如何使用 dplyr::mutate 改變兩個列表列

Question

我有以下 dataframe：

library(tidyverse)
dat <- structure(list(peptide_name = c(
  "foo", "foo", "foo",
  "foo", "foo", "foo", "bar", "bar", "bar",
  "bar", "bar", "bar"
), predicted = c(
  1, 0.965193935171986,
  1.002152924502, 1.13340754433401, 1.24280233366, 1.43442435500686,
  1, 1.07873571757982, 1.141383975916, 1.247359728244, 1.259245716526,
  1.23549751707385
), trueval = c(
  1, 1.174927114, 1.279883382, 1.752186589,
  1.994169096, 2.358600583, 1, 0.977742448, 1.305246423, 1.500794913,
  1.532591415, 1.197138315
)), row.names = c(NA, -12L), class = c(
  "tbl_df",
  "tbl", "data.frame"
))

dat

它看起來像這樣：

   peptide_name predicted trueval
   <chr>            <dbl>   <dbl>
 1 foo              1       1    
 2 foo              0.965   1.17 
 3 foo              1.00    1.28 
 4 foo              1.13    1.75 
 5 foo              1.24    1.99 
 6 foo              1.43    2.36 
 7 bar              1       1    
 8 bar              1.08    0.978
 9 bar              1.14    1.31 
10 bar              1.25    1.50 
11 bar              1.26    1.53 
12 bar              1.24    1.20 

每個foo和bar肽包含相同的行數。 我想要做的是在兩個肽之間執行 *Pearson correlation`。

以下代碼是我的嘗試：

dat %>%  
  group_by(peptide_name) %>% 
  # Here create list-columns
  nest() %>% 
  mutate(pn = row_number()) %>% 
  dplyr::select(pn, everything()) %>% 
  pivot_wider(-pn, names_from = peptide_name, values_from = data) %>% 
  # Attempt to calculate Pearson correlation
  mutate(pearson = cor(foo, bar, method = "pearson")) 

但它失敗了：

Error in `mutate()`:
! Problem while computing `pearson = cor(foo, bar, method =
  "pearson")`.
Caused by error in `cor()`:
! 'x' must be numeric

正確的做法是什么？

關聯的最終預期結果：

foo   bar  type
0.97 0.85  pearson_cor

Answer 1

問題似乎在於您如何將 arguments 傳遞給cor() function。我能夠使以下代碼正常工作：

 dat %>%  
  group_by(peptide_name) %>% 
  # Here create list-columns
  nest() %>% 
  mutate(pn = row_number()) %>% 
  dplyr::select(pn, everything()) %>% 
  pivot_wider(-pn, names_from = peptide_name, values_from = data) %>% 
  mutate(pearson_foo = cor(foo[[1]][[1]], foo[[1]][[2]], method = "pearson"),
         pearson_bar = cor(bar[[1]][[1]], bar[[1]][[2]], method = "pearson"))

但是，我很想知道是否有人對您的問題有更優雅的解決方案，因為我的解決方案涉及添加一個額外的列。 我會繼續玩弄它，看看我能不能想出更好的東西......

編輯：里奇對summarise()的回答要簡單得多！