R 操作特定序列的字符向量答案

【问题标题】：R manipulation a character vector for a specific sequenceR 操作特定序列的字符向量
【发布时间】：2020-07-28 21:27:52
【问题描述】：

我有一个项目 ID 列表如下：

ids <- c("12_a","23_b")

考虑到这些项目 ID，我想为三个组（G1、G2 和 G3）生成如下字符变量。

    #for the first item
Equal = (G1,12_a, Slope[0]), 
        (G2,12_a, Slope[0]), 
        (G3,12_a, Slope[0]);
Equal = (G1,12_a, Slope[1]), 
        (G2,12_a, Slope[1]), 
        (G3,12_a, Slope[1]);
Equal = (G1,12_a, Slope[2]), 
        (G2,12_a, Slope[2]), 
        (G3,12_a, Slope[2]);
Equal = (G1,12_a, Intercept[0]), 
        (G2,12_a, Intercept[0]), 
        (G3,12_a, Intercept[0]);
    
    #for the second item
Equal = (G1,23_b, Slope[0]), 
        (G2,23_b, Slope[0]), 
        (G3,23_b, Slope[0]);
Equal = (G1,23_b, Slope[1]), 
        (G2,23_b, Slope[1]), 
        (G3,23_b, Slope[1]);
Equal = (G1,23_b, Slope[2]), 
        (G2,23_b, Slope[2]), 
        (G3,23_b, Slope[2]);
Equal = (G1,23_b, Intercept[0]), 
        (G2,23_b, Intercept[0]), 
        (G3,23_b, Intercept[0]);

所需输出背后的逻辑是Slope[] 中的值对于三个组应该是0,1, and 3。并且Intercept[] 的值应该是0 三倍于三组。

以前有人有类似的吗？

谢谢！

【问题讨论】：

标签： r character

【解决方案1】：

我相信这就是你想要的。我为此做了一个函数：

#Vector of ids
ids <- c("12_a","23_b")
#Function to create data
#x is a vector of ids
#n is the number of times you want data repeated according to your specifications
create <- function(x,n)
{
  #Build vars
  groups <- rep(paste0('Group_',1:n),n)
  intercept <- rep(rep(0,n),n) 
  slope <- do.call(c,lapply(0:(n-1),rep,n))
  df <- data.frame(groups,intercept,slope)
  #Feed ids
  index <- length(x)
  #Use a loop
  List <- list()
  for(i in 1:index)
  {
    List[[i]] <- data.frame(id=x[i],df)
  }
  #Create final object
  DF <- do.call(rbind,List)
  return(DF)
}
#Use the function
create(x = ids,n = 3)

它产生：

     id  groups intercept slope
1  12_a Group_1         0     0
2  12_a Group_2         0     0
3  12_a Group_3         0     0
4  12_a Group_1         0     1
5  12_a Group_2         0     1
6  12_a Group_3         0     1
7  12_a Group_1         0     2
8  12_a Group_2         0     2
9  12_a Group_3         0     2
10 23_b Group_1         0     0
11 23_b Group_2         0     0
12 23_b Group_3         0     0
13 23_b Group_1         0     1
14 23_b Group_2         0     1
15 23_b Group_3         0     1
16 23_b Group_1         0     2
17 23_b Group_2         0     2
18 23_b Group_3         0     2

它也适用于其他设置：

#Another test
ids2 <- ids <- c("12_a","23_b","65_c")
create(x = ids2,n = 5)

更新： 我已经更新了函数来创建一个与你想要的结构相似的数据框：

create <- function(x,n)
{
  #Build vars
  groups <- rep(paste0('Group_',1:n),n)
  intercept <- data.frame(groups=paste0('Group_',1:n),var='intercept',val=0) 
  val <- do.call(c,lapply(0:(n-1),rep,n))
  df <- data.frame(groups,var='slope',val)
  #Bind all
  dfm <- rbind(df,intercept)
  #Feed ids
  index <- length(x)
  #Use a loop
  List <- list()
  for(i in 1:index)
  {
    List[[i]] <- data.frame(id=x[i],dfm)
  }
  #Create final object
  DF <- do.call(rbind,List)
  return(DF)
}

它将像这样工作：

DF <- create(x = ids,n = 3)

     id  groups       var val
1  12_a Group_1     slope   0
2  12_a Group_2     slope   0
3  12_a Group_3     slope   0
4  12_a Group_1     slope   1
5  12_a Group_2     slope   1
6  12_a Group_3     slope   1
7  12_a Group_1     slope   2
8  12_a Group_2     slope   2
9  12_a Group_3     slope   2
10 12_a Group_1 intercept   0
11 12_a Group_2 intercept   0
12 12_a Group_3 intercept   0
13 23_b Group_1     slope   0
14 23_b Group_2     slope   0
15 23_b Group_3     slope   0
16 23_b Group_1     slope   1
17 23_b Group_2     slope   1
18 23_b Group_3     slope   1
19 23_b Group_1     slope   2
20 23_b Group_2     slope   2
21 23_b Group_3     slope   2
22 23_b Group_1 intercept   0
23 23_b Group_2 intercept   0
24 23_b Group_3 intercept   0

【讨论】：

@amisos55 是的，您只需分配给一个新对象，例如mydf <- create(x = ids,n = 3) 让我知道它是否有效。
有没有办法将此结构转换为我想要的输出中的字符值？
@amisos55 我已经更新了函数以生成与您想要的结构相似的数据帧。让我知道这是否有帮助。

【解决方案2】：

这是我想出来的。

ids <- c("12_a","23_b")
group <- 3
Slope.0 <- c() # store slope vector
Intercept.0 <- c() # store intercept vector

for(i in 1:length(ids)) {

  for(j in 0:group) { # here with the length(State) I gained the sequqnece of 0,1,2,3
    slope.0 <- paste0(paste0("Equal = ",paste0(paste("(", "G1, ",ids[i], ","," Slope[",j,"])",collapse=", ", sep=""),", ",
                                        paste( "(", "G2, ",ids[i], ","," Slope[",j,"])",collapse=", ", sep=""),", ",
                                        paste( "(", "G3, ",ids[i], ","," Slope[",j,"])",collapse=", ", sep=""))), ";")
    Slope.0 <- c(Slope.0, slope.0)
    
  }
  
  
  
  intercept.0 <- paste0(paste0("Equal = ",paste0(paste("(", "G1, ",ids[i], ","," Intercept[0])",collapse=", ", sep=""),", ",
                                          paste( "(", "G2, ",ids[i], ","," Intercept[0])",collapse=", ", sep=""),", ",
                                          paste( "(", "G3, ",ids[i], ","," Intercept[0])",collapse=", ", sep=""))),";")
  Intercept.0 <- c(Intercept.0, intercept.0)}

【讨论】：