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我将制表符分隔的文本文件中的数据导入到 data.frame 中。现在我想调整和编辑一个名为 C1 的列的内容。为此目的向我推荐了正则表达式。起初我使用： 

for (rn in 1:length(C1))
C1s <- strsplit(as.character(C1[rn]), "; ", fixed = TRUE)[[1]]

将示例中的各个条目分开，例如：

  • [张俊玲] 中国农业大学, 科尔资源与环境科学, 北京 100094, 人民共和国; [Zhang, Junling] Univ Hohenheim, Inst Plant Nutr, D-7000 Stuttgart, Germany；[George, Eckhard] Humboldt Univ, Inst Crop Sci, Dept Plant Nutr, D-14979 Grossbeeren, Germany；[George, Eckhard] Leibniz Inst 蔬菜和观赏作物 Theodor, D-14979 Grossbeeren, Germany
  • UNIV DORTMUND，INST PHYS，D-44221 DORTMUND，德国；洪堡大学，INST PHYS，D-15738 ZEUTHEN，德国
  • UNIV KEELE，DEPT CHEM，KEELE ST5 5BG，STAFFS，英格兰；MT SINAI HOSP,MT SINAI SCH MED,DR AA FISHBERG CTR NEUROBIOL,NEW YORK,NY 10029; 柏林洪堡大学，慈善医院，病理学和临床生化部，柏林，德国
  • NIFAD，分子微生物实验室，BETHESDA，MD

然后我想

  • 将所有内容设置为大写
  • 将字符串“Humboldt...Germany”替换为“HUMBOLDT” 将字符串“England”、“Scotland”、“Wales”和“North Ireland”替换为“UNITED KINGDOM”
  • 用“USA”替换美国州缩写和邮政编码的组合
  • 用“USA”替换美国的州缩写

此外，我想删除除分号（包括它们后面的空格）、它们前面的单词以及上面示例中的最后一个单词之外的所有内容。
我用了 

gsub('[.*\\] ', ''(toupper(C1s))

和 

gsub(',\\s*', ','(toupper(C1s))

，例如，但无法使其正常工作。
我想得到以下输出：

  • 中国人民；德国; 洪堡；德国
  • 德国; 洪堡
  • 英国; 美国; 洪堡
  • 美国

所以我的问题是：我怎样才能达到我想要的结果？
非常感谢您的考虑！

更新和附加问题

感谢 mrdwab 的有用回复和评论，我取得了很大进展。

不幸的是，直到现在我才意识到，也有这样的地址，其中方括号中的作者不止一位。不幸的是，mrdwab 提出的算法不适用于这些。 

> test = c("[Bocquet, F. C.; Giovanelli, L.; Abel, M.; Porte, L.; Themlin, J. -M.] Aix Marseille Univ, Inst Mat Microelect & Nanosci Prov IM2NP, F-13397 Marseille 20, France; [Bocquet, F. C.; Giovanelli, L.; Abel, M.; Porte, L.; Themlin, J. -M.] CNRS, Inst Mat Microelect & Nanosci Prov IM2NP, UMR 6242, Marseille, France; [Amsalem, P.; Koch, N.] Humboldt Univ, Inst Phys, D-12489 Berlin, Germany; [Petaccia, L.; Topwal, D.; Gorovikov, S.; Goldoni, A.] Sincrotrone Trieste, I-34149 Trieste, Italy")

这是我得到的结果：

> test
[1] "[BOCQUET,  F. C." "GIOVANELLI,  L."  "ABEL,  M."        "PORTE,  L."      
[5] "FRANCE"           "[BOCQUET,  F. C." "GIOVANELLI,  L."  "ABEL,  M."       
[9] "PORTE,  L."       "FRANCE"           "[AMSALEM,  P."    "HUMBOLDT"        
[13] "[PETACCIA,  L."   "TOPWAL,  D."      "GOROVIKOV,  S."   "ITALY"

这是我想要得到的结果：

[1] "FRANCE"; "FRANCE"; "HUMBOLDT"; "ITALY"

我尝试使用它来单独删除每个方括号及其内容：

C1s = gsub("(.*)[(.*)]", "\\2", C1s)

但不是第一个左括号和最后一个右括号之间的所有内容都被删除了......如果我先用逗号替换方括号内的所有分号，也许它会起作用？我试过了

C1s = gsub("[(.*);(.*)]", "[(.*),(.*)]", C1s)

来实现这一点，但它没有奏效。
所以我很感激你在这方面的帮助！

除此之外，我仍然陷入另一个障碍，我似乎无法自己解决，不幸的是......
这是我目前的输出： 

> C1s
[1] "PEOPLES R CHINA" "GERMANY"         "HUMBOLDT"         "GERMANY"
[5] "GERMANY"         "HUMBOLDT"        "UNITED KINGDOM"   "USA"
[9] "HUMBOLDT"        "USA"
> dims
[1] 4 2 3 1
> is.list(C1)
[1] FALSE
> is.vector(C1)
[1] TRUE

