英文:
Applying a custom function to every row in r
问题
I created a function to calculate the rollmean of a row in a dataframe:
rollmean_circular <- function(x) {t(rollmean(t(cbind(x[9:10],x,x[1:2])),5))}
这是一个用来计算数据框中一行的滚动均值的函数：
rollmean_circular <- function(x) {t(rollmean(t(cbind(x[9:10],x,x[1:2])),5))}
What this function does is given a vector, it will append the last 2 element to the front and first 2 element to the back and then do a rollmean so there will not be any NAs at the front or back.
这个函数的作用是给定一个向量，它会将最后两个元素添加到前面，将前两个元素添加到后面，然后进行滚动均值计算，以确保前面和后面没有任何NA值。
It works perfectly when I apply to 1 row of a df.
当我将其应用于数据框的一行时，它可以完美运行。
r = df[1,]
rollmean_circular(r)
r = df[1,]
rollmean_circular(r)
However, when I use apply to apply this function to every row of my dataframe, it returns a logical(0).
然而，当我使用apply将此函数应用于数据框的每一行时，它返回一个logical(0)。
apply(df,1,rollmean_circular)
apply(df,1,rollmean_circular)
Can I know what I am missing?
我能知道我漏掉了什么吗？
When I apply another function that gives the same output for a single row, it works:
当我应用另一个函数，该函数对单行产生相同的输出时，它可以正常工作：
stdize <- function(x, na.rm=T) {(x - min(x, na.rm=T)) / (max(x, na.rm=T) - min(x, na.rm=T))}
stdize <- function(x, na.rm=T) {(x - min(x, na.rm=T)) / (max(x, na.rm=T) - min(x, na.rm=T))}
stdize(r)
stdize(r)
apply(df,1,stdize)
apply(df,1,stdize)
英文:
I created a function to calculate the rollmean of a row in a dataframe:
rollmean_circular <- function(x) {t(rollmean(t(cbind(x[9:10],x,x[1:2])),5))}
df <- structure(list(X1 = c(5L, 5L, 9L, 0L, 9L, 10L, 10L, 1L, 0L, 10L
), X2 = c(6L, 8L, 6L, 9L, 7L, 5L, 0L, 7L, 5L, 8L), X3 = c(10L,
7L, 2L, 1L, 2L, 10L, 2L, 9L, 6L, 4L), X4 = c(6L, 0L, 9L, 1L,
6L, 8L, 3L, 7L, 8L, 1L), X5 = c(0L, 9L, 8L, 3L, 1L, 8L, 3L, 9L,
5L, 2L), X6 = c(0L, 10L, 9L, 10L, 3L, 1L, 6L, 0L, 6L, 9L), X7 = c(9L,
10L, 0L, 10L, 10L, 9L, 0L, 1L, 10L, 2L), X8 = c(2L, 6L, 3L, 7L,
7L, 9L, 8L, 9L, 1L, 0L), X9 = c(0L, 8L, 8L, 9L, 0L, 5L, 9L, 9L,
4L, 8L), X10 = c(1L, 4L, 3L, 0L, 1L, 7L, 3L, 6L, 5L, 0L)), class = "data.frame", row.names = c(NA,
-10L))
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 5 6 10 6 0 0 9 2 0 1
2 5 8 7 0 9 10 10 6 8 4
3 9 6 2 9 8 9 0 3 8 3
4 0 9 1 1 3 10 10 7 9 0
5 9 7 2 6 1 3 10 7 0 1
6 10 5 10 8 8 1 9 9 5 7
7 10 0 2 3 3 6 0 8 9 3
8 1 7 9 7 9 0 1 9 9 6
9 0 5 6 8 5 6 10 1 4 5
10 10 8 4 1 2 9 2 0 8 0
What this function does is given a vector, it will append the last 2 element to the front and first 2 element to the back and then do a rollmean so there will not be any NAs at the front or back.
It works perfectly when I apply to 1 row of a df.
r = df[1,]
rollmean_circular[r]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
1 4.4 5.6 5.4 4.4 5 3.4 2.2 2.4 3.4 2.8
However, when I use apply to apply this function to every row of my dataframe, it returns a logical(0).
apply(df,1,rollmean_circular)
logical(0)
Can I know what I am missing?
