運行cor()時找不到物件

我試圖從 R 中的組合資料集中找到兩列（sunshine_in_hours 和 AgeGroup_30_to_34）之間的相關性。但是，每次我嘗試運行 cor() 函式時，我都會得到這個錯誤：

Error in pmatch(use, c("all.obs", "complete.obs", "pairwise.complete.obs",  : 
  object 'AgeGroup_30_to_34' not found

這是 dput(head) 代碼片段：

structure(list(Date = structure(c(18659, 18660, 18661, 18663, 
18665, 18666, 18667, 18668, 18669, 18670, 18671, 18673, 18674, 
18675, 18676, 18677, 18678, 18679, 18680, 18681, 18682, 18683, 
18684, 18685, 18686, 18687, 18688, 18689, 18690, 18691), class = "Date"), 
    Year = c(2021, 2021, 2021, 2021, 2021, 2021, 2021, 2021, 
    2021, 2021, 2021, 2021, 2021, 2021, 2021, 2021, 2021, 2021, 
    2021, 2021, 2021, 2021, 2021, 2021, 2021, 2021, 2021, 2021, 
    2021, 2021), Month = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
    2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3), AgeGroup_30_to_34 = c(0, 
    0, 0, 2, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 
    2, 0, 0, 1, 2, 0, 3, 0, 0, 0), Sunshine_in_hours = c(1.6, 
    3.4, 13.1, 8.9, 2, 1.7, 12.7, 11.6, 5.5, 5.6, 4.9, 9.2, 8.3, 
    11.9, 12.4, 12.4, 5.9, 0, 6.3, 8.5, 9.9, 8.7, 6.3, 1, 9.2, 
    6.3, 1.4, 2.1, 2.6, 3.6), City = c("Melbourne", "Melbourne", 
    "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", 
    "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", 
    "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", 
    "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", 
    "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", 
    "Melbourne", "Melbourne", "Melbourne")), row.names = c(NA, 
-30L), class = c("tbl_df", "tbl", "data.frame"))

我試圖運行代碼：

Combined <- inner_join(covidS, weatherS, by = 'Date')%>%
  mutate(Date = mdy(Date),
         Year = year(Date),
         Month = month(Date),
         Day = day(Date))%>%
  select(Date, Year, Month, AgeGroup_30_to_34, Sunshine_in_hours, City)%>%
  filter(City == 'Melbourne')%>%
  cor(Sunshine_in_hours, AgeGroup_30_to_34 )

我已經嘗試查找有關如何執行此操作的教程，但是我一直碰壁。任何幫助將不勝感激。

uj5u.com熱心網友回復：

cor需要兩個輸入，你給它 3，其中兩個它不理解。嘗試這個：

Combined <- inner_join(covidS, weatherS, by = 'Date')%>%
  mutate(Date = mdy(Date),
         Year = year(Date),
         Month = month(Date),
         Day = day(Date))%>%
  select(Date, Year, Month, AgeGroup_30_to_34, Sunshine_in_hours, City)%>%
  filter(City == 'Melbourne') 

corr = cor(Combined$Sunshine_in_hours, Combined$AgeGroup_30_to_34 )

請記住，當您使用管道時，您將最后一個物件作為您正在呼叫的函式的第一個引數提供。在這種情況下，您的代碼相當于：

cor(inner_join(covidS, weatherS, by = 'Date')%>%
  mutate(Date = mdy(Date),
         Year = year(Date),
         Month = month(Date),
         Day = day(Date))%>%
  select(Date, Year, Month, AgeGroup_30_to_34, Sunshine_in_hours, City)%>%
  filter(City == 'Melbourne'),
Sunshine_in_hours, AgeGroup_30_to_34 )

Sunshine_in_hours因此，AgeGroup_30_to_34如果函式不知道這些是來自另一個資料幀的列，那么兩者都沒有任何意義。問題是，這個函式是為基本 R 編碼的，而你的其余編程是dplyr，它們是不同的范例。如有疑問，請務必檢查檔案

uj5u.com熱心網友回復：

使用magrittrexposition 管道%$%代替%>%您可以執行以下操作：

library(magrittr)

dat %$%
  cor(Sunshine_in_hours, AgeGroup_30_to_34)
#> [1] -0.0006941058

標籤：r

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