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问如何将标题中的“日期”列格式化为日期
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Stack Overflow用户

提问于 2022-08-06 14:17:30

回答 3查看 44关注 0票数 0

我已经导入了一个excel电子表格到R和数据框架有许多列与日期在标题中。我可以将命名列格式化为aa日期如下：

df$date <- as.Date(as.numeric(df$date), origin = "1899-12-30")

对于标题中包含“日期”的所有列，我将如何做到这一点？这里有一个数据框架示例，尽管它没有真正的列那么多列。理想情况下，答案应该使用dplyr。

df <- structure(list(source = c("Track", "Track", "Track", "Track", 
"Track"), sample_type = c("SQC", "DNA", "PBMC", "PBMC", "PBMC"
), collection_date = c("39646", "39654", "39643", "39644", "40389"
), collection_date2 = c("39646", "39654", "39643", "39644", "40389"
), received_date = c("39651", "39660", "39685", "39685", "40421"
), storage_date = c("39653", "39744", "39685", "39685", "40421"
)), row.names = c(NA, -5L), class = c("tbl_df", "tbl", "data.frame"
))

dplyr

回答 3

Stack Overflow用户

回答已采纳

发布于 2022-08-06 14:26:12

我们可以使用across和contains来选择包含字符串"date“的所有变量。

library(tidyverse)
df <- structure(list(source = c("Track", "Track", "Track", "Track", 
                                "Track"), sample_type = c("SQC", "DNA", "PBMC", "PBMC", "PBMC"
                                ), collection_date = c("39646", "39654", "39643", "39644", "40389"
                                ), collection_date2 = c("39646", "39654", "39643", "39644", "40389"
                                ), received_date = c("39651", "39660", "39685", "39685", "40421"
                                ), storage_date = c("39653", "39744", "39685", "39685", "40421"
                                )), row.names = c(NA, -5L), class = c("tbl_df", "tbl", "data.frame"
                                ))

df <- df %>%
  mutate(across(contains("date"), ~as.Date(as.numeric(.x), origin = "1899-12-30")))

head(df)
#> # A tibble: 5 × 6
#>   source sample_type collection_date collection_date2 received_date storage_date
#>   <chr>  <chr>       <date>          <date>           <date>        <date>      
#> 1 Track  SQC         2008-07-17      2008-07-17       2008-07-22    2008-07-24  
#> 2 Track  DNA         2008-07-25      2008-07-25       2008-07-31    2008-10-23  
#> 3 Track  PBMC        2008-07-14      2008-07-14       2008-08-25    2008-08-25  
#> 4 Track  PBMC        2008-07-15      2008-07-15       2008-08-25    2008-08-25  
#> 5 Track  PBMC        2010-07-30      2010-07-30       2010-08-31    2010-08-31

票数 1

Stack Overflow用户

发布于 2022-08-06 14:32:01

以下是另一种方法。janitor包对此excel_numeric_to_date有自己的功能。

library(dplyr)
library(janitor)
df %>% 
  mutate(across(contains("date"), ~excel_numeric_to_date(as.numeric(.))))

  source sample_type collection_date collection_date2 received_date storage_date
  <chr>  <chr>       <date>          <date>           <date>        <date>      
1 Track  SQC         2008-07-17      2008-07-17       2008-07-22    2008-07-24  
2 Track  DNA         2008-07-25      2008-07-25       2008-07-31    2008-10-23  
3 Track  PBMC        2008-07-14      2008-07-14       2008-08-25    2008-08-25  
4 Track  PBMC        2008-07-15      2008-07-15       2008-08-25    2008-08-25  
5 Track  PBMC        2010-07-30      2010-07-30       2010-08-31    2010-08-31

票数 2

Stack Overflow用户

发布于 2022-08-06 14:25:48

为此，我将使用一个循环：

for (col in grep('date', names(df))) {
  df[[col]] <- as.Date(as.numeric(df[[col]]), origin="1899-12-30")
}

票数 0

页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持

原文链接：

https://stackoverflow.com/questions/73260540

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