问题
I have two lists with different structure:
listA <- list(c("a","b","c"), c("d","e"))
listB <- list(0.05, 0.5)
listA
[[1]]
[1] "a" "b" "c"
[[2]]
[1] "d" "e"
listB
[[1]]
[1] 0.05
[[2]]
[1] 0.5
I have an idea of how to use looping to combine both lists in a dataframe that looks like the one below but I'm sure that there is a more efficient way of doing this.
data.frame(A = c("a","b","c","d","e"), B = c(rep(0.05,3), rep(0.5,2)))
A B
1 a 0.05
2 b 0.05
3 c 0.05
4 d 0.50
5 e 0.50
回答1:
This is another option:
do.call(rbind, Map(data.frame, A=listA, B=listB))
# A B
# 1 a 0.05
# 2 b 0.05
# 3 c 0.05
# 4 d 0.50
# 5 e 0.50
回答2:
Maybe there is a more elegant way that keeps the class numeric of list2's elements... But this one works as well
df <- do.call(rbind,mapply(cbind, listA, listB))
df <- as.data.frame(df, stringsAsFactors = FALSE)
df[,2] <- as.numeric(df[,2])
EDIT Way better is Matthew Plourde's solution using Map aka mapply(data.frame, A=listA, B=listB, SIMPLIFY = FALSE)
回答3:
I'd prefer this:
do.call(rbind,
Map(function(...) setNames(cbind.data.frame(...),
c("A", "B")),
listA, listB))
# A B
#1 a 0.05
#2 b 0.05
#3 c 0.05
#4 d 0.50
#5 e 0.50
回答4:
Here is another way:
do.call(rbind,
lapply(1:length(listA),
function(i)
data.frame(A=unlist(listA[i]),
B=unlist(listB[i]))))
回答5:
If looking for a tidyverse solution, here is the analogue to the accepted answer. Using the dfr suffix to the map function family enables a very simple solution which should also be faster than do.call("rbind").
library(tidyverse)
listA <- list(c("a","b","c"), c("d","e"))
listB <- list(0.05, 0.5)
map2_dfr(listA, listB, ~ tibble(A = .x, B = .y))
#> # A tibble: 5 x 2
#> A B
#> <chr> <dbl>
#> 1 a 0.05
#> 2 b 0.05
#> 3 c 0.05
#> 4 d 0.5
#> 5 e 0.5
Created on 2019-02-12 by the reprex package (v0.2.1)
回答6:
Another way without using do.call:
cbind(data.frame(listA), data.frame(listB))
来源:https://stackoverflow.com/questions/28630024/combine-two-lists-in-a-dataframe-in-r