a****u
发帖数: 95 | 1 How about using rbind first and aggregate
x<-matrix(1:10,ncol=2,byrow=TRUE)
y<-matrix(1:20,ncol=2,byrow=TRUE)
z<-rbind(x,y)
aggregate(z,by=list(z[,1],z[,2]),FUN=tail,1)
If only duplicate for one column need to be removed, can try use match and
rbind
index<-match(x[,1],y[,1])
rbind(x[-index,],y) |
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t********m
发帖数: 939 | 2 【 以下文字转载自 DataSciences 讨论区 】
发信人: tulipdream (xiaohuaidan), 信区: DataSciences
标 题: 请教一个R问题：怎么rbind一系列data，如data1，data2,....data1000
发信站: BBS 未名空间站 (Thu Nov 13 17:35:36 2014, 美东)
有没有一个简单点的方法，类似于rbind(data1-data1000)的，求各位不吝赐教，谢谢
！ |
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d********t
发帖数: 837 | 3 Reduce(rbind, list(data1,data2,data3,data4))
example:
Reduce(rbind, list(data.frame(x1=c(1,2,3),x2=c(2,3,4)),data.frame(x1=c(5,6,7
),x2=c(7,5,4)),data.frame(x1=c(5,4,3),x2=c(7,6,5)))) |
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n*****3
发帖数: 1584 | 4 if there are patterns in the names, or you want to rbind all the datasets in
the namespace,
you can get/mget them to a list,
then Reduce(..., rbind( ), ).. |
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t*******i
发帖数: 742 | 5 有个叫rbind.fill或者cbind.fill的命令
是一个人写的package
搜一下好了 |
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k*******a
发帖数: 772 | 6 如果data1-data1000在一个list的话，就用do.call函数
否则可以试试
mycall = c("rbind", paste0("data", 1:1000))
mycall = lapply(mycall, as.name)
result = eval(as.call(mycall)) |
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n*****3
发帖数: 1584 | 7 mget the list of dataframe by regular expression
then Reduce this list with rbind
,7 |
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H*H
发帖数: 472 | 8 Reduce跟do.call都能实现，但是要是你的数据都比较大，这两个都比较慢
dplyr包的rbind_all是 do.call(rbind, list)的c++版本，
dplyr::rbind_all在数据量比较大时可以轻松提速十倍 |
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w*******9
发帖数: 1433 | 9 most intuitive way (similar to kirklanda's solution):
a <- paste("data",1:1000,sep="",collapse=",")
eval(parse(text=paste(c("rbind(",a,")"),sep=""))) |
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t********m
发帖数: 939 | 10 有没有一个简单点的方法，类似于rbind(data1-data1000)的，求各位不吝赐教，谢谢
！ |
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o**o
发帖数: 58 | 11 rbindlist
或者do.call(rbind, list()) |
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t*****w
发帖数: 254 | 12 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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I*****a
发帖数: 5425 | 13 你这个不算是吧。
n = 1000 # training size
ntest = 1000 # test size; make this big only for illustration
id.train = 1:n
id.test = (n + 1):(n + ntest)
ratio = 0.99
n0 = round(n * ratio)
n1 = n - n0
nsimu = 100
res = NULL
for (i in 1:nsimu){
p = c(runif(n0, 0, 0.5), runif(n1, 0.5, 1), runif(ntest, 0.6, 1) )
y = sapply(p, function(x){rbinom(n = 1, size = 1, prob = x)})
x = log(p / (1 - p)) # beta is c(0, 1)
dat = data.frame(x = x, y = y)
f... 阅读全帖 |
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z********u
发帖数: 1031 | 14
哥 你解开了我多年的疑惑
原来是rbind的函数有问题 我说怎么我用rbind简直像死机了一样
我不是专业写R的 偶尔用用 |
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k*******a
发帖数: 772 | 15 第一题（程序算得）
#马 #可能结果
2 3
3 13
4 75
5 541
6 4683
7 47293
8 545835
附我的R code:
total<-c()
for (n in 2:8)
{
x<-(n-2):0
x<-2^x
count<-c()
for (i in 0:(2^(n-1)-1))
{
y<-floor(i/x)%%2
pos<-c(0,(1:(n-1))[y==1],n)
dif<-pos[-1]-pos[-length(pos)]
dif<-prod(factorial(dif))
dif<-factorial(n)/dif
count<-c(count,dif)
}
new<-data.frame(n=n,count=sum(count))
total<-rbind(total,new)
} |
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l*******r
发帖数: 3799 | 16 in R:
library(quantmod) # A library for stock analysis
DJIA <- c("MMM", "AA", "AXP", "T", "BAC", "BA", "CAT", "CVX", "TRV", "KO", "
DD", "XOM", "GE", "CSCO", "HPQ", "HD", "IBM", "INTC", "JNJ", "JPM", "KFT", "
MCD", "MRK", "MSFT", "PFE", "PG", "UTX", "VZ", "WMT", "DIS")
getSymbols(DJIA) # Download historical data for Dow 30 stocks
getSymbols("^DJI") # Download historical data for Dow index
x <- do.call("rbind", lapply(DJIA, function(x) get(x)[, 4])) # Extract
close price
plot |
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C*****H
发帖数: 7927 | 17 法律的问题我是不太懂。
但是编程写code run simulation的事情稍微懂一点，
看到过一篇很好的分析牌例的文章，讲述在庄家有AA而自己没有K的情况下的过手，
感谢作者，但因为不确定他本人是不是愿意被提起，所以只引用内容在此：
If we assume cards are randomly dealt, my simulation shows, given that
zhuang does have AA and doesn't have any K, the probability of having at
least one k from his partner is around .55 with 100,000 times of simulation.
