r错误中的错误（）`：！计算'年=年度（用户_ year）````''。由`as.posixlt.numeric（）`：：！ ＆＃x27; rigins＆＃x27;必须提供
我知道这个问题已经很多次，但是由于某种原因，我仍然会遇到这个错误，尽管我应该根据这个论坛上的答案来起作用。也许我错过了一步。
基本上，我想将user_year从num更改为 date，因此传单>传单 MAP LEGEND不显示一年使用逗号。
示例数据（df）：
structure(list(USER_Date = structure(c(18996, 18633, 19011, 19012,
19016, 19019, 19020, 19011, 19023, 19023), class = "Date"), USER_Year = c(2022,
2021, 2022, 2022, 2022, 2022, 2022, 2022, 2022, 2022), USER_Month = c(1,
1, 1, 1, 1, 1, 1, 1, 1, 1), USER_Day = c(4, 6, 19, 20, 24, 27,
28, 19, 31, 31)), sfc_columns = c("x", "y"), class = "data.frame", row.names = c(NA,
10L))
代码：
library(tidyverse)
# User lubridate
df = df %>% mutate(Year = year(USER_Year))
错误：
Error in `mutate()`:
! Problem while computing `Year = year(USER_Year)`.
Caused by error in `as.POSIXlt.numeric()`:
! 'origin' must be supplied
I know this question has been asked a lot of times but for some reason I still get this error even I though it should work based on the answers on this forum. Maybe I am missing a step.
Basically, I would like to change USER_Year from num to date so the leaflet map legend does not show the year with commas.
Sample Data (df):
structure(list(USER_Date = structure(c(18996, 18633, 19011, 19012,
19016, 19019, 19020, 19011, 19023, 19023), class = "Date"), USER_Year = c(2022,
2021, 2022, 2022, 2022, 2022, 2022, 2022, 2022, 2022), USER_Month = c(1,
1, 1, 1, 1, 1, 1, 1, 1, 1), USER_Day = c(4, 6, 19, 20, 24, 27,
28, 19, 31, 31)), sfc_columns = c("x", "y"), class = "data.frame", row.names = c(NA,
10L))
Code:
library(tidyverse)
# User lubridate
df = df %>% mutate(Year = year(USER_Year))
Error:
Error in `mutate()`:
! Problem while computing `Year = year(USER_Year)`.
Caused by error in `as.POSIXlt.numeric()`:
! 'origin' must be supplied
如果你对这篇内容有疑问，欢迎到本站社区发帖提问 参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。
绑定邮箱获取回复消息
由于您还没有绑定你的真实邮箱，如果其他用户或者作者回复了您的评论，将不能在第一时间通知您！
发布评论
评论（2）
我建议改用年份转换为角色。这应该与传单一起使用。
输出
I'd recommend converting the year to character instead. This should work with leaflet.
Output
约会总是有一年，月和一天。因此，您将必须选择这样的随机日子：
或者更好，使用实际的月和日值：
然后使用
格式（DF $年，“％y”）仅从“日期A date always has a year, month and day. So you will have to pick a random day like this:
or better, use the actual month and day values:
Then use
format(df$Year, "%Y")to only get the year from the date