rowwise（）产生均值（）和sum（）的错误

Question

我正在尝试获取行中某些列的mean（）和sum（）。此代码将产生数据集：

library(tidyverse)

test_data <- tibble(part_id = 1:5,
                      a_1 = c("a", "b", "c", "d", "a"),
                      a_2 = c("b", NA, "b", "a", "d"),
                      a_3 = c("b", "b", "d", "d", "a"))


test_data <- test_data %>%
  mutate_at(vars(a_1, a_2), .funs = list(scored = ~case_when(
    . == "a" | . == "b" ~ 1,
    . == "c" ~ 0,
    . == "d" ~ -100)))

如果我尝试使用rowSums（）或rowMeans（），则会得到正确的答案：

library(tidyverse)

test_data <- test_data %>%
  mutate(a_total = rowSums(dplyr::select(., contains("scored")), na.rm = TRUE),
         a_mean = rowMeans(dplyr::select(., contains("scored")), na.rm = TRUE))

但是，如果尝试使用rowwise（）继之以sum（）或mean（），则将无法正常工作：

library(tidyverse)

test_data <- test_data %>%
  rowwise() %>%
  mutate(a_total = base::sum(dplyr::select(., contains("scored")), na.rm = TRUE),
         a_mean = base::mean(dplyr::select(., contains("scored")), na.rm = TRUE)) %>%
  ungroup()

对于sum（），它给出了总和，有效地忽略了rowwise（），而对于mean（），所有答案都是NA，我得到每一行的警告：

Warning messages:
1: In mean.default(dplyr::select(., contains("scored")), na.rm = TRUE) :
  argument is not numeric or logical: returning NA

我也尝试通过包括c（）函数来进行此修改，就像列出每个列一样。这导致以下错误：

library(tidyverse)

test_data <- test_data %>%
  rowwise() %>%
  mutate(a_total = base::sum(c(dplyr::select(., contains("scored"))), na.rm = TRUE),
         a_mean = base::mean(c(dplyr::select(., contains("scored"))), na.rm = TRUE)) %>%
  ungroup()

Error in base::sum(c(dplyr::select(., contains("scored"))), na.rm = TRUE) : 
  invalid 'type' (list) of argument

我如何使用rowwise（）进行这项工作？为什么这种行为与典型行为和rowSums（）或rowMeans（）如此不同？

我感谢任何见识！

Answer 1

问题是rowwise正在按行分组，sum，mean等在vector上起作用。它本质上适用于单行data.frame。通过用unlist换行，可以将其从data.frame转换为vector

library(dplyr)
test_data <- test_data %>%
                  rowwise() %>%
                  mutate(a_total = base::sum(unlist(dplyr::select(., 
                               contains("scored")), recursive = FALSE), na.rm = TRUE),
                         a_mean = base::mean(unlist(dplyr::select(., 
                               contains("scored")), recursive = FALSE), na.rm = TRUE)) %>%
                   ungroup()

---

或使用pmap

library(purrr)
test_data  %>%
   mutate(a_total = pmap_dbl(select(., contains("scored")),
                    ~ sum(c(...), na.rm = TRUE)),
          a_mean =  pmap_dbl(select(., contains("scored")),
                    ~ mean(c(...), na.rm = TRUE)))

Answer 2

如果您要坚持使用rowwise()来捕获要求和并求平均值的变量，这是另一种方法：

{rlang}

由

library(dplyr)

test_data <- tibble(part_id = 1:5,
                    a_1 = c("a", "b", "c", "d", "a"),
                    a_2 = c("b", NA, "b", "a", "d"),
                    a_3 = c("b", "b", "d", "d", "a"))


test_data <- test_data %>%
  mutate_at(vars(a_1, a_2), .funs = list(scored = ~case_when(
    . == "a" | . == "b" ~ 1,
    . == "c" ~ 0,
    . == "d" ~ -100)))


# Get the names of the variables you want
vars <- test_data %>% select(contains("scored")) %>% names()

# Use `rlang` so that `dplyr` will recognize the variable names
test_data %>%
  rowwise() %>%
  mutate(a_sum = sum(c(!!!rlang::syms(vars)), na.rm = TRUE),
         a_mean = mean(c(!!!rlang::syms(vars)), na.rm = TRUE)) %>% 
  ungroup()
#> # A tibble: 5 x 8
#>   part_id a_1   a_2   a_3   a_1_scored a_2_scored a_sum a_mean
#>     <int> <chr> <chr> <chr>      <dbl>      <dbl> <dbl>  <dbl>
#> 1       1 a     b     b              1          1     2    1  
#> 2       2 b     <NA>  b              1         NA     1    1  
#> 3       3 c     b     d              0          1     1    0.5
#> 4       4 d     a     d           -100          1   -99  -49.5
#> 5       5 a     d     a              1       -100   -99  -49.5

（v0.3.0）在2020-04-05创建