每隔n个单词将单词向量拆分(向量在列表中)

问题描述:

如何最好地分割列表中单词的向量?这是我目前正在做的事情(感谢geektrader的回答超过here),但它使RStudio颤抖并冻结了很多。这个问题与我以前的问题密切相关。这里有什么新的列表结构,这更接近我的实际使用情况。每隔n个单词将单词向量拆分(向量在列表中)

# reproducible data 
examp1 <- "When discussing performance with colleagues, teaching, sending a bug report or searching for guidance on mailing lists and here on SO, a reproducible example is often asked and always helpful. What are your tips for creating an excellent example? How do you paste data structures from r in a text format? What other information should you include? Are there other tricks in addition to using dput(), dump() or structure()? When should you include library() or require() statements? Which reserved words should one avoid, in addition to c, df, data, etc? How does one make a great r reproducible example?" 
examp2 <- "Sometimes the problem really isn't reproducible with a smaller piece of data, no matter how hard you try, and doesn't happen with synthetic data (although it's useful to show how you produced synthetic data sets that did not reproduce the problem, because it rules out some hypotheses). Posting the data to the web somewhere and providing a URL may be necessary. If the data can't be released to the public at large but could be shared at all, then you may be able to offer to e-mail it to interested parties (although this will cut down the number of people who will bother to work on it). I haven't actually seen this done, because people who can't release their data are sensitive about releasing it any form, but it would seem plausible that in some cases one could still post data if it were sufficiently anonymized/scrambled/corrupted slightly in some way. If you can't do either of these then you probably need to hire a consultant to solve your problem" 
examp3 <- "You are most likely to get good help with your R problem if you provide a reproducible example. A reproducible example allows someone else to recreate your problem by just copying and pasting R code. There are four things you need to include to make your example reproducible: required packages, data, code, and a description of your R environment. Packages should be loaded at the top of the script, so it's easy to see which ones the example needs. The easiest way to include data in an email is to use dput() to generate the R code to recreate it. For example, to recreate the mtcars dataset in R, I'd perform the following steps: Run dput(mtcars) in R Copy the output In my reproducible script, type mtcars <- then paste. Spend a little bit of time ensuring that your code is easy for others to read: make sure you've used spaces and your variable names are concise, but informative, use comments to indicate where your problem lies, do your best to remove everything that is not related to the problem. The shorter your code is, the easier it is to understand. Include the output of sessionInfo() as a comment. This summarises your R environment and makes it easy to check if you're using an out-of-date package. You can check you have actually made a reproducible example by starting up a fresh R session and pasting your script in. Before putting all of your code in an email, consider putting it on http://gist.github.com/. It will give your code nice syntax highlighting, and you don't have to worry about anything getting mangled by the email system." 
examp4 <- "Do your homework before posting: If it is clear that you have done basic background research, you are far more likely to get an informative response. See also Further Resources further down this page. Do help.search(keyword) and apropos(keyword) with different keywords (type this at the R prompt). Do RSiteSearch(keyword) with different keywords (at the R prompt) to search R functions, contributed packages and R-Help postings. See ?RSiteSearch for further options and to restrict searches. Read the online help for relevant functions (type ?functionname, e.g., ?prod, at the R prompt) If something seems to have changed in R, look in the latest NEWS file on CRAN for information about it. Search the R-faq and the R-windows-faq if it might be relevant (http://cran.r-project.org/faqs.html) Read at least the relevant section in An Introduction to R If the function is from a package accompanying a book, e.g., the MASS package, consult the book before posting. The R Wiki has a section on finding functions and documentation" 
examp5 <- "Before asking a technical question by e-mail, or in a newsgroup, or on a website chat board, do the following: Try to find an answer by searching the archives of the forum you plan to post to. Try to find an answer by searching the Web. Try to find an answer by reading the manual. Try to find an answer by reading a FAQ. Try to find an answer by inspection or experimentation. Try to find an answer by asking a skilled friend. If you're a programmer, try to find an answer by reading the source code. When you ask your question, display the fact that you have done these things first; this will help establish that you're not being a lazy sponge and wasting people's time. Better yet, display what you have learned from doing these things. We like answering questions for people who have demonstrated they can learn from the answers. Use tactics like doing a Google search on the text of whatever error message you get (searching Google groups as well as Web pages). This might well take you straight to fix documentation or a mailing list thread answering your question. Even if it doesn't, saying “I googled on the following phrase but didn't get anything that looked promising” is a good thing to do in e-mail or news postings requesting help, if only because it records what searches won't help. It will also help to direct other people with similar problems to your thread by linking the search terms to what will hopefully be your problem and resolution thread. Take your time. Do not expect to be able to solve a complicated problem with a few seconds of Googling. Read and understand the FAQs, sit back, relax and give the problem some thought before approaching experts. Trust us, they will be able to tell from your questions how much reading and thinking you did, and will be more willing to help if you come prepared. Don't instantly fire your whole arsenal of questions just because your first search turned up no answers (or too many). Prepare your question. Think it through. Hasty-sounding questions get hasty answers, or none at all. The more you do to demonstrate that having put thought and effort into solving your problem before seeking help, the more likely you are to actually get help. Beware of asking the wrong question. If you ask one that is based on faulty assumptions, J. Random Hacker is quite likely to reply with a uselessly literal answer while thinking Stupid question..., and hoping the experience of getting what you asked for rather than what you needed will teach you a lesson." 

