问题描述
请考虑以下小标题和以下矢量:
consider the following tibble and the follwing vector:
library(tidyverse)
a <- tibble(val1 = 10:15, val2 = 20:25)
params <- 1:3
我还有一个函数 myfun
,该函数将任意长度的向量和一个整数作为输入,并返回相同长度的向量。出于演示目的,您可以想到
Also I have a function myfun
which takes a vector of arbitrary length and an integer as input and returns an vector of the same length. For demonstration purposes you can think of
myfun <- function(x, k) dplyr::lag(x, k)
我想创建以下对象:对于 a $中的每一列c $ c>并为
params
中的每个元素创建一个由 myfun(col,params [i])$给定的新列c $ c>。
在上面的玩具示例中,例如可以这样实现:
I want to create the follwing: for each column in a
and for each element in params
I want to create a new column given by myfun(col, params[i])
.In the toy example above this could for example be achieved like this:
a %>% mutate_at(1:2, funs(run1 = myfun), k = params[1]) %>%
mutate_at(1:2, funs(run2 = myfun), k = params[2]) %>%
mutate_at(1:2, funs(run3 = myfun), k = params[3])
是否有更优雅的方法可以做到这一点?如果参数很长,那么此解决方案将变得不可行。当然,可以使用for循环来做到这一点,但我认为tidyverse中可能有解决方案(也许使用 purrr :: map
?)
Is there a more elegant approach to do this? If params is very long then this solution becomes infeasible. Of course one could do this with a for loop, but I thought that there might be a solution within the tidyverse (maybe using purrr::map
?)
谢谢!
推荐答案
以下是使用tidyverse的解决方案:
Here is a solution using tidyverse:
library(tidyverse)
a <- tibble(val1 = 10:15, val2 = 20:25)
params <- 1:3
#set the column names, add leading zeroes based om max(params)
run_names <- paste0("run", formatC(params, width = nchar(max(params)), flag = "0"))
#what functions to perform
lag_functions <- setNames(paste("dplyr::lag( ., ", params, ")"), run_names)
#perfporm functions
a %>% mutate_at(vars(1:2), funs_(lag_functions ))
# # A tibble: 6 x 8
# val1 val2 val1_run1 val2_run1 val1_run2 val2_run2 val1_run3 val2_run3
# <int> <int> <int> <int> <int> <int> <int> <int>
# 1 10 20 NA NA NA NA NA NA
# 2 11 21 10 20 NA NA NA NA
# 3 12 22 11 21 10 20 NA NA
# 4 13 23 12 22 11 21 10 20
# 5 14 24 13 23 12 22 11 21
# 6 15 25 14 24 13 23 12 22
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