您好我在这里有一个有趣的问题。假设我有一个长字符，其中包括其他人之间的城市名称。
test<-"Ucsd Medical Center, San Diego, California, USA|Yale Cancer Center, New Haven, Connecticut, USA|Massachusetts General Hospital., Boston, Massachusetts, USA|Dana Farber Cancer Institute, Boston, Massachusetts, USA|Washington University, Saint Louis, Missouri, USA|Mount SInai Medical Center, New York, New York, USA|Memorial Sloan Kettering Cancer Center, New York, New York, USA|Carolinas Healthcare System, Charlotte, North Carolina, USA|University Hospitals Case Medical Center; Seidman Cancer Center, Cleveland, Ohio, USA|Vanderbilt University Medical Center, Nashville, Tennessee, USA|Seattle Cancer Care Alliance, Seattle, Washington, USA|National Cancer Center, Gyeonggi-do, Korea, Republic of|Seoul National University Hospital, Seoul, Korea, Republic of|Severance Hospital, Yonsei University Health System, Seoul, Korea, Republic of|Korea University Guro Hospital, Seoul, Korea, Republic of|Asan Medical Center., Seoul, Korea, Republic of|VU MEDISCH CENTRUM; Dept. of Medical Oncology"
我的目标是提取它的所有城市名称。我通过以下五个步骤实现了它。
#replace | with ,
test2<-str_replace_all(test, "[|]", ", ")
# Remove punctuation from data
test3<-gsub("[[:punct:]\n]","",test2)
# Split data at word boundaries
test4 <- strsplit(test3, " ")
# Load data from package maps
data(world.cities)
# Match on cities in world.cities
citiestest<-lapply(test4, function(x)x[which(x %in% world.cities$name)])
结果可能是正确的
citiestest
[[1]]
[1] "San" "Boston" "Boston" "Washington" "York"
[6] "York" "Kettering" "York" "York" "Charlotte"
[11] "Carolina" "Cleveland" "Nashville" "Seattle" "Seattle"
[16] "Washington" "Asan"
但正如你所看到的那样，我不能处理具有两个单词的城市（纽约，圣地亚哥等），因为它们是分开的。当然，手动修复此问题不是一个选项，因为我的真实数据集非常大。
答案 0 :(得分：2)
以下是使用strsplit和sub的基本R选项：
terms <- unlist(strsplit(test, "\\s*\\|\\s*"))
cities <- sapply(terms, function(x) gsub("[^,]+,\\s*([^,]+),.*", "\\1", x))
cities[1:3]
Ucsd Medical Center, San Diego, California, USA
"San Diego"
Yale Cancer Center, New Haven, Connecticut, USA
"New Haven"
Massachusetts General Hospital., Boston, Massachusetts, USA
"Boston"
答案 1 :(得分：2)
一种相当不同的方法可能或多或少有用，具体取决于手头的数据：将每个地址传递给地理编码API，然后将城市拉出响应。
library(tidyverse)
places <- data_frame(string = "Ucsd Medical Center, San Diego, California, USA|Yale Cancer Center, New Haven, Connecticut, USA|Massachusetts General Hospital., Boston, Massachusetts, USA|Dana Farber Cancer Institute, Boston, Massachusetts, USA|Washington University, Saint Louis, Missouri, USA|Mount SInai Medical Center, New York, New York, USA|Memorial Sloan Kettering Cancer Center, New York, New York, USA|Carolinas Healthcare System, Charlotte, North Carolina, USA|University Hospitals Case Medical Center; Seidman Cancer Center, Cleveland, Ohio, USA|Vanderbilt University Medical Center, Nashville, Tennessee, USA|Seattle Cancer Care Alliance, Seattle, Washington, USA|National Cancer Center, Gyeonggi-do, Korea, Republic of|Seoul National University Hospital, Seoul, Korea, Republic of|Severance Hospital, Yonsei University Health System, Seoul, Korea, Republic of|Korea University Guro Hospital, Seoul, Korea, Republic of|Asan Medical Center., Seoul, Korea, Republic of|VU MEDISCH CENTRUM; Dept. of Medical Oncology") %>%
separate_rows(string, sep = '\\|')
places <- places %>%
