两个数字向量上的全对全 setdiff，具有接受匹配的数字阈值 [英] All-to-all setdiff on two numeric vectors with a numeric threshold for accepting matches
问题描述
我想要做的或多或少是以下两个线程中讨论的问题的组合:
What I want to do is more or less a combination of the problems discussed in the two following threads:
我有两个数字向量:
b_1 <- c(543.4591, 489.36325, 12.03, 896.158, 1002.5698, 301.569)
b_2 <- c(22.12, 53, 12.02, 543.4891, 5666.31, 100.1, 896.131, 489.37)
我想将 b_1 中的 所有 元素与 b_2 中的所有元素进行比较，反之亦然.
I want to compare all elements in b_1 against all elements in b_2 and vice versa.
如果 b_1 中的 element_i 是 NOT 等于 范围 中的 any 数字> element_j ± 0.045 in b_2 那么element_i 必须上报.
If element_i in b_1 is NOT equal to any number in the range element_j ± 0.045 in b_2 then element_i must be reported.
同样，如果 b_2 中的 element_j 是 NOT 等于 范围内的 any 数 element_i ± 0.045 in b_1 则 element_j 必须上报.
Likewise, if element_j in b_2 is NOT equal to any number in the range element_i ± 0.045 in b_1 then element_j must be reported.
因此，基于上面提供的向量的示例答案将是:
Therefore, example answer based on the vectors provided above will be:
### based on threshold = 0.045
in_b1_not_in_b2 <- c(1002.5698, 301.569)
in_b2_not_in_b1 <- c(22.12, 53, 5666.31, 100.1)
是否有 R 函数可以做到这一点?
Is there an R function that would do this?
推荐答案
如果你乐于使用非 base 包，data.table::inrange 是一个方便的功能.
If you are happy to use a non-base package, data.table::inrange is a convenient function.
x1[!inrange(x1, x2 - 0.045, x2 + 0.045)]
# [1] 1002.570 301.569
x2[!inrange(x2, x1 - 0.045, x1 + 0.045)]
# [1] 22.12 53.00 5666.31 100.10
<小时>
inrange 在更大的数据集上也很有效.例如在1e5 向量，inrange 是 >比其他两种替代方案快 700 倍:
inrange is also efficient on larger data sets. On e.g. 1e5 vectors, inrange is > 700 times faster than the two other alternatives:
n <- 1e5
b1 <- runif(n, 0, 10000)
b2 <- b1 + runif(n, -1, 1)
microbenchmark(
f1 = f(b1, b2, 0.045, 5000),
f2 = list(in_b1_not_in_b2 = b1[sapply(b1, function(x) !any(abs(x - b2) <= 0.045))],
in_b2_not_in_b1 = b2[sapply(b2, function(x) !any(abs(x - b1) <= 0.045))]),
f3 = list(in_b1_not_in_b2 = b1[!inrange(b1, b2 - 0.045, b2 + 0.045)],
in_b2_not_in_b1 = b2[!inrange(b2, b1 - 0.045, b1 + 0.045)]),
unit = "relative", times = 10)
# Unit: relative
# expr min lq mean median uq max neval
# f1 1976.931 1481.324 1269.393 1103.567 1173.3017 1060.2435 10
# f2 1347.114 1027.682 858.908 766.773 754.7606 700.0702 10
# f3 1.000 1.000 1.000 1.000 1.0000 1.0000 10
<小时>
是的，它们给出了相同的结果:
And yes, they give the same result:
n <- 100
b1 <- runif(n, 0, 10000)
b2 <- b1 + runif(n, -1, 1)
all.equal(f(b1, b2, 0.045, 5000),
list(in_b1_not_in_b2 = b1[sapply(b1, function(x) !any(abs(x - b2) <= 0.045))],
in_b2_not_in_b1 = b2[sapply(b2, function(x) !any(abs(x - b1) <= 0.045))]))
# TRUE
all.equal(f(b1, b2, 0.045, 5000),
list(in_b1_not_in_b2 = b1[!inrange(b1, b2 - 0.045, b2 + 0.045)],
in_b2_not_in_b1 = b2[!inrange(b2, b1 - 0.045, b1 + 0.045)]))
# TRUE
<小时>
当 搜索 inrange 时，有几个相关的、可能有用的答案所以.
Several related, potentially useful answers when searching for inrange on SO.
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