将函数应用于 data.frame 的每一列并组织输出 [英] Apply a function to each column of a data.frame and organize the output

问题描述

我有这个向量:

 x <- c(5,2,-4,-6,-2,1,4,2,-3,-6,-1,8,9,5,-6,-11)

我使用这个功能:

myfunction <- function(x){n <- 长度(x)fx <- 数字(n)fx[1] <- min(x[1],0)for(i in 2:n){fx[i] <- min(0,fx[i-1]+x[i])}外汇x_min <-min(x)fx_min <- min(fx)fx_05 <- 数字(n)fx_05[1] <- min(fx[1],0)for (i in 2:n) {如果(总和(fx_05[i-1]+x[i])>0){fx_05[i] <- 0} else if ((sum(fx_05[i-1]+x[i]))<(fx_min*0.5)) {fx_05[i] <- (fx_min*0.5)} else { fx_05[i] <- sum(fx_05[i-1]+x[i]) }}FX_05as.data.frame(矩阵(c(x，fx_05)，ncol = 2))}xx <- 我的函数(x)

数据帧xx是

 V1 V21 5 0.02 2 0.03 -4 -4.04 -6 -8.55 -2 -8.s6 1 -7.57 4 -3.58 2 -1.59 -3 -4.510 -6 -8.511 -1 -8.512 8 -0.513 9 0.014 5 0.015 -6 -6.016 -11 -8.5`

我想将此函数应用于 data.frame :

df <- data.frame(x <- c(5,2,-4,-6,-2,1,4,2,-3,-6,-1,8,9,5,-6,-11),y <- c(5,2,-4,-6,-2,1,4,2,-3,-6,-1,8,9,5,-6,-11),z <- c(5,2,-4,-6,-2,1,4,2,-3,-6,-1,8,9,5,-6,-11))

使用:

输出 <- myfunction(df)

它不起作用，并且使用:

输出 <- data.frame(sapply(df, myfunction))

data.frame 输出的形式不正确.data.frame 的每个原始列应该是 2 列.

解决方案

在这种情况下，您希望使用 lapply.它将处理 data.frame 的每一列，因为它实际上是一个等长向量的列表，并且每列返回一个两列的 data.frame.

x <- lapply(df, myfunction)

此外，sapply 工作得很好.唯一的区别是它在开始时看起来不同.有关所有解决方案之间的差异，请参阅 print(x).

x <- sapply(df, myfunction)

之后您可能希望再次将它们从列表中组合到一个 data.frame 中.你可以用 do.call

做到这一点

df2 <- do.call(cbind, x)

这会弄乱列名.您可以使用 names

更改这些

names(df2) <- NULLdf2# 1 5 0.0 5 0.0 5 0.0# 2 2 0.0 2 0.0 2 0.0# 3 -4 -4.0 -4 -4.0 -4 -4.0# 4 -6 -8.5 -6 -8.5 -6 -8.5# ....

旁注:

如果您没有 data.frame 而是一个矩阵作为输入，另一个选项是 apply 和 MARGIN = 2.

x <- apply(df, MARGIN = 2, myfunction)

虽然在这个例子中，它也能工作，但当你的向量有不同的数据类型时你会遇到麻烦，因为它在应用函数之前将 data.frame 转换为矩阵.因此不推荐.有关更多信息，请参阅这篇详细且易于理解的帖子！

进一步阅读:
Hadley Wickham 的高级 R.另请查看此站点上有关数据类型的部分.
Peter Werner 的博文<小时>

我非常感谢 @Gregor 对这篇文章的贡献.

I have this vector:

 x <- c(5,2,-4,-6,-2,1,4,2,-3,-6,-1,8,9,5,-6,-11)

I use this function:

myfunction <- function(x){
     n <- length(x)
     fx <- numeric(n)
     fx[1] <- min(x[1],0)
     for(i in 2:n){fx[i] <- min(0,fx[i-1]+x[i])}
     fx

     x_min <-min(x)
     fx_min <- min(fx)

     fx_05 <- numeric(n)
     fx_05[1] <- min(fx[1],0)
     for (i in 2:n) {
       if (sum(fx_05[i-1]+x[i])>0) {  
          fx_05[i] <- 0
       } else if ((sum(fx_05[i-1]+x[i]))<(fx_min*0.5)) {
          fx_05[i] <- (fx_min*0.5)
       } else { fx_05[i] <- sum(fx_05[i-1]+x[i]) }
     }
     fx_05
     as.data.frame(matrix(c(x, fx_05), ncol = 2 ))
}
xx <- myfunction(x)

The dataframe xx is

    V1   V2
1    5  0.0
2    2  0.0
3   -4 -4.0
4   -6 -8.5
5   -2 -8.s
6    1 -7.5
7    4 -3.5
8    2 -1.5
9   -3 -4.5
10  -6 -8.5
11  -1 -8.5
12   8 -0.5
13   9  0.0
14   5  0.0
15  -6 -6.0
16 -11 -8.5`

I would like to apply this function to a data.frame :

df <- data.frame(x <- c(5,2,-4,-6,-2,1,4,2,-3,-6,-1,8,9,5,-6,-11),
                   y <- c(5,2,-4,-6,-2,1,4,2,-3,-6,-1,8,9,5,-6,-11),
                   z <- c(5,2,-4,-6,-2,1,4,2,-3,-6,-1,8,9,5,-6,-11))

Using:

output <- myfunction(df) 

It doesn't work, and using:

outputs <- data.frame(sapply(df, myfunction))

the form of the data.frame output is not correct. It should be 2 columns for each original column of the data.frame.

解决方案

In this case, you would like to use lapply. It will handle each column of the data.frame, as it actually is a list of equal-length vectors, and return a two column data.frame each.

x <- lapply(df, myfunction)

Also, sapply works just fine. The only difference is that it looks different at the beginning. See print(x) for the difference between all solutions.

x <- sapply(df, myfunction)

Afterwards you probably want to combine them from a list to a data.frame again. You can do this with do.call

df2 <- do.call(cbind, x)

This will mess up the column names. You can change these using names

names(df2) <- NULL
df2
# 1    5  0.0   5  0.0   5  0.0
# 2    2  0.0   2  0.0   2  0.0
# 3   -4 -4.0  -4 -4.0  -4 -4.0
# 4   -6 -8.5  -6 -8.5  -6 -8.5
# ....

Side Note:

If you don't have a data.frame but a matrix as input, another option would be apply with the with MARGIN = 2.

x <- apply(df, MARGIN = 2, myfunction)

Although in this example, it works as well, you will run into trouble when having differing data types across your vectors as it converts the data.frame to a matrix before applying the function. Therefore it is not recommended. More info on that can be found in this detailed and easy-to-understand post!

Further reading on this:
Hadley Wickham's Advanced R. Also check out the section on data types on this site.
Peter Werner's blog post

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I greatly appreciate the input of @Gregor on this post.

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