# make a big list of character vectors containing words 
list_examps <- lapply(1:5, function(i) eval(parse(text=paste0("examp",i)))) 
list_examps <- rep(list_examps, 2000) 

# my current method 
n <- 30 # number of words in each chunk 
temp1 <- vector("list", length(list_examps)) 
temp2 <- vector("list", length(list_examps)) 
for(i in 1:length(list_examps)) 
{ 
    temp1[[i]] <- unlist(strsplit(list_examps[[i]], " ")) 
    temp2[[i]] <- split(unlist(strsplit(temp1[[i]] , " ")), 
         seq_along(unlist(strsplit(temp1[[i]], " ")))%/%n) 
} 
listofnwords <- unlist(temp2, recursive = FALSE) # desired output 

有没有更有效的方法来做到这一点?

UPDATE 1加入(相当温和的)机器规格

> sessionInfo() 
    R version 3.0.0 (2013-04-03) 
    Platform: x86_64-w64-mingw32/x64 (64-bit) 

    locale: 
    [1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United States.1252 
    [3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C       
    [5] LC_TIME=English_United States.1252  

    attached base packages: 
    [1] stats  graphics grDevices utils  datasets methods base  

    loaded via a namespace (and not attached): 
    [1] tools_3.0.0 

    > gc() 
       used (Mb) gc trigger (Mb) max used (Mb) 
    Ncells 434928 23.3  818163 43.7 667722 35.7 
    Vcells 7086406 54.1 12291671 93.8 12291671 93.8 

> Sys.getenv() # relevant exerpts, far off to the right for some reason... 

                                                                                                                                                      NUMBER_OF_PROCESSORS 
                                                                                                                                                          "4" 
                                                                                                                                                          OS 
                                                                                                                                                        "Windows_NT" 

                                                                                                                                                     PROCESSOR_ARCHITECTURE 
                                                                                                                                                         "AMD64" 
                                                                                                                                                      PROCESSOR_IDENTIFIER 
                                                                                                                                              "Intel64 Family 6 Model 42 Stepping 7, GenuineIntel" 
                                                                                                                                                       PROCESSOR_LEVEL 
                                                                                                                                                          "6" 
                                                                                                                                                      PROCESSOR_REVISION 
                                                                                                                                                         "2a07" 

UPDATE 2次答案的速度测试,使用上述数据上的EC2实例(*层)。赢家是... SimonO101!感谢大家的帮助。

# make a big list of character vectors containing words 
list_examps <- lapply(1:5, function(i) eval(parse(text=paste0("examp",i)))) 
list_examps <- rep(list_examps, 200) 

# my current method 
n <- 30 # number of words in each chunk 
temp1 <- vector("list", length(list_examps)) 
temp2 <- vector("list", length(list_examps)) 

me <- function(list_examps){ 
    for(i in 1:length(list_examps)) 
    { 
    temp1[[i]] <- unlist(strsplit(list_examps[[i]], " ")) 
    temp2[[i]] <- split(unlist(strsplit(temp1[[i]] , " ")), 
         seq_along(unlist(strsplit(temp1[[i]], " ")))%/%n) 
    } 
    listofnwords <- unlist(temp2, recursive = FALSE) # desired output 
} 

dr <- function(list_examps){ 
    f <- function(list_examps) { 
    y <- unlist(strsplit(list_examps, " ")) 
    ly <- length(y) 
    split(y, gl(ly%/%n+1, n, ly)) 
    } 

    listofnwords <- sapply(list_examps, f) 
    listofnwords <- unlist(listofnwords, recursive=F) 
} 

si <- function(list_examps){ 

    words <- unlist((sapply(list_examps , strsplit , " "))) 
    results <- lapply(seq(0, length(words) , by = n) , function(x) c(words[(x+1):(x+n)])) 

