R中具有数据框的每一行的最小值和最大值

Posted 2023-03-17

【中文标题】R中具有数据框的每一行的最小值和最大值

【英文标题】：Min and max for each row with dataframe in R

【发布时间】：2020-12-23 16:40:26

【问题描述】：

我需要计算几个参加在线测试的人的处理时间。因此，对于每个人来说，都有许多时间戳（每个任务一个时间戳）。处理的持续时间是根据最小和最大日期值之间的时间差来计算的。以下示例有效 (student_1)，但仅在没有缺失值时有效（student_2 和 student_3）。有什么想法吗？

library(anytime)

number <- c(1, 2, 3)
uniquename <- c("student_1", "student_2",  "student_3")
timestamp_1 <- c(anytime("2020-02-25T12:42:56.476Z"),NA,anytime("2020-02-25T10:05:22.388Z"))
timestamp_2 <- c(anytime("2020-02-25T12:51:22.388Z"),anytime("2020-02-25T12:51:22.388Z"),NA)
timestamp_3 <- c(anytime("2020-02-25T13:00:45.042Z"),anytime("2020-02-25T13:00:45.042Z"),NA)
timestamp_4 <- c(anytime("2020-02-25T13:31:48.073Z"),anytime("2020-02-25T13:31:48.073Z"),NA)
timestamp_5 <- c(anytime("2020-02-25T14:22:57.103Z"),anytime("2020-02-25T15:00:00Z"),anytime("2020-02-25T14:05:00Z"))

df3 <- data.frame(number,
                  uniquename,
                  timestamp_1,
                  timestamp_2,
                  timestamp_3,
                  timestamp_4,
                  timestamp_5)

df3$date_min <- apply(df3[3:7], 1, FUN=min)
df3$date_max <- apply(df3[3:7], 1, FUN=max)

df3$date_min <- anytime(df3$date_min)
df3$date_max <- anytime(df3$date_max)

df3$diff <- difftime(df3$date_min, df3$date_max, units = "mins")
df3$diff <- round(df3$diff,0)
df3$diff <- as.numeric(df3$diff)*(-1)

View(df3)

【参考方案1】：

这是data.table 方法

library( data.table )
setDT(df3)
#get columns with timestamps
cols = grep( "^timestamp_", names(df3), value = TRUE )
#check if at least two timestampa are given, then calculate min and max
df3[ df3[, rowSums( !is.na(.SD) ), .SDcols = cols ] >= 2, 
     date_min := do.call( pmin, c( .SD, list( na.rm = TRUE ) ) ), .SDcols = cols ]
df3[ df3[, rowSums( !is.na(.SD) ), .SDcols = cols ] >= 2, 
     date_max := do.call( pmax, c( .SD, list( na.rm = TRUE ) ) ), .SDcols = cols ]

#    number uniquename         timestamp_1         timestamp_2
# 1:      1  student_1 2020-02-25 12:42:56 2020-02-25 12:51:22
# 2:      2  student_2                <NA> 2020-02-25 12:51:22
# 3:      3  student_3 2020-02-25 10:05:22                <NA>
#            timestamp_3         timestamp_4         timestamp_5
# 1: 2020-02-25 13:00:45 2020-02-25 13:31:48 2020-02-25 14:22:57
# 2: 2020-02-25 13:00:45 2020-02-25 13:31:48 2020-02-25 15:00:00
# 3:                <NA>                <NA> 2020-02-25 14:05:00
#               date_min            date_max
# 1: 2020-02-25 12:42:56 2020-02-25 14:22:57
# 2: 2020-02-25 12:51:22 2020-02-25 15:00:00
# 3: 2020-02-25 10:05:22 2020-02-25 14:05:00

【讨论】：

已更新，每行至少检查两个不同的时间戳

【参考方案2】：

据我所知，您可以使用当前方法为min() 和max() 添加na.rm 参数：

df3$date_min <- apply(df3[3:7], 1, min, na.rm = TRUE)
df3$date_max <- apply(df3[3:7], 1, max, na.rm = TRUE)

df3[c("number", "uniquename", "date_min", "date_max")]
  number uniquename            date_min            date_max
1      1  student_1 2020-02-25 12:42:56 2020-02-25 14:22:57
2      2  student_2 2020-02-25 12:51:22 2020-02-25 15:00:00
3      3  student_3 2020-02-25 10:05:22 2020-02-25 14:05:00

【讨论】：

'na.rm = TRUE ' 是缺少的选项。现在考虑所有时间戳并忽略缺失值。谢谢。