When I apply another function that gives the same output for a single row, it works:
stdize <- function(x, na.rm=T) {(x - min(x, na.rm=T)) / (max(x, na.rm=T) - min(x, na.rm=T))}
stdize(r)
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 0.5 0.6 1 0.6 0 0 0.9 0.2 0 0.1
apply(df,1,stdize)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
X1 0.5 0.5 1.0000000 0.0 0.9 1.0000000 1.0 0.1111111 0.0 1.0
X2 0.6 0.8 0.6666667 0.9 0.7 0.4444444 0.0 0.7777778 0.5 0.8
X3 1.0 0.7 0.2222222 0.1 0.2 1.0000000 0.2 1.0000000 0.6 0.4
X4 0.6 0.0 1.0000000 0.1 0.6 0.7777778 0.3 0.7777778 0.8 0.1
X5 0.0 0.9 0.8888889 0.3 0.1 0.7777778 0.3 1.0000000 0.5 0.2
X6 0.0 1.0 1.0000000 1.0 0.3 0.0000000 0.6 0.0000000 0.6 0.9
X7 0.9 1.0 0.0000000 1.0 1.0 0.8888889 0.0 0.1111111 1.0 0.2
X8 0.2 0.6 0.3333333 0.7 0.7 0.8888889 0.8 1.0000000 0.1 0.0
X9 0.0 0.8 0.8888889 0.9 0.0 0.4444444 0.9 1.0000000 0.4 0.8
X10 0.1 0.4 0.3333333 0.0 0.1 0.6666667 0.3 0.6666667 0.5 0.0
答案1
得分: 2
看起来你在你的函数中混淆了向量和矩阵。你可以在函数中使用 unlist，然后稍后再进行转置。
rollmean_circular <- function(x) zoo::rollmean(unlist(c(x[9:10], x, x[1:2])), 5)
使用 apply 函数和上述函数来处理数据框 df，然后进行转置：
t(apply(df, 1, rollmean_circular))
这将产生与你提供的示例相似的结果，不过这只是翻译了你的代码，没有其他内容。
英文:
Seems you're confusing vectors and matrices in your function. You could unlist in the function and transpose later.
rollmean_circular <- function(x) zoo::rollmean(unlist(c(x[9:10], x, x[1:2])),5)
t(apply(df, 1, rollmean_circular))
# X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
# [1,] 4.4 5.6 5.4 4.4 5.0 3.4 2.2 2.4 3.4 2.8
# [2,] 6.4 4.8 5.8 6.8 7.2 7.0 8.6 7.6 6.6 6.2
# [3,] 5.6 5.8 6.8 6.8 5.6 5.8 5.6 4.6 4.6 5.8
# [4,] 3.8 2.2 2.8 4.8 5.0 6.2 7.8 7.2 5.2 5.0
# [5,] 3.8 5.0 5.0 3.8 4.4 5.4 4.2 4.2 5.4 4.8
# [6,] 7.4 8.0 8.2 6.4 7.2 7.0 6.4 6.2 8.0 7.2
# [7,] 4.8 3.6 3.6 2.8 2.8 4.0 5.2 5.2 6.0 6.0
# [8,] 6.4 6.0 6.6 6.4 5.2 5.2 5.6 5.0 5.2 6.4
# [9,] 4.0 4.8 4.8 6.0 7.0 6.0 5.2 5.2 4.0 3.0
# [10,] 6.0 4.6 5.0 4.8 3.6 2.8 4.2 3.8 4.0 5.2
This can also be done in base R (w/ most of the credits to @MattiPastell):
fun <- function(x, n=5) na.omit(filter(c(tail(x, 2), x, head(x, 2)), rep(1 / n, n), sides=2))
t(apply(df, 1, fun))
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,] 4.4 5.6 5.4 4.4 5.0 3.4 2.2 2.4 3.4 2.8
