r - 如何将地址(格式不均匀)拆分为R中的各个字段

Question

我想将这些地址分成各自的类别(街道号码、街道名称、城市、州和邮政编码)以最终检查哪些是相同的。任何人都可以帮助了解如何在 R 中执行此操作的基本想法吗？

    Company                          Address

 1. A                    1 NE 1 Street Miami,FL 33132
 2. B                     1 1st Street Miami,FL 33132
 3. C                      1 NE 1st St Miami,FL 33132
 4. D                     1 1st Street Miami,FL 33134
 5. E               100 Biscayne Blvd. Miami,FL 33132
 6. F               100 Biscayne Blvd Miami ,FL 33132
 7. G 100 Biscayne Boulevard Suite 604 Miami,FL 33132
 8. H     100 Biscayne Blvd. Suite 604 Miami,FL 33132
 9. I            100 N. Biscayne Blvd. Miami,FL 33132

Answer 1

尝试 gsubfn 包中的 read.pattern 。如果行位于文件中，则将 text = Lines 替换为给出文件名的字符串。这可能相当脆弱，一旦您有更多数据可供尝试，您可能需要稍微调整正则表达式。

Lines <- "Company                          Address
 1. A                    1 NE 1 Street Miami,FL 33132
 2. B                     1 1st Street Miami,FL 33132
 3. C                      1 NE 1st St Miami,FL 33132
 4. D                     1 1st Street Miami,FL 33134
 5. E               100 Biscayne Blvd. Miami,FL 33132
 6. F               100 Biscayne Blvd Miami ,FL 33132
 7. G 100 Biscayne Boulevard Suite 604 Miami,FL 33132
 8. H     100 Biscayne Blvd. Suite 604 Miami,FL 33132
 9. I            100 N. Biscayne Blvd. Miami,FL 33132"

library(gsubfn)
DF <- read.pattern(text = Lines, 
  pattern = "\\S+ \\S+ *(\\d+) (.*) (\\S+) ?,(\\S+) (\\d+)$",
  skip = 1, 
  as.is = TRUE,
  col.names = c("No", "Street", "City", "State", "Zip"))

给予:

> DF
   No                       Street  City State   Zip
1   1                  NE 1 Street Miami    FL 33132
2   1                   1st Street Miami    FL 33132
3   1                    NE 1st St Miami    FL 33132
4   1                   1st Street Miami    FL 33134
5 100               Biscayne Blvd. Miami    FL 33132
6 100                Biscayne Blvd Miami    FL 33132
7 100 Biscayne Boulevard Suite 604 Miami    FL 33132
8 100     Biscayne Blvd. Suite 604 Miami    FL 33132
9 100            N. Biscayne Blvd. Miami    FL 33132

这是可视化的正则表达式:

\S+ \S+ *(\d+) (.*) (\S+) ?,(\S+) (\d+)$

Debuggex Demo