string 计算字符串中所有单词的数量
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Count the number of all words in a string
提问by John
Is there a function to count the number of words in a string? For example:
是否有计算字符串中单词数的函数？例如：
str1 <- "How many words are in this sentence"
to return a result of 7.
返回 7 的结果。
采纳答案by AVSuresh
You can use strsplitand sapplyfunctions
您可以使用strsplit和sapply功能
sapply(strsplit(str1, " "), length)
回答by Martin Morgan
Use the regular expression symbol \\Wto match non-word characters, using +to indicate one or more in a row, along with gregexprto find all matches in a string. Words are the number of word separators plus 1.
使用正则表达式符号\\W匹配非单词字符，+用于表示一行中的一个或多个，以及gregexpr查找字符串中的所有匹配项。单词是单词分隔符的数量加 1。
lengths(gregexpr("\W+", str1)) + 1
This will fail with blank strings at the beginning or end of the character vector, when a "word" doesn't satisfy \\W's notion of non-word (one could work with other regular expressions, \\S+, [[:alpha:]], etc., but there will always be edge cases with a regex approach), etc. It is likely more efficient than strsplitsolutions, which will allocate memory for each word. Regular expressions are described in ?regex.
这将在字符向量的开头或结尾处出现空白字符串而失败，当“单词”不满足\\W的非单词概念时（可以使用其他正则表达式\\S+，[[:alpha:]], 等，但总会有是使用正则表达式方法的边缘情况）等。它可能比strsplit解决方案更有效，后者将为每个单词分配内存。中描述了正则表达式?regex。
UpdateAs noted in the comments and in a different answer by @Andri the approach fails with (zero) and one-word strings, and with trailing punctuation
更新如评论和@Andri 的不同答案中所述，该方法失败，带有（零）和一个单词字符串，并带有尾随标点符号
str1 = c("", "x", "x y", "x y!" , "x y! z")
lengths(gregexpr("[A-z]\W+", str1)) + 1L
# [1] 2 2 2 3 3
Many of the other answers also fail in these or similar (e.g., multiple spaces) cases. I think my answer's caveat about 'notion of one word' in the original answer covers problems with punctuation (solution: choose a different regular expression, e.g., [[:space:]]+), but the zero and one word cases are a problem; @Andri's solution fails to distinguish between zero and one words. So taking a 'positive' approach to finding words one might
许多其他答案在这些或类似（例如，多个空格）的情况下也失败了。我认为我的答案在原始答案中关于“一个词的概念”的警告涵盖了标点问题（解决方案：选择不同的正则表达式，例如，[[:space:]]+），但零和一个词的情况是一个问题；@Andri 的解决方案无法区分零个单词和一个单词。因此，采取一种“积极”的方法来寻找可能的词
sapply(gregexpr("[[:alpha:]]+", str1), function(x) sum(x > 0))
Leading to
导致
sapply(gregexpr("[[:alpha:]]+", str1), function(x) sum(x > 0))
# [1] 0 1 2 2 3
Again the regular expression might be refined for different notions of 'word'.
同样，正则表达式可能会针对“单词”的不同概念进行细化。
I like the use of gregexpr()because it's memory efficient. An alternative using strsplit()(like @user813966, but with a regular expression to delimit words) and making use of the original notion of delimiting words is
我喜欢使用，gregexpr()因为它的内存效率很高。另一种使用strsplit()（如@user813966，但使用正则表达式来分隔单词）并利用分隔单词的原始概念是
lengths(strsplit(str1, "\W+"))
# [1] 0 1 2 2 3
This needs to allocate new memory for each word that is created, and for the intermediate list-of-words. This could be relatively expensive when the data is 'big', but probably it's effective and understandable for most purposes.
这需要为创建的每个单词以及中间单词列表分配新的内存。当数据“大”时，这可能相对昂贵，但对于大多数目的来说它可能是有效且易于理解的。
回答by petermeissner
Most simple waywould be:
最简单的方法是：
require(stringr)
str_count("one, two three 4,,,, 5 6", "\S+")
... counting all sequences on non-space characters (\\S+).
... 计算非空格字符 ( \\S+)上的所有序列。
But what about a little function that lets us also decide which kind of wordswe would like to count and which works on whole vectorsas well?
但是关于一个小功能，让我们也决定什么哪一种的话，我们想计算和对整个向量工作呢？
require(stringr)
nwords <- function(string, pseudo=F){
ifelse( pseudo,
pattern <- "\S+",
pattern <- "[[:alpha:]]+"
)
str_count(string, pattern)
}
nwords("one, two three 4,,,, 5 6")
# 3
nwords("one, two three 4,,,, 5 6", pseudo=T)
# 6
回答by arekolek
I use the str_countfunction from the stringrlibrary with the escape sequence \wthat represents:
我str_count将stringr库中的函数与\w表示的转义序列一起使用：
any ‘word' character (letter, digit or underscore in the current locale: in UTF-8 mode only ASCII letters and digits are considered)
任何“单词”字符（当前语言环境中的字母、数字或下划线：在 UTF-8 模式下仅考虑 ASCII 字母和数字）
Example:
例子：
> str_count("How many words are in this sentence", '\w+')
[1] 7
Of all other 9 answers that I was able to test, only two (by Vincent Zoonekynd, and by petermeissner) worked for all inputs presented here so far, but they also require stringr.