} 

er <- function(x){ 
    x <- do.call(paste, x) 
    x <- strsplit(x, ' ')[[1]] 
    result <- split(x, cut(seq_along(x), breaks = seq(0, by = n, length(x)) , include.lowest = TRUE)) 
} 


library(rbenchmark) 

benchmark(
    me(list_examps), 
    dr(list_examps), 
    si(list_examps), 
    er(list_examps), 
    replications = 10) 

而且这里的结果:

   test replications elapsed relative user.self sys.self user.child sys.child 
2 dr(list_examps)   10 48.104 1.119 47.907 0.000   0   0 
4 er(list_examps)   10 71.316 1.660 70.645 0.568   0   0 
1 me(list_examps)   10 48.156 1.121 48.543 0.000   0   0 
3 si(list_examps)   10 42.971 1.000 42.875 0.000   0   0 
+0

什么是您的系统规格?在我的桌面上,我能够在6.44秒内运行上面的所有代码,几乎没有明显的冻结 – 2013-04-26 09:05:49

+0

也许我应该编辑我的示例数据,使其更像我的真实数据,这些数据是我想要的5000-7000个字左右的矢量分成1000个字块。 – Ben 2013-04-26 09:22:27

+0

@本人编辑我的答案,现在它应该工作。您的示例数据运行时间为1.5秒,而您的方法运行时间为7秒。 – 2013-04-26 10:25:12

一个可能的解决方案:

f <- function(x) { 
    y <- unlist(strsplit(x, " ")) 
    ly <- length(y) 
    split(y, gl(ly%/%n+1, n, ly)) 
    } 

listofnwords <- sapply(list_examps, f) 
listofnwords <- unlist(listofnwords, recursive=F) 
+0

尽管这稍微慢一点,但它会返回带有项目名称的列表我需要的格式,所以这是一个更完整的答案。 – Ben 2013-04-27 08:06:33

我会使用sapplyunlist得到的话,那么使用lapply将它们串联到列表中的每30个字的项目(这需要1.6秒在您的示例):

f1 <- function(x){ 

    words <- unlist((sapply(list_examps , strsplit , " "))) 
    results <- lapply(seq(0, length(words) , by = 30) , function(x) c(words[(x+1):(x+30)])) 

} 

而在与其他解决方案,它被定义为比较:

f2 <- function(x){ 
    x <- do.call(paste, x) 
    x <- strsplit(x, ' ')[[1]] 
    result <- split(x, cut(seq_along(x), breaks = seq(1, by = 30, length(x)) , include.lowest = TRUE)) 
} 

我们看到,在这个例子中是使用lapply和sapply 将运行约两倍的速度(但取决于你的使用情况可能发生变化):

microbenchmark(f1(list_examps) , f2(list_examps) , times = 10L) 
#Unit: seconds 
#   expr  min  lq median  uq  max neval 
# f1(list_examps) 1.463855 1.531033 1.596762 1.667241 1.815084 10 
# f2(list_examps) 2.787036 2.884537 2.945287 3.019572 3.357706 10 
+0

我不太明白在哪里放置'n'在代码中......块在哪里产生? – Ben 2013-04-26 09:05:24

+1

啊,对不起 - 我有罪DR(你的帖子不是TL)。让我重新考虑.... :-) – 2013-04-26 09:07:47

+0

感谢您花时间编辑其他答案并进行比较! – Ben 2013-04-26 23:42:23

这个怎么样?

x <- do.call(paste, list_examps) 
x <- strsplit(x, ' ')[[1]] 
result <- split(x, cut(seq_along(x), breaks = seq(0, by = 30, length(x)))) 
result[[1]] 
+0

+1不错的解决方案,但你需要添加'include.lowest = TRUE'来获得正确的结果 – 2013-04-26 09:27:50

+1

而且我注意到结果中的第一个列表项是31个项目(因为需要使用include.lowest = TRUE“,但是你可以通过将'seq'的开始更改为0来解决这个问题,因为R从1开始索引,然后删除'include.lowest = TRUE' – 2013-04-26 09:47:45