# [2,] 6.4 4.8 5.8 6.8 7.2 7.0 8.6 7.6 6.6 6.2
# [3,] 5.6 5.8 6.8 6.8 5.6 5.8 5.6 4.6 4.6 5.8
# [4,] 3.8 2.2 2.8 4.8 5.0 6.2 7.8 7.2 5.2 5.0
# [5,] 3.8 5.0 5.0 3.8 4.4 5.4 4.2 4.2 5.4 4.8
# [6,] 7.4 8.0 8.2 6.4 7.2 7.0 6.4 6.2 8.0 7.2
# [7,] 4.8 3.6 3.6 2.8 2.8 4.0 5.2 5.2 6.0 6.0
# [8,] 6.4 6.0 6.6 6.4 5.2 5.2 5.6 5.0 5.2 6.4
# [9,] 4.0 4.8 4.8 6.0 7.0 6.0 5.2 5.2 4.0 3.0
# [10,] 6.0 4.6 5.0 4.8 3.6 2.8 4.2 3.8 4.0 5.2
答案2
得分: 0
rollmean将自动在其输入的每一列上工作，因此可以直接进行操作，而不需要使用apply：
library(zoo)
t(rollmean(t(cbind(df[9:10], df, df[1:2])), 5))
或者在R的基础上使用stats::filter，它也可以在每一列上工作：
t(filter(t(df), rep(1, 5)/5, circular = TRUE))
这两种方法都会得到这个矩阵：
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 4.4 5.6 5.4 4.4 5.0 3.4 2.2 2.4 3.4 2.8
[2,] 6.4 4.8 5.8 6.8 7.2 7.0 8.6 7.6 6.6 6.2
[3,] 5.6 5.8 6.8 6.8 5.6 5.8 5.6 4.6 4.6 5.8
[4,] 3.8 2.2 2.8 4.8 5.0 6.2 7.8 7.2 5.2 5.0
[5,] 3.8 5.0 5.0 3.8 4.4 5.4 4.2 4.2 5.4 4.8
[6,] 7.4 8.0 8.2 6.4 7.2 7.0 6.4 6.2 8.0 7.2
[7,] 4.8 3.6 3.6 2.8 2.8 4.0 5.2 5.2 6.0 6.0
[8,] 6.4 6.0 6.6 6.4 5.2 5.2 5.6 5.0 5.2 6.4
[9,] 4.0 4.8 4.8 6.0 7.0 6.0 5.2 5.2 4.0 3.0
[10,] 6.0 4.6 5.0 4.8 3.6 2.8 4.2 3.8 4.0 5.2
根据您的应用需求，您可以考虑将这些序列存储在列中，而不是行，这样就不需要进行转置操作。
英文:
rollmean will automatically work on every column of its input so this can be done directly eliminating the apply:
library(zoo)
t(rollmean(t(cbind(df[9:10], df, df[1:2])), 5))
or using stats::filter in the base of R which will also work on every column:
t(filter(t(df), rep(1, 5)/5, circular = TRUE))
Either of tehse give this matrix:
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 4.4 5.6 5.4 4.4 5.0 3.4 2.2 2.4 3.4 2.8
[2,] 6.4 4.8 5.8 6.8 7.2 7.0 8.6 7.6 6.6 6.2
[3,] 5.6 5.8 6.8 6.8 5.6 5.8 5.6 4.6 4.6 5.8
[4,] 3.8 2.2 2.8 4.8 5.0 6.2 7.8 7.2 5.2 5.0
[5,] 3.8 5.0 5.0 3.8 4.4 5.4 4.2 4.2 5.4 4.8
[6,] 7.4 8.0 8.2 6.4 7.2 7.0 6.4 6.2 8.0 7.2
[7,] 4.8 3.6 3.6 2.8 2.8 4.0 5.2 5.2 6.0 6.0
[8,] 6.4 6.0 6.6 6.4 5.2 5.2 5.6 5.0 5.2 6.4
[9,] 4.0 4.8 4.8 6.0 7.0 6.0 5.2 5.2 4.0 3.0
[10,] 6.0 4.6 5.0 4.8 3.6 2.8 4.2 3.8 4.0 5.2
Depending on the needs of your application you could consider storing these series in columns rather than rows in which case the transposes would not be needed.
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