在 R 中将字符串拆分为固定长度元素的最快方法

问：

如何在 R 中将字符串拆分为固定长度的元素是一个常见问题，典型的答案要么依赖 ，要么后跟 。 例如，可以通过指定 3 个字符的固定长度来将字符串切入。substring(x)strsplit(x, sep="")paste(y, collapse = "")"azertyuiop""aze", "rty","uio", "p"

我正在寻找最快的方法。 在对长字符串（> 1000 个字符）进行了一些测试后，我发现这太慢了。因此，策略是将字符串拆分为单个字符，然后通过应用一些巧妙的方法将它们粘贴回所需长度的组中。substring()

这是我能想到的最快的功能。这个想法是将字符串拆分为单独的字符，然后在字符向量的正确位置散布一个分隔符，将字符（和分隔符）折叠回字符串，然后再次拆分字符串，但这次指定分隔符。

splitInParts <- function(string, size) {              #can process a vector of strings. "size" is the length of desired substrings
    chars <- strsplit(string,"",T)
    lengths <- nchar(string)
    nFullGroups <- floor(lengths/size)                #the number of complete substrings of the desired size

    #here we prepare a vector of separators (comas), which we will replace by the characters, except at the positions that will have to separate substring groups of length "size". Assumes that the string doesn't have any comas.
    seps  <-  Map(rep, ",", lengths + nFullGroups)     #so the seps vector is longer than the chars vector, because there are separators (as may as they are groups)
    indices <- Map(seq, 1, lengths + nFullGroups)      #the positions at which separators will be replaced by the characters
    indices <- lapply(indices, function(x) which(x %% (size+1) != 0))  #those exclude the positions at which we want to retain the separators (I haven't found a better way to generate such vector of indices)

    temp <- function(x,y,z) {        #a fonction describing the replacement, because we call it in the Map() call below
        x[y] <- z
        x
    }
    res <- Map(temp, seps, indices, chars)             #so now we have a vector of chars with separators interspersed
    res <- sapply(res, paste, collapse="", USE.NAMES=F)  #collapses the characters and separators
    res <- strsplit(res, ",", T)                        #and at last, we can split the strings into elements of the desired length
}

这看起来很乏味，但我试图简单地将向量放入具有足够行数的矩阵中，然后用 折叠矩阵列。这要慢得多。而将字符向量拆分为适当长度的向量列表，以便折叠元素，甚至会更慢。charsapply(mat, 2, paste, collapse="")split()

因此，如果您能更快地找到东西，请告诉我。如果没有，那么我的函数可能会有一些用处。:)