但是我究竟如何使用dims中的信息来创建所需的输出？： 

[1] "PEOPLES R CHINA"; "GERMANY"; "HUMBOLDT"; "GERMANY"
[2] "GERMANY"; "HUMBOLDT"
[3] "UNITED KINGDOM"; "USA"; "HUMBOLDT"
[4] "USA"

非常感谢您的支持！

4

1 回答 1

3

Assuming your data looks something like this:

test = c("[Zhang, Junling] China Agr Univ, Coll Resources & Environm Sci, Beijing 100094, Peoples R China", 
         " [Zhang, Junling] Univ Hohenheim, Inst Plant Nutr, D-7000 Stuttgart, Germany", 
         " [George, Eckhard] Humboldt Univ, Inst Crop Sci, Dept Plant Nutr, D-14979 Grossbeeren, Germany", 
         " [George, Eckhard] Leibniz Inst Vegetable & Ornamental Crops Theodor, D-14979 Grossbeeren, Germany"
)

First, let's make it all upper case:

test = toupper(test)

Then, let's change HUMBODT...GERMANY to HUMBOLDT

test = gsub("(.*)(HUMBOLDT)(.*)", "\\2", test)

Then, let's just extract the last part of the string.

test = gsub("(.*), +([A-Z ]+$)", "\\2", test)
> test
[1] "PEOPLES R CHINA" "GERMANY"         "HUMBOLDT"        "GERMANY"

If I had a more complete example from you, I might be able to help you with the other parts too.

Update

Here's a mostly worked solution that should help you figure out what to do.

Using "scan" to copy your data from your post, we can get data that looks like this:

test = c("[Zhang, Junling] China Agr Univ, Coll Resources & Environm Sci, Beijing 100094, Peoples R China; [Zhang, Junling] Univ Hohenheim, Inst Plant Nutr, D-7000 Stuttgart, Germany; [George, Eckhard] Humboldt Univ, Inst Crop Sci, Dept Plant Nutr, D-14979 Grossbeeren, Germany; [George, Eckhard] Leibniz Inst Vegetable & Ornamental Crops Theodor, D-14979 Grossbeeren, Germany", 
         "UNIV DORTMUND,INST PHYS,D-44221 DORTMUND,GERMANY; HUMBOLDT UNIV,INST PHYS,D-15738 ZEUTHEN,GERMANY", 
         "UNIV KEELE,DEPT CHEM,KEELE ST5 5BG,STAFFS,ENGLAND; MT SINAI HOSP,MT SINAI SCH MED,DR A A FISHBERG CTR NEUROBIOL,NEW YORK,NY 10029; HUMBOLDT UNIV BERLIN,CHARITE HOSP,DEPT PATHOL & CLIN BIOCHEM,BERLIN,GERMANY", 
         "NIFAD,MOLEC MICROBIOL LAB,BETHESDA,MD")

You'll need to go step-by-step to be able to make the transformations you want. Here's how I did it.

test.orig = test # A backup, just in case
test = toupper(test) # To uppercase
test = strsplit(test, ";") # Split everything up into a list
dims = sapply(test, length) # We might need this later
test = unlist(test) # Now, a single vector
test = gsub(",", " , ", test) # Pad those commas with some space
# Replace any 'HUMBOLDT'...'GERMANY' stuff with 'HUMBOLDT'
test = gsub("(.*)(HUMBOLDT)(.*)(GERMANY$)", "\\2", test)
# Replace a bunch of stuff with 'UNITED KINGDOM'
test = gsub("ENGLAND|SCOTLAND|WALES|NORTH IRELAND", "UNITED KINGDOM", test)
# Search for a pair of letters followed by a space followed by five digits
# and replace it with USA
test = gsub(" [A-Z]{2} [0-9]{5}", " USA", test)
# Find the item following the last comma
test = gsub("(.*), ([A-Z ]+)$", "\\2", test)
# Remove any whitespace
test = gsub("^ | $", "", test)
> test
[1] "PEOPLES R CHINA" "GERMANY"         "HUMBOLDT"        "GERMANY"        
[5] "GERMANY"         "HUMBOLDT"        "UNITED KINGDOM"  "USA"            
[9] "HUMBOLDT"        "MD"                  
# Unlist it if you need to
> split(test, rep(dims, dims))
$`1`
[1] "MD"

$`2`
[1] "GERMANY"  "HUMBOLDT"

$`3`
[1] "UNITED KINGDOM" "USA"            "HUMBOLDT"      

$`4`
[1] "PEOPLES R CHINA" "GERMANY"         "HUMBOLDT"        "GERMANY"

But, I'll leave you the task of figuring out how to replace state abbreviations with USA, and how to put it back in the same line order.

于 2012-05-07T09:33:43.957 回答