使用stringdist查找多個單詞的匹配項

我有如下測驗資料。我正在嘗試為單詞向量查找（接近）匹配項，stringdist因為實際資料庫很大：

library(stringdist)
test_data <- structure(list(Province = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 
2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3), Year = c(2000, 
2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000, 
2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000, 2001, 2001, 
2001, 2002, 2002, 2002), Municipality = c("Some", "Anything", 
"Nothing", "Someth.", "Anything", "Not", "Something", "Anything", 
"None", "Some", "Anything", "Nothing", "Someth.", "Anything", 
"Not", "Something", "Anything", "None", "Some", "Anything", "Nothing", 
"Someth.", "Anything", "Not", "Something", "Anything", "None"
), `Other Values` = c(0.41, 0.42, 0.34, 0.47, 0.0600000000000001, 
0.8, 0.14, 0.15, 0.01, 0.41, 0.42, 0.34, 0.47, 0.0600000000000001, 
0.8, 0.14, 0.15, 0.01, 0.41, 0.42, 0.34, 0.47, 0.0600000000000001, 
0.8, 0.14, 0.15, 0.01)), row.names = c(NA, -27L), class = c("tbl_df", 
"tbl", "data.frame"))

# A tibble: 27 x 4
   Province  Year Municipality `Other Values`
      <dbl> <dbl> <chr>                 <dbl>
 1        1  2000 Some                 0.41  
 2        1  2000 Anything             0.42  
 3        1  2000 Nothing              0.34  
 4        1  2001 Someth.              0.47  
 5        1  2001 Anything             0.0600
 6        1  2001 Not                  0.8   
 7        1  2002 Something            0.14  
 8        1  2002 Anything             0.15  
 9        1  2002 None                 0.01  
10        2  2000 Some                 0.41  
# ... with 17 more rows

我試著跑：

test_match_out <- amatch(c("Anything","Something"),test_data[,3],maxDist=2)

編輯：

按照 zx8754 的評論，我嘗試了：

test_match_out <- amatch(c("Anything","Something"),test_data[[3]],maxDist=2)

和：

test_match_out <- amatch(c("Anything","Something"),test_data$Municipality,maxDist=2)

我的印象是前一行 ( amatch) 會給我一些類似于索引向量的東西，其中會有匹配。但它只是給了我一個帶有兩個NA值的向量。我誤解了什么amatch，還是語法有問題？

我想要得到values的這amatch是一個匹配，并且匹配的單詞。

期望的輸出：

test_data_2 <- structure(list(Province = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 
2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3), Year = c(2000, 
2000, 2000, 2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000, 
2001, 2001, 2001, 2002, 2002, 2002, 2000, 2000, 2000, 2001, 2001, 
2001, 2002, 2002, 2002), Municipality = c("Some", "Anything", 
"Nothing", "Someth.", "Anything", "Not", "Something", "Anything", 
"None", "Some", "Anything", "Nothing", "Someth.", "Anything", 
"Not", "Something", "Anything", "None", "Some", "Anything", "Nothing", 
"Someth.", "Anything", "Not", "Something", "Anything", "None"
), `Other Values` = c(0.41, 0.42, 0.34, 0.47, 0.0600000000000001, 
0.8, 0.14, 0.15, 0.01, 0.41, 0.42, 0.34, 0.47, 0.0600000000000001, 
0.8, 0.14, 0.15, 0.01, 0.41, 0.42, 0.34, 0.47, 0.0600000000000001, 
0.8, 0.14, 0.15, 0.01), `Matched Values` = c(NA, 0.42, NA, NA, 0.06000, 
NA, 0.14, 0.15, NA, NA, 0.42, NA, NA, 0.0600000000000001, 
NA, 0.14, 0.15, NA, NA, 0.42, NA, NA, 0.0600000000000001, 
NA, 0.14, 0.15, NA), `Matched Values` = c(NA, "Anything", NA, NA, "Anything", 
NA, "Something", "Anything", NA, NA, "Anything", NA, NA, "Anything", 
NA, "Something", "Anything", NA, NA, "Anything", NA, NA, "Anything", 
NA, "Something", "Anything", NA)), row.names = c(NA, -27L), class = c("tbl_df", 
"tbl", "data.frame"))

uj5u.com熱心網友回復：

獲取匹配的索引，然后更新所有匹配的行：

ix <- amatch(c("Anything","Something"), test_data[[ 3 ]], maxDist = 2)
# [1] 2 7

ifelse(test_data$Municipality %in% test_data$Municipality[ ix ], 
       test_data$`Other Values`, NA)
#  [1]   NA 0.42   NA   NA 0.06   NA 0.14 0.15   NA   NA 0.42
# [12]   NA   NA 0.06   NA 0.14 0.15   NA   NA 0.42   NA   NA
# [23] 0.06   NA 0.14 0.15   NA

標籤：r 细绳 模糊搜索 字符串分配器

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