r - 对两个数据帧中的两列进行条件 setdiff(全部到全部)，并具有用于进行匹配的数字范围

Question

以下是两个示例数据框:

df1 <- data.frame(Time1v1 = c(55.25, 59.36, 40.26, 786.008, 980.569, 11.2, 10.11, 23.11),
                  Time2v1 = c(81, 12, 13, 11.0112, 93.9, 14.8, 15.3, 78.91))

df2 <- data.frame(Time1v2 = c(10.13, 980.659, 14.42, 90.1, 40.3298, 9234, 59.35),
                  Time2v2 = c(25.1, 88.9, 120, 911, 22.1253, 81, 15.1))


> df1
  Time1v1 Time2v1
1  55.250 81.0000
2  59.360 12.0000
3  40.260 13.0000
4 786.008 11.0112
5 980.569 93.9000
6  11.200 14.8000
7  10.110 15.3000
8  23.110 78.9100

> df2
    Time1v2 Time2v2
1   10.1300   25.1000
2  980.6590   88.9000
3   14.4200  120.0000
4   90.1000  911.0000
5   40.3298   22.1253
6 9234.0000   81.0000
7   59.3500   15.1000

我想比较 df1 的每一行每一行 df2 。如果 Time1 之间的差异df1 和 df2 的范围为 [-0.1,+0.1] AND 差异 Time2范围为 [-10,+10]那么必须删除 df1 中的该特定行。

---

尝试解决

这里尝试解决这个问题。有更好的办法吗？

df1$remove <- rep("No", nrow(df1))
for(i in 1:nrow(df1)){
    for(j in 1:nrow(df2)){
        if(data.table::inrange(df1$Time1v1[i], df2$Time1v2[j] - 0.1, df2$Time1v2[j] + 0.1) && data.table::inrange(df1$Time2v1[i], df2$Time2v2[j] - 10, df2$Time2v2[j] + 10)) {df1$remove[i] <- "remove"}
    }
}

这给了我:

> df1
      Time1v1 Time2v1 remove
    1  55.250 81.0000     No
    2  59.360 12.0000 remove
    3  40.260 13.0000 remove
    4 786.008 11.0112     No
    5 980.569 93.9000 remove
    6  11.200 14.8000     No
    7  10.110 15.3000 remove
    8  23.110 78.9100     No

---

预期最终结果

最终预期的输出将是:

> df1[which(df1$remove != "remove"),-3]

  Time1v1 Time2v1
1  55.250 81.0000
4 786.008 11.0112
6  11.200 14.8000
8  23.110 78.9100

---

相关

Perform non-pairwise all-to-all comparisons between two unordered character vectors --- The opposite of intersect --- all-to-all setdiff

All-to-all setdiff on two numeric vectors with a numeric threshold for accepting matches

Answer 1

这是一个手动(手动声明列)方法来执行此操作，

 m1 <- outer(df1$Time1v1, df2$Time1v2, `-`)
 m2 <- outer(df1$Time2v1, df2$Timev2, `-`)

i1 <- intersect(which(m1 >= -0.1 & m1 <= 0.1, arr.ind = TRUE)[,1], 
                which(m2 >= -10 & m2 <= 10, arr.ind = TRUE)[,1])
df1[-i1,]

#  Time1v1 Time2v1
#1  55.250 81.0000
#4 786.008 11.0112
#6  11.200 14.8000
#8  23.110 78.9100