使用difftime时出现错误，“较长的对象长度不是较短的对象长度的倍数”

Question

我用下面的代码计算连续两行的秒数差

set.seed(79)

library(outbreaks)
library(lubridate)

# Import data
disease_df <- measles_hagelloch_1861[, 3, drop = FALSE]

# Generate a random time for each day
disease_df$time <- sample(1:86400, nrow(disease_df), replace = TRUE)
disease_df$time <- hms::as.hms(disease_df$time)

# Combine date and time
disease_df$time1 <- with(disease_df, ymd(date_of_prodrome) + hms(time))

# Sort data
disease_df <- disease_df[order(disease_df$time1), ]

# Difference in days of two consecutive row
disease_df$diff <- as.numeric(difftime(disease_df$date_of_prodrome,
                                       dplyr::lag(disease_df$date_of_prodrome, 1), units = 'days'))

# Difference in seconds of two consecutive row
disease_df$diff1 <- as.numeric(difftime(disease_df$time1,
                                       dplyr::lag(disease_df$time1, 1), units = 'secs'))

这里是结果数据框

和错误消息longer object length is not a multiple of shorter object length。

[能否请您解释为什么difftime可以在几天内正常工作，但会在几秒钟内导致错误？非常感谢！

Answer 1

time1列的类型为"POSIXlt"。我不确定如何将difftime与units = 'secs'一起使用不起作用，但是如果将其转换为POSIXct，则不会出现任何错误。

disease_df$time1 <- as.POSIXct(disease_df$time1)
disease_df$diff1 <- as.numeric(difftime(disease_df$time1,
                               dplyr::lag(disease_df$time1, 1), units = 'secs'))

Answer 2

显然dplyr不满意dplyr::lag(disease_df$time1, 1)，因为disease_df$time1的格式。

将其转换为POSIXct即可，因此只需更新代码的这一部分：

# Combine date and time and convert to POSIXct
disease_df$time1 <- as.POSIXct(with(disease_df, ymd(date_of_prodrome) + hms(time)))