使用 refinr 包比较和优化单独列中的字符串

Question

我的很多时间都花在合并国家、城市、名称或政党列的两个数据框上。现在，它是 refinr package，一个 OpenRefine 的 R 端口，派上用场了。只是我还没有弄清楚如何比较 'the same' 列中的两个并命名字符串，就像我在单个向量上使用 refinr 一样。我在 R 方面没有那么丰富的经验，所以这听起来可能有点含糊。也许我的例子让事情更清楚一些。

library(tidyverse)
library(refinr)

# I would like to add the values (and the right name's) of this example df...
df1 <- tribble(
  ~uid, ~name, ~value,
  "A", "Red", 13,
  "A", "violet", 145,
  "B", "Blue", 3,
  "B", "yellow", 56,
  "C", "yellow-purple", 789,
  "C", "green", 17
  )

# ...to the following df
df2 <- tribble(
  ~uid, ~name,
  "A", "red",
  "B", "blu",
  "C", "YellowPurple",
  "C", "green"
  )

# The following code of course produces NA values
df3 <- left_join(df1, df2, by = c("uid", "name"))

# While the following is the desired outcome

# A tibble: 4 x 3
  uid   name           value
  <chr> <chr>          <dbl>
1 A     Red             13 
2 B     Blue             3
3 C     yellow-purple  789   
4 C     green           17

key_collision_merge() 和 n_gram_merge() 处理单个向量中的字符串。我的问题是，我可以在两列而不是一列之间比较和更改字符串吗？

如果这是可能的，我会节省很多时间！

提前致谢。

Answer 1

你可以试试

library(refinr)
library(tidyverse)
df1 %>% 
  bind_rows(df2, .id = "id") %>% 
  mutate(key=key_collision_merge(name)) %>% 
  split(.$id) %>% 
  inner_join(x=select(.[[1]],-id), y=select(.[[2]], uid, key), by=c("uid", "key"))
# A tibble: 3 x 4
  uid   name          value key         
  <chr> <chr>         <dbl> <chr>       
1 A     Red             13. Red         
2 C     yellow-purple  789. YellowPurple
3 C     green           17. green

但是 "blu" 没有被两个 refiner 函数识别为 "blue"。因此，您可以通过添加此行 mutate(name=gsub("blu","blue",name))

来包含一个 gsub 来更改此特定字符串

Answer 2

我不确定这是 refinr 的最佳用途，它主要用于协调单个列中的单词拼写。你想做的事情看起来像一个模糊连接，并且有一个R package for that。使用示例可以是：

library(tidyverse)
library(fuzzyjoin)


df1 <- tribble(
  ~uid, ~name, ~value,
  "A", "Red", 13,
  "A", "violet", 145,
  "B", "Blue", 3,
  "B", "yellow", 56,
  "C", "yellow-purple", 789,
  "C", "green", 17
)

# ...to the following df
df2 <- tribble(
  ~uid, ~name,
  "A", "red",
  "B", "blu",
  "C", "YellowPurple",
  "C", "green"
)

df3 <- df2 %>%
  stringdist_left_join(df1,
                       distance_col = "dist", 
                       method='soundex') %>% 
  select(uid=uid.x, name=name.y, value)

df3
  # A tibble: 4 x 3
  uid   name          value
  <chr> <chr>         <dbl>
1 A     Red              13
2 B     Blue              3
3 C     yellow-purple   789
4 C     green            17

我用的是soundex算法，但是还有其他方法，都是基于stringdist package.

使用 refinr 包比较和优化单独列中的字符串

Compare and refine strings in separate columns with refinr package

r

string-matching

openrefine