我正在进行网络抓取/制图项目,我从餐馆网站上删除了地址数据,并将结果存储为列表 - 在此示例中称为{{1 }}。
问题是,如何最好地将这些列表项转换为单个data.frame / tibble(目前使用loc_list
),但是在新的data.frame中,还有一个标题为bind_rows( )
的列对应于每个列表项名称。在我的示例中,输出将有3 metro
s,然后是3 alpharetta
,然后是1 atlanta
。
buford
目标输出:
loc_list
$alpharetta
# A tibble: 3 x 2
names address
<chr> <chr>
1 East Roswell US 2630 Holcomb Bridge Rd Alpharetta, GA 30022
2 Old Milton US 4305 Old Milton Parkway Ste 101 Alpharetta, GA 30022
3 Windward US 875 N Main Street Ste 306 Alpharetta, GA 30009
$atlanta
# A tibble: 3 x 2
names address
<chr> <chr>
1 Philips Arena US 100 Techwood Drive Atlanta, GA 30303
2 Virginia Highlands US 1006 N Highland Ave Atlanta, GA 30306
3 Perimeter US 1211 Ashford Crossing Atlanta, GA 30346
$buford
# A tibble: 1 x 2
names address
<chr> <chr>
1 Woodward US 3250 Woodward Crossing Blvd Buford, GA 30519
答案 0 :(得分:1)
alistaire指出bind_rows
对.id
已经足够了。以下是示例数据:
alpharetta <- tibble(names=c("East Roswell", "Old Milton"),
address = c("US 2630 Holcomb Bridge Rd Alpharetta, GA 30022", "4305 Old Milton Parkway Ste 101 Alpharetta, GA 30022"))
atlanta <- tibble(names=c("Philips Arena", "Virginia Highlands"),
address = c("US 100 Techwood Drive Atlanta, GA 30303", "US 1006 N Highland Ave Atlanta, GA 30306"))
loc_list <- list(alpharetta = alpharetta, atlanta = atlanta)
bind_rows(loc_list, .id="metro")
# A tibble: 4 x 3
metro names address
<chr> <chr> <chr>
1 alpharetta East Roswell US 2630 Holcomb Bridge Rd Alpharetta, GA 30022
2 alpharetta Old Milton 4305 Old Milton Parkway Ste 101 Alpharetta, GA 30022
3 atlanta Philips Arena US 100 Techwood Drive Atlanta, GA 30303
4 atlanta Virginia Highlands US 1006 N Highland Ave Atlanta, GA 30306