多列数据并获得平均R程序 [英] Multiple columns of data and getting average R program

问题描述

我之前问过这样的问题，但是我决定简化数据格式，因为我是R的新手，不了解发生了什么....这是问题

I asked a question like this before but I decided to simplify my data format because I'm very new at R and didnt understand what was going on....here's the link for the question How to handle more than multiple sets of data in R programming?

但是我编辑了数据的外观，并决定以这种格式保留它.

But I edited what my data should look like and decided to leave it like this..in this format...

X1.0   X X2.0 X.1
   0.9 0.9  0.2 1.2
  1.3 1.4  0.8 1.4

如您所见，我有四列数据，我正在处理的实际数据最多为2000个数据点.....列"X1.0"和"X2.0"指的是时间". ..so我想要的是基于我的2列时间分别为"X1.0"和"X2.0"的每100秒的"X"和"X.1"的平均值...我可以使用此命令

As you can see I have four columns of data, The real data I'm dealing with is up to 2000 data points.....Columns "X1.0" and "X2.0" refer "Time"...so what I want is the average of "X" and "X.1" every 100 seconds based on my 2 columns of time which are "X1.0" and "X2.0"...I can do it using this command

cuts <- cut(data$X1.0, breaks=seq(0, max(data$X1.0)+400, 400))
   by(data$X, cuts, mean)

但是，这只会给我一组数据的平均值....是"X1.0"和"X" ....我将如何做，以便我可以从多个数据中获取平均值一个数据集....我也想停止这种输出

But this will only give me the average from one set of data....which is "X1.0" and "X".....How will I do it so that I could get averages from more than one data set....I also want to stop having this kind of output

cuts: (0,400]
[1] 0.7
------------------------------------------------------------ 
cuts: (400,800]
[1] 0.805

请注意，输出每400秒完成一次....我真的想要列出这些切割的列表，这些切割是不同时间间隔的平均值...请帮助...我只是使用data=read.delim("clipboard")来获得我的数据进入程序

Note that the output was done every 400 s....I really want a list of those cuts which are the averages at different intervals...please help......I just used data=read.delim("clipboard") to get my data into the program

推荐答案

您想要获得什么输出有点令人困惑.

It is a little bit confusing what output do you want to get.

首先，我更改名字，但这是可选的

First I change colnames but this is optional

colnames(dat) <- c('t1','v1','t2','v2')

然后我将使用ave，它与by相似，但输出更好.我正在使用矩阵的技巧来索引列:

Then I will use ave which is like by but with better output. I am using a trick of a matrix to index column:

matrix(1:ncol(dat),ncol=2)  ## column1 is col1 adn col2...
     [,1] [,2]
[1,]    1    3
[2,]    2    4

然后我将此矩阵与apply一起使用.这里是整个解决方案:

Then I am using this matrix with apply. Here the entire solution:

cbind(dat,
      apply(matrix(1:ncol(dat),ncol=2),2,
     function(x,by=10){      ## by 10 seconds! you can replace this 
                             ## with 100 or 400 in you real data
     t.col <- dat[,x][,1]   ## txxx
     v.col <- dat[,x][,2]   ## vxxx
     ave(v.col,cut(t.col, 
                   breaks=seq(0, max(t.col),by)),
         FUN=mean)})
  )

编辑更正剪切并简化代码

cbind(dat,
     apply(matrix(1:ncol(dat),ncol=2),2,
           function(x,by=10)ave(dat[,x][,1], dat[,x][,1] %/% by)))
   X1.0   X X2.0 X.1       1         2
1   0.9 0.9  0.2 1.2  3.3000  3.991667
2   1.3 1.4  0.8 1.4  3.3000  3.991667
3   2.0 1.7  1.6 1.1  3.3000  3.991667
4   2.6 1.9  2.2 1.6  3.3000  3.991667
5   9.7 1.0  2.8 1.3  3.3000  3.991667
6  10.7 0.8  3.5 1.1 12.8375  3.991667
7  11.6 1.5  4.1 1.8 12.8375  3.991667
8  12.1 1.4  4.7 1.2 12.8375  3.991667
9  12.6 1.8  5.4 1.2 12.8375  3.991667
10 13.2 2.1  6.3 1.3 12.8375  3.991667
11 13.7 1.6  6.9 1.1 12.8375  3.991667
12 14.2 2.2  9.4 1.3 12.8375  3.991667
13 14.6 1.8 10.0 1.5 12.8375 10.000000

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