英文:
Creating a function to apply to every row of several different DFs in r
问题
我有一堆看起来像这样的巨大数据框：
a b c d e f g
--------------
1| 4 3 6 3 7 3 7
2| 4 6 2 5 6 2 1
3| 3 6 2 6 8 1 5
我需要一个可以应用于每个数据框的函数，但创建一个行和函数需要包括数据框的名称：
apf <- function(rowSums(**df**[, c(1, 3, 4, 7)]))
如何使数据框在函数中成为一个变量，以便我可以为每个数据框运行以下操作：
apply(dfn, 1, apf(dfn))
然后将输出放入新列中：
dfn$new_c <- apply(dfn, 1, apf(dfn))
英文:
I have a bunch of gigantic data frames that look something like:
a b c d e f g
--------------
1| 4 3 6 3 7 3 7
2| 4 6 2 5 6 2 1
3| 3 6 2 6 8 1 5
I need a function that I can apply to every df but making a rowsum function requires you to include the df name:
apf <- function(rowSums(**df**[, c(1, 3, 4, 7)]))
How can I make df a variable in the function such that I can run the following for each df:
apply(dfn, 1, apf(dfn))
and then put the output in a new colun
dfn$new_c <- apply(dfn, 1, apf(dfn))
答案1
得分: 1
创建好你的函数之后（参见下面的我的示例玩具），你可以将其应用于你的帧列表：
apf <- function(d) {
d$sums = rowSums(d[, c(1,3,4,7)])
return(d)
}
dfs = lapply(list(a=a, b=b, c=c), apf)
英文:
Once you create your function (See my toy example below), you can apply it to a list of your frames:
apf <- function(d) {
d$sums = rowSums(d[, c(1,3,4,7)])
return(d)
}
dfs = lapply(list(a=a, b=b, c=c),apf)
答案2
得分: -1
在将数据框放入列表后：
df_list <- list(A = df, B = df, C = df)
使用任何代替 rowSums 的函数，该函数接受一个参数。
数据
df_list <- list(A = structure(list(a = c(4L, 4L, 3L), b = c(3L, 6L, 6L),
c = c(6L, 2L, 2L), d = c(3L, 5L, 6L), e = c(7L, 6L, 8L),
f = 3:1, g = c(7L, 1L, 5L)), class = "data.frame", row.names = c(NA,
-3L)), B = structure(list(a = c(4L, 4L, 3L), b = c(3L, 6L, 6L
), c = c(6L, 2L, 2L), d = c(3L, 5L, 6L), e = c(7L, 6L, 8L), f = 3:1,
g = c(7L, 1L, 5L)), class = "data.frame", row.names = c(NA,
-3L)), C = structure(list(a = c(4L, 4L, 3L), b = c(3L, 6L, 6L
), c = c(6L, 2L, 2L), d = c(3L, 5L, 6L), e = c(7L, 6L, 8L), f = 3:1,
g = c(7L, 1L, 5L)), class = "data.frame", row.names = c(NA,
-3L)))
英文:
After putting the data frames in a list
df_list <- list(A = df, B = df, C = df)
sapply(df_list, rowSums)
A B C
[1,] 33 33 33
[2,] 26 26 26
[3,] 31 31 31
Place any function instead rowSums that takes one argument.
Data
df_list <- list(A = structure(list(a = c(4L, 4L, 3L), b = c(3L, 6L, 6L),
c = c(6L, 2L, 2L), d = c(3L, 5L, 6L), e = c(7L, 6L, 8L),
f = 3:1, g = c(7L, 1L, 5L)), class = "data.frame", row.names = c(NA,
-3L)), B = structure(list(a = c(4L, 4L, 3L), b = c(3L, 6L, 6L
), c = c(6L, 2L, 2L), d = c(3L, 5L, 6L), e = c(7L, 6L, 8L), f = 3:1,
g = c(7L, 1L, 5L)), class = "data.frame", row.names = c(NA,
-3L)), C = structure(list(a = c(4L, 4L, 3L), b = c(3L, 6L, 6L
), c = c(6L, 2L, 2L), d = c(3L, 5L, 6L), e = c(7L, 6L, 8L), f = 3:1,
g = c(7L, 1L, 5L)), class = "data.frame", row.names = c(NA,
-3L)))
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