我以前也问过类似的问题。我的问题比上一个问题复杂一点。对于我的问题,y参数不是固定的。
在函数(X,Y({SOME function}中,X是字符列表,Y是数据帧列表。基本上,我希望函数分别按顺序处理X和Y,并将输出作为一个列表。例如,X列表的第一元素和Y列表的第一个元素、X列表的第二元素和Y名单的第二个元素、Y列表的第三个元素、,。。。
X,Y 示例
X <- c("1", "2")
y1 <- data.frame("person.1" = "Amy", "bestfood..1" = "fish", "bestthing..1" = "book",
"person.2" = "Mike", "bestfood..2" = "fish", "bestthing..2" = "book")
y2 <- data.frame("person.1" = "Amy","bestfood..1" = "carrot", "bestthing..1" = "cloth",
"person.2" = "Mike","bestfood..2" = "carrot", "bestthing..2" = "cloth")
Y <- list(y1,y2)
功能:
addID <- function(X, Y) {
rowlength <- length(Y)
df <- as.data.frame(matrix(NA, nrow = rowlength, ncol = 3))
colnames(df) <- c("ID", "Person", "Food")
df[1:nrow(df), 1] <- X
# name
namecols <-grep("person",colnames(Y))
for (i in 1:length(namecols)) {
name <- Y[1, namecols[i]]
df[i, 2] <- as.character(name)
}
# food
foodcols <-
grep("bestfood",colnames(Y))
for (i in 1:length(foodcols)) {
food <- Y[1, foodcols[i]]
df[i, 3] <- as.character(foodcols)
}
return(df)
}
}
我试着使用lapply,但不知道如何将X列表包括在内。当我尝试这个:
lapply(Y, function, X=X)
该功能无法正常工作。我想知道是否有其他方法可以将X包含在其中(我在单个字符和数据帧上尝试了这个函数,它运行得很好。(
我希望这是清楚的。如果没有,请指出,我会尽力澄清。提前谢谢。
更新:
我按照评论的建议尝试了Map。它返回:不正确的维度数。我在函数中添加了一些细节。似乎R卡在了最后一行。
outcome <- Map(addID, Y, X)
我得到
error in Y[1, namecols[i]] : incorrect number of dimensions
In addition: Warning message:
In `[<-.data.frame`(`*tmp*`, 1:nrow(df), 1, value = list(person.1 = 1L, :
provided 6 variables to replace 1 variables
结果应该是:
z1 <- data.frame(ID = c(1,2), Person = c("Amy","Mike"), Food = c("fish", "fish"))
z2 <- data.frame(ID = c(1,2), Person = c("Amy","Mike"), Food = c("carrot", "carrot"))
outcome <- list(z1,z2)
我们可以在tidyverse
中轻松做到这一点
library(dplyr)
library(tidyr)
bind_rows(Y, .id = 'ID') %>%
select(ID, starts_with('person'), contains('food')) %>%
pivot_longer(cols = -ID, names_to = c(".value"),
names_pattern = "([^.]+)\.+\d+")
-输出
# A tibble: 4 x 3
ID person bestfood
<chr> <chr> <chr>
1 1 Amy fish
2 1 Mike fish
3 2 Amy carrot
4 2 Mike carrot
使用OP的功能,如果我们修改,它将工作
addID <- function(X, Y) {
rowlength <- length(Y)
df <- as.data.frame(matrix(NA, nrow = rowlength, ncol = 3))
colnames(df) <- c("ID", "Person", "Food")
df[1:nrow(df), 1] <- X
namecols <- grep("person",colnames(Y))
df[, 2] <- unlist(Y[namecols])
foodcols <- grep("bestfood", colnames(Y))
df[,3] <- unlist(Y[foodcols])
return(unique(df))
}
-测试
Map(addID, X, Y)
$`1`
ID Person Food
1 1 Amy fish
2 1 Mike fish
$`2`
ID Person Food
1 2 Amy carrot
2 2 Mike carrot