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我正在尝试左连接两个数据框（df1，df2）。数据框共有两列：区域和坡度。区域是一个因子列，斜率是数字。

    df1 = data.frame(slope = c(1:6), zone = c(rep("Low", 3), rep("High", 3)))
    df2 = data.frame(slope = c(2.4, 2.4,6.2), zone = c(rep("Low", 1), rep("High", 2)), other = c(rep("a", 1), rep("b", 1), rep("c", 1)))
    df1
    df2

我想加入数据框，以便它们首先在区域上完全匹配，然后是坡度最接近的匹配。如果有两个等距的斜率值，只要一致地应用规则并且不会导致重复行，连接向上还是向下舍入都没有关系。

我更喜欢用fuzzy_join 或dplyr 而不是data.table 来做到这一点。

结果应该类似于：

    df3 = data.frame(slope = c(1:6), zone = c(rep("Low", 3), rep("High", 3)), other = c(rep("a", 3), rep("b",1), rep("c",2)))
    df3

其中“其他”的值首先由区域确定，然后是最近的坡度。

我试过了：

    distance_left_join(df, df2, by=c("zone"= "zone", "slope"="slope"))

以及其他类型的模糊连接，但我认为它们可能无法正常工作，因为列的类型不同。我怀疑有一个fuzzy_left_join 解决方案，但我不明白如何创建一个匹配函数。

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1 回答 1

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以下是如何使用多个 match_fun 进行模糊连接。如果你想混合复杂的 match_fun，你必须自己定义一个函数，就像我在这里所做的那样：Passing arguments into multiple match_fun functions in R blurjoin::fuzzy_join

df1 = data.frame(slope = c(1:6), zone = c(rep("Low", 3), rep("High", 3)))
df2 = data.frame(slope = c(2.4, 2.4,6.2), zone = c(rep("Low", 1), rep("High", 2)), other = c(rep("a", 1), rep("b", 1), rep("c", 1)))

library(fuzzyjoin); library(dplyr)

# First, need to define match_fun_distance. 
# This is copied from the source code for distance_join in https://github.com/dgrtwo/fuzzyjoin
match_fun_distance <- function(v1, v2) {
  
  # settings for this method
  method = "manhattan"
  max_dist = 99
  distance_col = "dist"
  
  if (is.null(dim(v1))) {
    v1 <- t(t(v1))
    v2 <- t(t(v2))
  }
  if (method == "euclidean") {
    d <- sqrt(rowSums((v1 - v2)^2))
  }
  else if (method == "manhattan") {
    d <- rowSums(abs(v1 - v2))
  }
  ret <- tibble::tibble(instance = d <= max_dist)
  if (!is.null(distance_col)) {
    ret[[distance_col]] <- d
  }
  ret
}

(joined_result <- fuzzy_join(df1, df2, 
                             by=c("zone"= "zone", "slope"="slope"), 
                             match_fun = list(`==`, match_fun_distance),
                             mode = "left"))
#>   slope.x zone.x slope.y zone.y other slope.dist zone.dist
#> 1       1    Low     2.4    Low     a        1.4        NA
#> 2       2    Low     2.4    Low     a        0.4        NA
#> 3       3    Low     2.4    Low     a        0.6        NA
#> 4       4   High     2.4   High     b        1.6        NA
#> 5       4   High     6.2   High     c        2.2        NA
#> 6       5   High     2.4   High     b        2.6        NA
#> 7       5   High     6.2   High     c        1.2        NA
#> 8       6   High     2.4   High     b        3.6        NA
#> 9       6   High     6.2   High     c        0.2        NA

joined_result %>%
  group_by(slope.x, zone.x) %>%
  top_n(1, -slope.dist)
#> # A tibble: 6 x 7
#> # Groups:   slope.x, zone.x [6]
#>   slope.x zone.x slope.y zone.y other slope.dist zone.dist
#>     <int> <fct>    <dbl> <fct>  <fct>      <dbl>     <dbl>
#> 1       1 Low        2.4 Low    a          1.4          NA
#> 2       2 Low        2.4 Low    a          0.400        NA
#> 3       3 Low        2.4 Low    a          0.6          NA
#> 4       4 High       2.4 High   b          1.6          NA
#> 5       5 High       6.2 High   c          1.2          NA
#> 6       6 High       6.2 High   c          0.2          NA

由reprex 包于 2020-10-20 创建(v0.3.0)

于 2020-10-20T06:46:05.783 回答