mutate(geodata = map(string, ~{Sys.sleep(1); ggmap::geocode(.x, output = 'all')}))
places <- places %>%
mutate(address_components = map(geodata, list('results', 1, 'address_components')),
address_components = map(address_components,
~as_data_frame(transpose(.x)) %>%
unnest(long_name, short_name)),
city = map(address_components, unnest),
city = map_chr(city, ~{
l <- set_names(.x$long_name, .x$types);
coalesce(l['locality'], l['administrative_area_level_1'])
}))
比较结果和原始
places %>% select(city, string)
#> # A tibble: 17 x 2
#> city string
#> <chr> <chr>
#> 1 San Diego Ucsd Medical Center, San Diego, California, USA
#> 2 New Haven Yale Cancer Center, New Haven, Connecticut, USA
#> 3 Boston Massachusetts General Hospital., Boston, Massachusetts, USA
#> 4 Boston Dana Farber Cancer Institute, Boston, Massachusetts, USA
#> 5 St. Louis Washington University, Saint Louis, Missouri, USA
#> 6 New York Mount SInai Medical Center, New York, New York, USA
#> 7 New York Memorial Sloan Kettering Cancer Center, New York, New York, USA
#> 8 Charlotte Carolinas Healthcare System, Charlotte, North Carolina, USA
#> 9 Cleveland University Hospitals Case Medical Center; Seidman Cancer Center, Cleveland, Ohio, USA
#> 10 Nashville Vanderbilt University Medical Center, Nashville, Tennessee, USA
#> 11 Seattle Seattle Cancer Care Alliance, Seattle, Washington, USA
#> 12 Goyang-si National Cancer Center, Gyeonggi-do, Korea, Republic of
#> 13 서울특별시 Seoul National University Hospital, Seoul, Korea, Republic of
#> 14 Seoul Severance Hospital, Yonsei University Health System, Seoul, Korea, Republic of
#> 15 Seoul Korea University Guro Hospital, Seoul, Korea, Republic of
#> 16 Seoul Asan Medical Center., Seoul, Korea, Republic of
#> 17 Amsterdam VU MEDISCH CENTRUM; Dept. of Medical Oncology
......好吧，它并不完美。最大的问题是，对于美国城市，城市被归类为localities，而对于韩国，城市被归类为administrative_area_level_1（在美国是国家）。与其他韩国行不同，12实际上有一个地点，这不是列出的城市（作为行政区域的响应）。此外，＆＃34;首尔＆＃34;第13行被莫名其妙地翻译成韩文。
好消息是＆＃34;圣路易斯＆＃34;已被缩短为＆＃34; St.路易斯＆＃34;，这是一个更标准化的形式，最后一排位于阿姆斯特丹。
扩展这种方法可能需要向Google支付一些使用其API的费用。
答案 2 :(得分：1)
我会做什么：
test2 <- str_replace_all(test, "[|]", ", ") #Same as you did
test3 <- unlist(strsplit(test2, split=", ")) #Turns string into a vector
check <- test3 %in% world.cities$name #Check if element vectors match list of city names
test3[check == TRUE] #Select vector elements that match list of city names
[1] "San Diego" "New Haven" "Boston" "Boston" "Saint Louis" "New York" "New York" "New York"
[9] "New York" "Charlotte" "Cleveland" "Nashville" "Seattle" "Washington"
答案 3 :(得分：1)
另一种无循环方式
pat="(,.\\w+,)|(,.\\w+.\\w+,)"
gsub("(,\\s)|,","",regmatches(m<-strsplit(test,"\\|")[[1]],regexpr(pat,m)))
[1] "San Diego" "New Haven" "Boston" "Boston" "Saint Louis" "New York" "New York"
[8] "Charlotte" "Cleveland" "Nashville" "Seattle" "Gyeonggi-do" "Seoul" "Seoul"
[15] "Seoul" "Seoul"
此页面中给出的其他结果确实失败：例如，有一个名为Greonggi-do的城镇，其他解决方案中没有给出。还有一些代码将整个字符串作为城镇
答案 4 :(得分：1)
要扩展上面的@ hrbrmstr评论，您可以使用Stanford CoreNLP库对每个字符串执行命名实体识别（NER）。对这样一项事业的一个重要警告是，大多数NER注释者只能将一个标记注释为“位置”或等同物，这在城市与州和国家混在一起时并不是很有用。除了通常的NER注释器之外，CoreNLP确实包含一个额外的正则表达式NER注释器，可以将NER粒度提高到城市级别。
在R中，您可以使用coreNLP包来运行注释器。它需要rJava，在某些情况下可能很难配置。您还需要下载实际（非常大）的库，可以使用coreNLP::downloadCoreNLP来完成，如果需要，可以将CORENLP_HOME中的~/.Renviron环境变量设置为安装路径。
另请注意，这种方法相当慢且资源密集，因为它在Java中做了很多工作。
library(tidyverse)