在我能够测试的所有其他 9 个答案中，只有两个（由 Vincent Zoonekynd 和 petermeissner 提供）适用于迄今为止此处提供的所有输入，但它们也需要stringr.
But only this solution works with all inputs presented so far, plus inputs such as "foo+bar+baz~spam+eggs"or "Combien de mots sont dans cette phrase ?".
但只有此解决方案适用于迄今为止提供的所有输入，以及诸如"foo+bar+baz~spam+eggs"或 之类的输入"Combien de mots sont dans cette phrase ?"。
Benchmark:
基准：
library(stringr)
questions <-
c(
"", "x", "x y", "x y!", "x y! z",
"foo+bar+baz~spam+eggs",
"one, two three 4,,,, 5 6",
"How many words are in this sentence",
"How many words are in this sentence",
"Combien de mots sont dans cette phrase ?",
"
Day after day, day after day,
We stuck, nor breath nor motion;
"
)
answers <- c(0, 1, 2, 2, 3, 5, 6, 7, 7, 7, 12)
score <- function(f) sum(unlist(lapply(questions, f)) == answers)
funs <-
c(
function(s) sapply(gregexpr("\W+", s), length) + 1,
function(s) sapply(gregexpr("[[:alpha:]]+", s), function(x) sum(x > 0)),
function(s) vapply(strsplit(s, "\W+"), length, integer(1)),
function(s) length(strsplit(gsub(' {2,}', ' ', s), ' ')[[1]]),
function(s) length(str_match_all(s, "\S+")[[1]]),
function(s) str_count(s, "\S+"),
function(s) sapply(gregexpr("\W+", s), function(x) sum(x > 0)) + 1,
function(s) length(unlist(strsplit(s," "))),
function(s) sapply(strsplit(s, " "), length),
function(s) str_count(s, '\w+')
)
unlist(lapply(funs, score))
Output:
输出：
6 10 10 8 9 9 7 6 6 11
回答by mathematical.coffee
str2 <- gsub(' {2,}',' ',str1)
length(strsplit(str2,' ')[[1]])
The gsub(' {2,}',' ',str1)makes sure all words are separated by one space only, by replacing all occurences of two or more spaces with one space.
该gsub(' {2,}',' ',str1)品牌确保所有单词都只有一个空格分开，用一个空格替换两个或多个空格的所有出现。
The strsplit(str,' ')splits the sentence at every space and returns the result in a list. The [[1]]grabs the vector of words out of that list. The lengthcounts up how many words.
将strsplit(str,' ')在每一个空间分割句子并返回结果列表中。该[[1]]抓住的话的矢量指出，名单中。该length计数多少字。
> str1 <- "How many words are in this sentence"
> str2 <- gsub(' {2,}',' ',str1)
> str2
[1] "How many words are in this sentence"
> strsplit(str2,' ')
[[1]]
[1] "How" "many" "words" "are" "in" "this" "sentence"
> strsplit(str2,' ')[[1]]
[1] "How" "many" "words" "are" "in" "this" "sentence"
> length(strsplit(str2,' ')[[1]])
[1] 7
回答by Vincent Zoonekynd
You can use str_match_all, with a regular expression that would identify your words.
The following works with initial, final and duplicated spaces.
您可以将str_match_all, 与可以识别您的单词的正则表达式一起使用。以下适用于初始、最终和重复空间。
library(stringr)
s <- "
Day after day, day after day,
We stuck, nor breath nor motion;
"
m <- str_match_all( s, "\S+" ) # Sequences of non-spaces
length(m[[1]])
回答by bartektartanus
Try this function from stringipackage
从stringi包中试试这个功能
require(stringi)
> s <- c("Lorem ipsum dolor sit amet, consectetur adipisicing elit.",
+ "nibh augue, suscipit a, scelerisque sed, lacinia in, mi.",
+ "Cras vel lorem. Etiam pellentesque aliquet tellus.",
+ "")
> stri_stats_latex(s)
CharsWord CharsCmdEnvir CharsWhite Words Cmds Envirs
133 0 30 24 0 0
回答by yuqian
You can use wcfunction in library qdap:
您可以在库qdap 中使用wc函数：
> str1 <- "How many words are in this sentence"
> wc(str1)
[1] 7
回答by Murali Menon
You can remove double spaces and count the number of " "in the string to get the count of words. Use stringrand rm_white{qdapRegex}
您可以删除双空格并计算" "字符串中的数量以获得单词数。使用 stringr和rm_white{ qdapRegex}
str_count(rm_white(s), " ") +1
回答by Sangram
Try this
尝试这个
length(unlist(strsplit(str1," ")))
