames.title) |
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s********1
发帖数: 54 | 47 data hh1;
input ID year;
cards;
1 1998
1 2001
2 1994
2 1995
2 1999
2 2001
3 1997
3 1999
3 2000
;
run;
proc univariate data=hh1;
var year;
by ID;
output out=dout max=dmax min=dmin;
run;
proc iml;
use dout;
read all into X;
do i = 1 to 3;
n=X[i,2]-X[i,3]+1;
xx_temp=repeat(i,n,1)||(X[i,3]:X[i,2])`;
xx=xx//xx_temp;
end;
print xx;
create hh1fix from xx[colname={ID year}];
append from xx;
quit; |
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t*****w
发帖数: 254 | 48 answer is the following;
your final result is the following:
student teacher senior
2732 3465 1
3347 3837 1
1179 1693 1
3875 1711 1
3875 2059 1
2032 1784 1
2848 3921 1
2148 1416 1
3038 1434 1
3530 2037 1
2585 3811 1
1481 3954 1
... 阅读全帖 |
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c***z
发帖数: 6348 | 49 ## build data frame
work <- c(12, 14, 4, 16, 12, 20, 25, 8, 24, 28, 4, 15)
edu <- c(6,3,8,8,4,4,1,3,12,9,11,4)
income <- c(34.7, 17.9, 22.7, 63.1, 33.0, 41.4, 20.7, 14.6, 97.3, 72.1, 49.1
, 52.0)
studay.df <- data.frame(cbind(work, edu, income))
## linear model
model_3 <- lm(income ~ ., data = studay.df) # OLS
summary_table <- data.frame(summary(model_3)$coefficients)
colnames(summary_table) <- c("coef", "std.error", "t_value", "p_value")
summary_table$regressor <- row.names(summary_table)
s... 阅读全帖 |
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t*****w
发帖数: 254 | 50 When I had my job interview, they always tested my SAS skill.However I use R
all the time. To help your preparation, read my R codes to see how much you
can understand it.
%in%
?keyword
a<-matrix(0,nrow=3,ncol=3,byrow=T)
a1 <- a1/(t(a1)%*%spooled%*%a1)^.5 #standadization in discrim
a1<- a>=2; a[a1]
abline(h = -1:5, v = -2:3, col = "lightgray", lty=3)
abline(h=0, v=0, col = "gray60")
abs(r2[i])>r0
aggregate(iris[,1:4], list(iris$Species), mean)
AND: &; OR: |; NOT: !
anova(lm(data1[,3]~data1[,1... 阅读全帖 |
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