如何通過組合每個第三列來創建表？

我有一個看起來像這樣的數據：如何通過組合每個第三列來創建表？

  SMA SMA.1  EMA SMA.2 SMA.1.1 EMA.1 
2016-08-30 1.12411 1.114418 42.50321 100.9875 102.7288 61.25217

SMA，SMA.1和EMA是一個類型的變量所做的一切計算。我們稱之爲A1。同樣，SMA.2 SMA.1.1，EMA屬於A2。這一直持續到第29個變量A29。我想創建一個表，看起來像：

Var SMA  SMA.1  EMA 
A1 1.12411 1.114418 42.50321 
A2 100.9875 102.7288 61.25217

我一直在試圖弄清楚這一點與一對（）函數天，但它不工作。這裏的dput，如果它是有幫助的：

An ‘xts’ object on 2016-08-30/2016-08-30 containing: 
    Data: num [1, 1:87] 1.12 1.11 42.5 100.99 102.73 ... 
- attr(*, "dimnames")=List of 2 
    ..$ : NULL 
    ..$ : chr [1:87] "SMA" "SMA.1" "EMA" "SMA.2" ... 
    Indexed by objects of class: [POSIXct,POSIXt] TZ: 
    xts Attributes: 
NULL 
> length(new.trend(MergedData)) 
[1] 87 
> dput(new.trend(MergedData)) 
structure(c(1.12411, 1.114418, 42.5032053202923, 100.9875, 102.7288, 
61.2521659094113, 1.30729, 1.312984, 48.6820468363914, 0.96929, 
0.976116, 63.3550259067348, 0.762599999999999, 0.75914, 36.9475885995277, 
1.29322, 1.301952, 58.0919871676451, 0.72508, 0.717614, 50.7849595338026, 
8.431315, 8.518538, 60.1081873107905, 8.22937, 8.378464, 56.4004957184729, 
24.03834, 24.251418, 57.4224323964315, 275.5605, 279.5882, 51.0296673272519, 
3.7801, 3.814314, 49.384958216963, 3.82987000000001, 3.88755, 
66.0031408704831, 64.45802, 64.756234, 52.7346708652452, 2.94456, 
2.979396, 51.219478918934, 13.79045, 13.967152, 73.6027751334041, 
3.2168, 3.231082, 54.8620717312901, 660.3725, 657.1164, 63.4249734975412, 
2903.875, 2961.2792, 47.0703023915683, 18.34358, 18.526388, 59.4661365407602, 
0.7604, 0.7558, 42.3745163443544, 6.65051000000001, 6.660192, 
62.9781410895429, 13173.25, 13135.84, 73.4189545097276, 66.9848050000001, 
66.9980439999999, 51.0467264878632, 1111.974, 1120.1316, 52.8805945280585, 
4.01933, 4.022258, 59.7858956387081, 1.3497, 1.348972, 70.0217831036426, 
46.48875, 46.79366, 45.2292753580903, 34.6466, 34.816, 37.0539681977138 
), .Dim = c(1L, 87L), .Dimnames = list(NULL, c("SMA", "SMA.1", 
"EMA", "SMA.2", "SMA.1.1", "EMA.1", "SMA.3", "SMA.1.2", "EMA.2", 
"SMA.4", "SMA.1.3", "EMA.3", "SMA.5", "SMA.1.4", "EMA.4", "SMA.6", 
"SMA.1.5", "EMA.5", "SMA.7", "SMA.1.6", "EMA.6", "SMA.8", "SMA.1.7", 
"EMA.7", "SMA.9", "SMA.1.8", "EMA.8", "SMA.10", "SMA.1.9", "EMA.9", 
"SMA.11", "SMA.1.10", "EMA.10", "SMA.12", "SMA.1.11", "EMA.11", 
"SMA.13", "SMA.1.12", "EMA.12", "SMA.14", "SMA.1.13", "EMA.13", 
"SMA.15", "SMA.1.14", "EMA.14", "SMA.16", "SMA.1.15", "EMA.15", 
"SMA.17", "SMA.1.16", "EMA.16", "SMA.18", "SMA.1.17", "EMA.17", 
"SMA.19", "SMA.1.18", "EMA.18", "SMA.20", "SMA.1.19", "EMA.19", 
"SMA.21", "SMA.1.20", "EMA.20", "SMA.22", "SMA.1.21", "EMA.21", 
"SMA.23", "SMA.1.22", "EMA.22", "SMA.24", "SMA.1.23", "EMA.23", 
"SMA.25", "SMA.1.24", "EMA.24", "SMA.26", "SMA.1.25", "EMA.25", 
"SMA.27", "SMA.1.26", "EMA.26", "SMA.28", "SMA.1.27", "EMA.27", 
"SMA.29", "SMA.1.28", "EMA.28")), index = structure(1472533200, tzone = "", tclass = c("POSIXct", 
"POSIXt")), .indexTZ = "", class = c("xts", "zoo"), .indexCLASS = c("POSIXct", 
"POSIXt"), tclass = c("POSIXct", "POSIXt"), tzone = "")

來源

2017-01-20 Nikitau

你有這個工程太下面2個很好的答案，如果滿足您的需求，請將該答案旁邊的複選標記集合起來，這樣就可以認爲這個問題是關閉的 – JustGettinStarted

簡單：

new_data_mat <- matrix(data_mat[1,], ncol=3, byrow=T) 
colnames(new_data_mat) <- c("SMA", "SMA2", "EMA")

來源

2017-01-20 19:59:55 thc

@ THC的答案肯定是簡單，但如果你想留在一個循環

# take vector of your data 
Data <- Data[1,] 
names(Data) 
New.Data <- NULL 
EMAc <- 3; SMAc <- 1; SMA.1c <- 2 
for (i in 1:(length(Data)/3)){ 
    i.New.Data <- cbind(Var=paste("A",i,sep=""), 
         EMA=Data[i*EMAc], 
         SMA=Data[i*SMAc], 
         SMA.1=Data[i*SMA.1c]) 
    New.Data <- rbind(New.Data,i.New.Data) 
} 
New.Data

來源

2017-01-20 21:07:36 JustGettinStarted

如何通過組合每個第三列來創建表？

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