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这看起来很简单，但由于某种原因，我不理解agrep涉及替换的模糊匹配的行为。当指定时，两个替换会产生预期的匹配all=2，但不会在指定时产生匹配substitutions=2。为什么是这样？

# Finds a match as expected
agrep("abcdeX", "abcdef", value = T,
      max.distance = list(sub=1, ins=0, del=0))
#> [1] "abcdef"


# Doesn't find a match as expected
agrep("abcdXX", "abcdef", value = T,
      max.distance = list(sub=1, ins=0, del=0))
#> character(0)


# Finds a match as expected
agrep("abcdXX", "abcdef", value = T,
      max.distance = list(all=2))
#> [1] "abcdef"
      

# Doesn't find a match UNEXPECTEDLY
agrep("abcdXX", "abcdef", value = T,
      max.distance = list(sub=2, ins=0, del=0))
#> character(0)

由reprex 包于 2021-06-03 创建 (v2.0.0 )

4

1 回答 1

2

all是一个始终适用的上限，与其他max.distance控件无关（除了cost）。默认为 10%。

# one characters can change
agrep(pattern = "abcdXX", x = "abcdef", value = TRUE,
     max.distance = list(sub = 2, ins = 0, del = 0, all = 0.1))
# character(0)

# two characters can change
agrep(pattern = "abcdXX", x = "abcdef", value = TRUE,
     max.distance = list(sub = 2, ins = 0, del = 0, all = 0.2))
# [1] "abcdef"

# one character can change
agrep(pattern = "abcdXX", x = "abcdef", value = TRUE,
    max.distance = list(sub = 1, ins = 1, del = 0, all = 0.1))
# character(0)

# two characters can change
agrep(pattern = "abcdXX", x = "abcdef", value = TRUE,
    max.distance = list(sub = 1, ins = 1, del = 0, all = 0.2))
# [1] "abcdef"

设置all的小数模式在 1 处切换到整数模式有一点问题。

# 8 insertions allowed
agrep(pattern = "abcdXXef", x = "abcdef", value = TRUE,
    max.distance = list(sub = 0, ins = 2, del = 0, all = 1 - 1e-9))
# [1] "abcdef"

# 1 insertion allowed
agrep(pattern = "abcdXXef", x = "abcdef", value = TRUE,
    max.distance = list(sub = 0, ins = 2, del = 0, all = 1))
# character(0)

当您all通过将其设置为小于 1 来抑制时，将应用距离模式的限制。

# two substitutions allowed
agrep(pattern = "abcdXX", 
    x = c("abcdef", "abcXdef", "abcefg"), value = TRUE,
    max.distance = list(sub = 2, ins = 0, del = 0, all = 1 - 1e-9))
# [1] "abcdef"

设置成本的目的是允许您以不同的速率在不同的方向上在突变空间中移动。这将取决于您的用例。例如，某些语言方言可能更可能添加字母。您可能会选择让删除花费两次插入。默认情况下，所有的权重均相等costs = NULL，即costs = c(ins = 1, del = 1, sub = 1)。

编辑：关于您关于为什么某些模式匹配而其他模式不匹配的评论，10% 是指模式中的字符数，向上取整。

agrep(pattern = "01234567XX89", x = "0123456789", value = TRUE, 
    max.distance = list(sub = 0, ins = 2, del = 0))
# [1] "0123456789"
agrep(pattern = "01234567XX", x = "0123456789", value = TRUE, 
    max.distance = list(sub = 2, ins = 0, del = 0))
# character(0)
num_mutations <- nchar(c("01234567XX89", "01234567XX")) * 0.1
num_mutations
# [1] 1.2 1.0
ceiling(num_mutations)
[1] 2 1

第二个模式只有 10 个字符，所以只允许替换一个。

于 2021-06-03T21:30:16.807 回答