Histogram shows this prob is stable.The following is my code and you're
welcome to test and give comments.
R code:
count<-NA
count1<-0
count2<-0
for(j in 1:1000... 阅读全帖 |
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C*****H
发帖数: 7927 | 18 法律的问题我是不太懂。
但是编程写code run simulation的事情稍微懂一点，
看到过一篇很好的分析牌例的文章，讲述在庄家有AA而自己没有K的情况下的过手，
感谢作者，但因为不确定他本人是不是愿意被提起，所以只引用内容在此：
If we assume cards are randomly dealt, my simulation shows, given that
zhuang does have AA and doesn't have any K, the probability of having at
least one k from his partner is around .55 with 100,000 times of simulation.
Histogram shows this prob is stable.The following is my code and you're
welcome to test and give comments.
R code:
count<-NA
count1<-0
count2<-0
for(j in 1:1000... 阅读全帖 |
|
C*****H
发帖数: 7927 | 19 ☆─────────────────────────────────────☆
Grandmaster (Garry Kasparov) 于 (Tue Sep 24 08:54:56 2013, 美东) 提到:
其实我不赞同举报作弊的做法，但我想了下，还是这么做了，回头谈谈我的想法。
证据证言推论我都有，没来得及说已经被人一通辱骂威胁。版主不喜欢我继续在这里说
话，我就此打住。如果有兴趣知道我的证据证言证人推论过程的，梦版见。
☆─────────────────────────────────────☆
ninedays (九天) 于 (Tue Sep 24 09:13:19 2013, 美东) 提到:
Evidence, please. Also better to first state the facts, i.e., what happened,
then the reason you think they were cheating. Two parts basically: 1) facts
; 2) your opinion.
☆─────────... 阅读全帖 |
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w*m
发帖数: 1806 | 20 mount舒服点，不然每次一进目录就跳到那个指定的目录，难受啊
我记得linux里是这样的，mount --rbind XX YY
solairs 下搞不定， 是不是无解？ |
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m*****n
发帖数: 3575 | 21 关键的问题在于循环里不能用这三个函数
c（）
cbind（）
rbind（）
但凡你用了，慢得出奇。
解决办法是刚开始就建好纯零矩阵，然后填数。
快百倍不止。 |
|
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s******n
发帖数: 63 | 23 谢谢。不知道为什么用rbind时，前面的identity变量就全乱了。 |
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a********a
发帖数: 346 | 24 I have a program as following,
beta1=2
beta2=8
#for (i in 1:2){
obs=2
x1=matrix(NA,obs,1)
x2=matrix(NA,obs,1)
for (g in 1:obs){
x1[g,]=runif(1,1,2)
x2[g,]=rnorm(1,1)
}
data=cbind(x1,x2)
data
#d=rbind(data[i])
#}
I run the program 2 times, and get the data like following each time,
> data
[,1] [,2]
[1,] 1.452696 0.2718456
[2,] 1.514587 1.2971293
> data
[,1] [,2]
[1,] 1.172726 -0.9674824
[2,] 1.159079 0.5483838
how can I combine these data by row, i.e. I want to get d |
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l*****k
发帖数: 587 | 25 I guess you can wrap it into a function that returns data when
called, then rbind your returned data |
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s*****n
发帖数: 2174 | 26 beta1=2
beta2=8
result <- NULL
for (i in 1:2){
obs=2
x1=matrix(NA,obs,1)
x2=matrix(NA,obs,1)
for (g in 1:obs){
x1[g,]=runif(1,1,2)
x2[g,]=rnorm(1,1)
}
data=cbind(x1,x2)
result <- rbind(result, data)
}
result |
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g********r
发帖数: 8017 | 27 而且你要用的是rbind.如果错用了cbind，可能导致部分矩阵被多次复制粘在一起。那
内存可就大了。 |
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a********a
发帖数: 346 | 28 Thanks goldmember again,
Sorry, I used rbind not cbind to combine all the matrix.
