需要帮助在 R 中实现功能 [英] Need help implementing a function in R

问题描述

我正在尝试以一种利用脚本的方式进行编码，因为我经常以这种格式获得输出，并且不得不在 Excel 中一一复制/粘贴它是一项真正的苦差事.但是，我在实现功能方面遇到了困难.

I'm trying to code in such a way as to take advantage of scripting, as I regularly get output in this format and having to copy/paste it in Excel one by one is a real chore. However, I'm stuck when it comes to implementing the functions.

因此，我的数据采用以下形式:

So, my data is in the following form:

Condition Sample1 Sample2 .... Sample n
T1        6.99    5.80    ....  n_1      
T2        2.05    3.04    ....  n_1      
T3        4.50    4.69    ....  n_1      
T4        4.71    5.22    ....  n_1      
T5        5.66    3.65    ....  n_1      
T6        9.76    2.89    ....  n_1      

我需要应用以下等式:，其中 x 是单个条目，n 是系数，这样整个方程看起来像这样:

I need to apply the following equation: , where x is the individual entry and n is a coefficient, such that the full equation looks something like this:

.

基本上，每列，我需要按顺序考虑每个元素，并将其乘以序列系数(来自 1 的奇数:长度条件)以获得每个样本的答案 S.我的数据集的大小不会改变——它总是 T1:T6，改变的是样本 1...n.理想情况下，S 的值将附加在列的底部，或者保存在一个单独的数据集中，参考它所属的样本.

Basically, per column, I need to consider each element in sequence and multiply it by a sequential coefficient (odd numbers from 1: length Condition) to get the answer S, for each Sample. The size of my dataset will not change - it will always be T1:T6, what will change is Sample 1...n. Ideally the value of S will be appended at the bottom of the column, or saved in a separate dataset with reference to the sample that it belongs to.

我尝试了许多解决方案，包括移调，但似乎无法解决它.

I've tried a number of solutions, including transposing, but can't seem to wrap my head around it.

我目前尝试在数据集的一部分上实现更简单的函数没有成功.

My current attempt at implementing a simpler function on a portion of the dataset yielded no success.

 for (i in 2:8){dT[7,i] <-
     ((1*dT[1,i])+(3*dT[2,i])+(5*dT[3,i])+(7*dT[4,i])+(9*dT[5,i]))+(11*dT[6,i])
 }

我认为正确的解决方案涉及某种*apply，但我完全不知道如何正确使用它们.

I think the right solution involves some kind of *apply but I'm completely clueless as to how to use those properly.

添加可重现的示例:

N   Condition   Sample A    Sample B    Sample C    Sample D

1   T1          91.323      78.758      70.298      66.765
3   T2          -3.737      -1.5        -7.744      -9.247
5   T3          5.205       4.533       2.284       2.178
7   T4          -0.486      -0.068      -1.386      -0.927
9   T5          0.337       -0.139      0.087       0.055
    S        -0.046296296   -0.123654391    0.394039047 0.445258425

推荐答案

看起来很简单.由于 R 是向量化的，所以使用向量的乘法.然后sum.

Seems simple. Since R is vectorized use vectors' multiplication. And then sum.

zirconium <- function(x){
    n <- 2*seq_along(x) - 1
    sum(x * (-1)^((n - 1)/2))
}

sapply(dT[-1], zirconium)
#Sample1 Sample2 
#   0.63    2.99

数据.

dT <-
structure(list(Condition = structure(1:6, .Label = c("T1", "T2", 
"T3", "T4", "T5", "T6"), class = "factor"), Sample1 = c(6.99, 
2.05, 4.5, 4.71, 5.66, 9.76), Sample2 = c(5.8, 3.04, 4.69, 5.22, 
3.65, 2.89)), .Names = c("Condition", "Sample1", "Sample2"), class = "data.frame", row.names = c(NA, 
-6L))

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