如何将值中的字符串转换为属性和值？

Question

我有 3mio 观察，属性为“other_tags”。 “other_tags”的值必须转换为新的属性和值。

输入（）

structure(list(osm_id = c(105093, 107975, 373652), other_tags = structure(c(2L, 
3L, 1L), .Label = c("\"addr:city\"=>\"Neuenegg\",\"addr:street\"=>\"Stuberweg\",\"building\"=>\"school\",\"building:levels\"=>\"2\"", 
"\"building\"=>\"commercial\",\"name\"=>\"Pollahof\",\"type\"=>\"multipolygon\"", 
"\"building\"=>\"yes\",\"amenity\"=>\"sport\",\"type\"=>\"multipolygon\""
), class = "factor")), class = "data.frame", row.names = c(NA, 
-3L))

这是数据的子样本：

osm_id  other_tags
105093  "building"=>"commercial","name"=>"Pollahof","type"=>"multipolygon"
107975  "building"=>"yes","amenity"=>"sport","type"=>"multipolygon"
373652  "addr:city"=>"Neuenegg","addr:street"=>"Stuberweg","building"=>"school","building:levels"=>"2"

这是所需的数据格式：创建新属性（仅用于建筑和便利设施）并添加值。

osm_id  building    amenity
105093  commercial
107975  yes         sport
373652  school

感谢您的帮助！

Answer 1

没那么难。

• other_tags 是因子列，所以我们必须在上面使用as.charachter
• 提取结果会生成一个中间列表，例如 s，其中所有变量都是分开的；使用strsplit 将这些从split = ',' 拆分后
• 将这些属性存储在 seaparte rwo 中，为新数据框中的每个属性，比如 df2
• 使用tidyr 中的separate() 将属性名称和值分成两个单独的列。分隔符sep 这次用作=>
• 使用str_remove_all 删除多余的引号
• 可选filter数据集
• pivot_wider 转换成所需的格式。

library(tidyverse)

s <- strsplit(as.character(df$other_tags), split = ",")

df2 <- data.frame(osm_id = rep(df$osm_id, sapply(s, length)), other_tags = unlist(s))

df2 %>% separate(other_tags, into = c("Col1", "Col2"), sep = "=>") %>%
  mutate(across(starts_with("Col"), ~str_remove_all(., '"'))) %>%
  filter(Col1 %in% c("amenity", "building")) %>%
  pivot_wider(id_cols = osm_id, names_from = Col1, values_from = Col2)

# A tibble: 3 x 3
  osm_id building   amenity
   <dbl> <chr>      <chr>  
1 105093 commercial NA     
2 107975 yes        sport  
3 373652 school     NA    

如果不使用过滤器

df2 %>% separate(other_tags, into = c("Col1", "Col2"), sep = "=>") %>%
  mutate(across(starts_with("Col"), ~str_remove_all(., '"'))) %>%
  pivot_wider(id_cols = osm_id, names_from = Col1, values_from = Col2)

# A tibble: 3 x 8
  osm_id building   name     type         amenity `addr:city` `addr:street` `building:levels`
   <dbl> <chr>      <chr>    <chr>        <chr>   <chr>       <chr>         <chr>            
1 105093 commercial Pollahof multipolygon NA      NA          NA            NA               
2 107975 yes        NA       multipolygon sport   NA          NA            NA               
3 373652 school     NA       NA           NA      Neuenegg    Stuberweg     2 

单管道语法

df %>% mutate(other_tags = as.character(other_tags),
              other_tags = str_split(other_tags, ",")) %>%
  unnest(other_tags) %>%
  mutate(other_tags = str_remove_all(other_tags, '"')) %>%
  separate(other_tags, into = c("Col1", "Col2"), sep = "=>") %>%
  filter(Col1 %in% c("amenity", "building")) %>%
  pivot_wider(id_cols = osm_id, names_from = Col1, values_from = Col2)

# A tibble: 3 x 3
  osm_id building   amenity
   <dbl> <chr>      <chr>  
1 105093 commercial NA     
2 107975 yes        sport  
3 373652 school     NA   

Answer 2

我们可以使用(g)sub 和str_extract 以及环视（只需两行代码）：

library(stringr)
df$building <- str_extract(gsub('"','', df$other_tags),'(?<=building=>)\\w+(?=,)')
df$amenity <- str_extract(gsub('"','', df$other_tags),'(?<=amenity=>)\\w+(?=,)')

如果出于某种原因您想删除列other_tags：

df$other_tags <- NULL

结果：

df
  osm_id   building amenity
1 105093 commercial    <NA>
2 107975        yes   sport
3 373652     school    <NA>