英文:
error while filtering the date from data frame
问题
我有一个包含多个日期列的大数据框，我试图在日期列中过滤出今天的日期，但无法过滤，因为我收到了下面的错误。
由as.POSIXlt.character()中的错误引起：
!字符字符串不符合标准的明确格式
运行rlang::last_trace()以查看错误发生的位置。
以下是日期格式，列的类别显示为字符
dd <- data.frame(c1 =c("1/16/2023 1:26", "1/16/2023 1:26", "1/16/2023 1:40", "1/16/2023 2:05","1/16/2023 2:05", "1/16/2023 2:10"))dd %>% filter(as.date(c1) == lubridate::today())dd %>% filter(as.date(as.character(c1) == lubridate::today()))dd %>% filter(c1 == lubridate::today())
英文:
I have a big data frame with multiple date columns and i am trying filter date with todays date in date column but unable to filter as i was getting some error below.
Caused by error in as.POSIXlt.character():
! character string is not in a standard unambiguous format
Run rlang::last_trace() to see where the error occurred.
below is the date format and class of the column is showing as character
dd <- data.frame(c1 =c("1/16/2023 1:26", "1/16/2023 1:26", "1/16/2023 1:40", "1/16/2023 2:05","1/16/2023 2:05", "1/16/2023 2:10"))dd %>% filter(as.date(c1) == lubridate::today())dd %>% filter(as.date(as.character(c1) == lubridate::today()))dd %>% filter(c1 == lubridate::today())
答案1
得分: 2
以下是代码的翻译部分：
library(tidyverse)dd <-tibble(c1 = c("1/16/2023 1:26","1/16/2023 1:26","1/16/2023 1:40","1/16/2023 2:05","1/16/2023 2:05","1/16/2023 2:10","5/25/2023 2:10"))dd %>%mutate(across(c1, mdy_hm)) %>%filter(date(c1) == today())
希望这有所帮助。
英文:
library(tidyverse)dd <-tibble(c1 = c("1/16/2023 1:26","1/16/2023 1:26","1/16/2023 1:40","1/16/2023 2:05","1/16/2023 2:05","1/16/2023 2:10","5/25/2023 2:10"))dd %>%mutate(across(c1, mdy_hm)) %>%filter(date(c1) == today())# A tibble: 1 x 1c1<dttm>1 2023-05-25 02:10:00
答案2
得分: 2
你可以提供一个格式：
library(tidyverse)library(lubridate)dd <- data.frame(c1 =c("1/16/2023 1:26", "1/16/2023 1:26", "1/16/2023 1:40", "1/16/2023 2:05","1/16/2023 2:05", "1/16/2023 2:10"))dd %>% filter(as.Date(c1, format = "%m/%d/%Y") == as.Date("2023-01-16"))#> c1#> 1 1/16/2023 1:26#> 2 1/16/2023 1:26#> 3 1/16/2023 1:40#> 4 1/16/2023 2:05#> 5 1/16/2023 2:05#> 6 1/16/2023 2:10dd %>% filter(as.Date(c1, format = "%m/%d/%y") == lubridate::today())#> [1] c1#> <0 rows> (or 0-length row.names)
创建于2023-05-25，使用 reprex v2.0.2
顺便提一下，as.date() 不存在，请尝试使您的可重现示例可重现。
英文:
You can provide a format:
library(tidyverse)library(lubridate)dd <- data.frame(c1 =c("1/16/2023 1:26", "1/16/2023 1:26", "1/16/2023 1:40", "1/16/2023 2:05","1/16/2023 2:05", "1/16/2023 2:10"))dd %>% filter(as.Date(c1, format = "%m/%d/%Y") == as.Date("2023-01-16"))#> c1#> 1 1/16/2023 1:26#> 2 1/16/2023 1:26#> 3 1/16/2023 1:40#> 4 1/16/2023 2:05#> 5 1/16/2023 2:05#> 6 1/16/2023 2:10dd %>% filter(as.Date(c1, format = "%m/%d/%y") == lubridate::today())#> [1] c1#> <0 rows> (or 0-length row.names)
<sup>Created on 2023-05-25 with reprex v2.0.2</sup>
btw as.date() doesn't exist, try to make your reproducible examples reproducible.
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