Question

您好我有两个表（下面的table1和table2），并希望根据最接近的时间戳加入它们以形成expected_output。如果可能的话，涉及dplyr的某种解决方案会很好，但如果它进一步使事情变得复杂则不会。

table1 = 
structure(list(date = structure(c(1437051300, 1434773700, 1431457200
), class = c("POSIXct", "POSIXt"), tzone = ""), val1 = c(94L, 
33L, 53L)), .Names = c("date", "val1"), row.names = c(NA, -3L
), class = "data.frame")

table2 = 
structure(list(date = structure(c(1430248288, 1435690482, 1434050843
), class = c("POSIXct", "POSIXt"), tzone = ""), val2 = c(67L, 
90L, 18L)), .Names = c("date", "val2"), row.names = c(NA, -3L
), class = "data.frame")

expected_output = 
structure(list(date = structure(c(1437051300, 1434773700, 1431457200
), class = c("POSIXct", "POSIXt"), tzone = ""), val1 = c(94L,
33L, 53L), val2 = c(90L, 18L, 67L)), .Names = c("date", "val1", 
"val2"), row.names = c(NA, -3L), class = "data.frame")

Answer 1

使用data.table与roll = "nearest"的滚动连接功能：

require(data.table) # v1.9.6+
setDT(table1)[, val2 := setDT(table2)[table1, val2, on = "date", roll = "nearest"]]

此处，val2列是通过使用date选项对roll = "nearest"列执行加入来创建的。对于table1$date的每一行，计算table2$date中距离最近的匹配行，并提取相应行的val2。

Answer 2

这可能会很慢，但是......

d   <- function(x,y) abs(x-y) # define the distance function
idx <- sapply( table1$date, function(x) which.min( d(x,table2$date) )) # find matches

cbind(table1,table2[idx,-1,drop=FALSE])
#                  date val1 val2
# 2 2015-07-16 08:55:00   94   90
# 3 2015-06-20 00:15:00   33   18
# 1 2015-05-12 15:00:00   53   67

构建idx的另一种方法是max.col(-outer(table1$date, table2$date, d))。

根据最接近的时间戳在R中连接两个数据帧

2 个答案: