使用 lubridate 和 dplyr 从 data.table 创建的持续时间错误

Question

我正在尝试聚合存储在 data.table 中的一些数据，然后根据聚合数据创建持续时间（来自 lubridate）。但是，当我尝试这样做时，我得到了一个错误。这是一个可重现的示例：

library(lubridate)
library(data.table)
library(dplyr)

data(lakers)
lakers.dt <- data.table(lakers, key = "player")

durations <- lakers.dt %>%
  mutate(better.date = ymd(date)) %>%
  group_by(player) %>%
  summarize(min.date = min(better.date), max.date = max(better.date)) %>%
  mutate(duration = interval(min.date, max.date))

# Source: local data table [371 x 4]
# 
# player   min.date   max.date
# 1                2008-10-28 2009-04-14
# 2   Aaron Brooks 2008-11-09 2009-04-03
# 3     Aaron Gray 2008-11-18 2008-11-18
# 4       Acie Law 2009-02-17 2009-02-17
# 5  Adam Morrison 2009-02-17 2009-04-12
# 6  Al Harrington 2008-12-16 2009-02-02
# 7     Al Horford 2009-02-17 2009-03-29
# 8   Al Jefferson 2008-12-14 2009-01-30
# 9    Al Thornton 2008-10-29 2009-04-05
# 10 Alando Tucker 2009-02-26 2009-02-26
# ..           ...        ...        ...
# Variables not shown: duration (dbl)
# Warning messages:
#   1: In unclass(e1) + unclass(e2) :
#   longer object length is not a multiple of shorter object length
# 2: In format.data.frame(df, justify = "left") :
#   corrupt data frame: columns will be truncated or padded with NAs

知道这个错误是什么意思，或者它来自哪里？

编辑：

当您忽略 dplyr 并在 data.table 中执行所有操作时，仍然会发生这种情况。这是我使用的代码：

lakers.dt[, better.date := ymd(date)]
durations <- lakers.dt[, list(min.date = min(better.date),
                              max.date = max(better.date)), by = player]
(durations[, duration := interval(min.date, max.date)])
# Error in `rownames<-`(`*tmp*`, value = paste(format(rn, right = TRUE),  : 
#   length of 'dimnames' [1] not equal to array extent
# In addition: Warning messages:
# 1: In unclass(e1) + unclass(e2) :
#   longer object length is not a multiple of shorter object length
# 2: In cbind(player = c("", "Aaron Brooks", "Aaron Gray", "Acie Law",  :
#   number of rows of result is not a multiple of vector length (arg 1)

Answer 1

您可以尝试将interval 输出转换为character 类（因为interval 输出不是vector）或用as.duration 包装（来自@Jake Fisher）

durations <- lakers.dt %>%
        mutate(better.date = ymd(date)) %>%
        group_by(player) %>%
        summarize(min.date = min(better.date), max.date = max(better.date)) %>%
        mutate(duration= as.duration(interval(min.date, max.date))
     )

或使用 as.vector 将其强制转换为 numeric 类。