library(coreNLP)
# set which annotators to use
writeLines('annotators = tokenize, ssplit, pos, lemma, ner, regexner\n', 'corenlp.properties')
initCoreNLP(libLoc = Sys.getenv('CORENLP_HOME'), parameterFile = 'corenlp.properties')
unlink('corenlp.properties') # clean up
places <- data_frame(string = "Ucsd Medical Center, San Diego, California, USA|Yale Cancer Center, New Haven, Connecticut, USA|Massachusetts General Hospital., Boston, Massachusetts, USA|Dana Farber Cancer Institute, Boston, Massachusetts, USA|Washington University, Saint Louis, Missouri, USA|Mount SInai Medical Center, New York, New York, USA|Memorial Sloan Kettering Cancer Center, New York, New York, USA|Carolinas Healthcare System, Charlotte, North Carolina, USA|University Hospitals Case Medical Center; Seidman Cancer Center, Cleveland, Ohio, USA|Vanderbilt University Medical Center, Nashville, Tennessee, USA|Seattle Cancer Care Alliance, Seattle, Washington, USA|National Cancer Center, Gyeonggi-do, Korea, Republic of|Seoul National University Hospital, Seoul, Korea, Republic of|Severance Hospital, Yonsei University Health System, Seoul, Korea, Republic of|Korea University Guro Hospital, Seoul, Korea, Republic of|Asan Medical Center., Seoul, Korea, Republic of|VU MEDISCH CENTRUM; Dept. of Medical Oncology") %>%
separate_rows(string, sep = '\\|') # separate strings
places_ner <- places %>%
mutate(annotations = map(string, annotateString),
tokens = map(annotations, 'token'),
tokens = map(tokens, group_by, token_id = data.table::rleid(NER)),
city = map(tokens, filter, NER == 'CITY'),
city = map(city, summarise, city = paste(token, collapse = ' ')),
city = map_chr(city, ~if(nrow(.x) == 0) NA_character_ else .x$city))
返回
places_ner %>% select(city, string)
#> # A tibble: 17 x 2
#> city string
#> <chr> <chr>
#> 1 San Diego Ucsd Medical Center, San Diego, California, USA
#> 2 New Haven Yale Cancer Center, New Haven, Connecticut, USA
#> 3 Boston Massachusetts General Hospital., Boston, Massachusetts, USA
#> 4 Boston Dana Farber Cancer Institute, Boston, Massachusetts, USA
#> 5 NA Washington University, Saint Louis, Missouri, USA
#> 6 NA Mount SInai Medical Center, New York, New York, USA
#> 7 NA Memorial Sloan Kettering Cancer Center, New York, New York, USA
#> 8 Charlotte Carolinas Healthcare System, Charlotte, North Carolina, USA
#> 9 Cleveland University Hospitals Case Medical Center; Seidman Cancer Center, Cleveland, Ohio, USA
#> 10 Nashville Vanderbilt University Medical Center, Nashville, Tennessee, USA
#> 11 Seattle Seattle Cancer Care Alliance, Seattle, Washington, USA
#> 12 NA National Cancer Center, Gyeonggi-do, Korea, Republic of
#> 13 Seoul Seoul National University Hospital, Seoul, Korea, Republic of
#> 14 Seoul Severance Hospital, Yonsei University Health System, Seoul, Korea, Republic of
#> 15 Seoul Korea University Guro Hospital, Seoul, Korea, Republic of
#> 16 Seoul Asan Medical Center., Seoul, Korea, Republic of
#> 17 NA VU MEDISCH CENTRUM; Dept. of Medical Oncology
失败：
regexner注释器的粒度是多少，但是（正如其名称所示）它由正则表达式起作用，有一个大小/熟悉度阈值，在该阈值下它不包含正则表达式。 You can add your own regex to it但是，如果它值得的话。 cleanNLP package还支持Stanford CoreNLP（以及其他一些后端），界面更易于使用（设置仍然很难），但据我所知，不允许使用{ {1}}目前由于它初始化CoreNLP的原因。
答案 5 :(得分：1)
您可以使用tidytext来提取二元组 - ＆gt;单词 - ＆gt;相交以获得共同部分
if