Here is the situation,inside each sample, I have a loop for i in 1to 15,
In each iteration of the loop, I get estimates like the following ( here I
only list 4 columns instead of 8 columns, they are fake numbers),
sample category beta beta sigma sigma
1 1 2.0 1.2 2.0 0.8
1 2 1.0 0.8 1.2 0.1
1 2 1.5 0.7 1.3 0.8
1 3 2.0 0.4 1.5 0.2
1 |
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s*****n
发帖数: 2174 | 29 a <- c(3,6,2,5,7,9)
K <- 4
## Question 1
sum(a[a > K])
## Question 2
sum(apply(rbind(K, a), 2, max)) |
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o****o
发帖数: 8077 | 30 use rbind to stack the two vectors:
#--------------------
> x = read.table("clipboard",sep=' ', as.is=T, header=F) #read in sample
data
> c1<-c(0.010313070, -0.02893566)
> c2<-c(0.245346732, 0.31654765)
> kmeans(x, centers=c(c1, c2))
K-means clustering with 2 clusters of sizes 3, 1
Cluster means:
V1 V2
1 0.01569253 0.1957229
2 0.54928243 -0.1534170
Clustering vector:
[1] 1 1 2 1
Within cluster sum of squares by cluster:
[1] 0.1789207 0.0000000
Available components:
[1] "clust |
|
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f***a
发帖数: 329 | 32 可以先把要处理的数组rbind()成一个matrix，再用apply(A,1,tt)，其中tt是你需要用
的函数。注意，把tt返回的值变成list。
例子：
tt <- function(t) { result <- matrix(t,2,); list(result) }
A <- matrix(1:16,4,)
r <- apply(A,1,tt) |
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t**i
发帖数: 688 | 33 ## suppose the third column is X
new.data = do.call("rbind",by(data, data$firm, function(dd)
{
data.frame(dd[,1:2],c(NA,dd$X[-1]),dd[,4])
})) |
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S******y
发帖数: 1123 | 34 I am wondering what data structure is "{ } " in R?
for example -
theta ={}
...
theta = rbind(theta, result)
Is theta a list or array?
Thanks. |
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t**i
发帖数: 688 | 35 do.call("rbind", by(...)) |
|
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s******e
发帖数: 841 | 37 maybe it works
rbind(A,B[,match(names(A),names(B))] |
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D******n
发帖数: 2836 | 38 dat<-as.matrix(read.table('rawfile.text',sep="",col.names=1:5,fill=T));
dat[is.na(dat)]<-0;
dat<-rbind(0,dat);
result<-dat+t(dat)+diag(5);
upstream |
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x***x
发帖数: 3401 | 39 I have a matrix M and a array L. L contains a list of integers.
The integers are the indexes of the row numbers in the matrix M.
I want to make a new matrix N where the ith row corresponds to the L[i]th
row of the M matrix.
Is there an easy way of doing this?
I can only think of a loop with rbind() function. Any other simpler ideas or
functions?
Thanks |
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t**i
发帖数: 688 | 40 do.call("rbind", strsplit(....))[,2] |
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t**i
发帖数: 688 | 41 do.call("rbind",by(dat, dat$ID, function(dd){dd[1,]})) |
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R******o
发帖数: 83 | 42 可以用table做n-waytable,用rbind,cbind之类灵活组合成自己想要的表，修饰一下保
存成tex 文件 用 latex 可以生成很漂亮的表。 或者存成 txt 文件，让SAS读进去再
print出来。 |
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d*********k
发帖数: 1239 | 43 比如
A<-c(1,2,3,4,5)
B<-c(1,2,3)
如果直接用cbind：
> cbind(A,B)
A B
[1,] 1 1
[2,] 2 2
[3,] 3 3
[4,] 4 1
[5,] 5 2
怎么得到下面这样的形式呢？
A B
[1,] 1 1
[2,] 2 2
[3,] 3 3
[4,] 4
[5,] 5
谢谢啦啊~~ |
|
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d*********k
发帖数: 1239 | 45 谢谢啦啊~
不过我要是有很多vector呢？要是一个一个的数element的个数的话，还是蛮麻烦的呢~
有没有什么
比较通用的方法呢？
谢谢了啊~ |
|
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d*********k
发帖数: 1239 | 47 我也想这用list，可是不是很熟悉
你能不能详细点？比如这个简单的例子？ 谢谢啦啊 |
|
b*******g
发帖数: 513 | 48 Here is an example:
> a<-c(1,2,3)
> b<-c(1,2,3,4)
> c<-list(a,b)
> c
[[1]]
[1] 1 2 3
[[2]]
[1] 1 2 3 4 |
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B**W
发帖数: 2273 | 49 B <- c(B, rep(NA,length(A)-length(B)))
cbind(A,B) |
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