R函式來bind_rows另一個函式的結果

我嘗試為具有一個或多個列變數的自定義表撰寫一個函式。我對具有一個變數的表實作了這一點： .

現在我嘗試實作一個函式來獲取一組變數的自定義表，例如均值和多串列。我的問題是將它們系結在一起。

這就是我所擁有的：

library(tidyverse)

## at first some example data:

dv1 <- c(1, 0, 1, 0, 1) # dependent variable 1
dv2 <- c(1, 0, 1, 1, 1) # dependent variable 2
iv1 <- c("m", "f", "f", "m", "m") # independent variable 1
iv2 <- c(30, 40, 30, 40, 40) # independent variable 2
iv3 <- c("b", "c", "b", "a", "a") # ...

DATA <- data_frame(iv1, iv2, iv3, dv1, dv2) # build data frame

# the help function

cross_fun <- function(.data, DV, IV = IVs, fn = ~ mean(.x)) {

   df <- .data   %>% 
      select(all_of({{ IV }}), {{ DV }})  %>% 
    mutate(var = "dv") %>% # here I would like to have the {{ DV }} Argument as values of var, but mutate(var = {{ DV }}) or mutate (var = quote(DV)) does'nt work
    mutate(across(all_of({{IV}}), as.character)) # for using it in "names_from" in pivot_wider

  LIST <- list() # define a list

  for (i in 1:(ncol(df)-2)) { # -1 for the DV
    LIST[[i]] <- df %>% select(i, {{ DV }}, var)
  }
  dt <- purrr::map(
  .x = LIST,
  .f = ~ tidyr::pivot_wider(.x, names_from = 1, values_from = 2, values_fn = fn)
                  ) %>%
    purrr::reduce(left_join, by ="var")

  return(dt)
}

# What I can do
## simple custom table

DATA %>% cross_fun(dv1, IV = c('iv3', 'iv1', 'iv2'))  

## or I use a set (IVs is standard in cross_fun) in multiple tables 
IVs <- c('iv3', 'iv1', 'iv2')

DATA %>% cross_fun(dv2)

## I can change the Variables for the columns and the function
DATA %>% 
  cross_fun(dv2, IV = c('iv3', 'iv1', 'iv2'), fn = ~sum(.x))

## now I try to bind them together in a way, that I can use it later in another function
List_2 <- list()

## I could write it in a List_2 ...
List_2[[1]] <- DATA %>% cross_fun(dv1)

# ... for every variable ...
List_2[[2]] <- DATA %>% cross_fun(dv2)

# ... and bind the rows
List_2 %>% 
  bind_rows()

# here comes my Problem, it doesn't work in my try with for loop ...
for (i in c('dv1', 'dv2')) { 
  Liste2[[i]] <- DATA %>% 
    cross_fun(DATA[[i]])
}

# or with map
DATA %>% 
  map(.x = c(dv1:dv2), .f = ~cross_fun(.x)) %>% # the cross_fun-function for more than one dependent variable
  bind_rows() 

抱歉代碼亂七八糟。我是 R 函式的初學者。

問候，本

uj5u.com熱心網友回復：

解決方案

首先，一個答案：purrr::map_dfr自動行系結它的結果，你應該像這樣指定它：

map_dfr(c("dv1", "dv2"), cross_fun, .data = DATA)

# # A tibble: 2 x 8
#   var       b     c     a     m     f  `30`  `40`
#   <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 dv        1     0   0.5 0.667   0.5     1 0.333
# 2 dv        1     0   1   1       0.5     1 0.667

批判

您的代碼中有一些問題，其中一些與傳遞不正確的引數有關。例如，您的for回圈傳遞DATA[[i]]給DV而不是簡單地i；這會傳遞來自 的實際值向量DATA[["dv1"]]，而不僅僅是"dv1"函式所期望的名稱。

以下解決了這個問題：

List_2 <- list()

# `i` is confusing because the loop iterates over characters, not integers;
# use something like `varname` instead
for (varname in c('dv1', 'dv2')) { 
  List_2[[varname]] <- DATA %>% 
    cross_fun(varname)
}
bind_rows(List_2)

# # A tibble: 2 x 8
#   var       b     c     a     m     f  `30`  `40`
#   <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 dv        1     0   0.5 0.667   0.5     1 0.333
# 2 dv        1     0   1   1       0.5     1 0.667

你的map電話有兩個問題。一，你沒有傳遞DATA給cross_function(); 您只是傳遞.x給.data引數，而沒有傳遞給其他引數。第二，您試圖DV作為符號而不是字符傳遞。雖然這是可能的，但它很棘手（并且嘗試使用它來迭代符號map會使它變得更棘手），并且您的代碼沒有設定為正確處理它。

以下解決了這個問題：

map(c("dv1", "dv2"), .f = ~ cross_fun(DATA, .x)) %>% 
  bind_rows() 

# # A tibble: 2 x 8
#   var       b     c     a     m     f  `30`  `40`
#   <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 dv        1     0   0.5 0.667   0.5     1 0.333
# 2 dv        1     0   1   1       0.5     1 0.667

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