拆分字符列并获取字符串中的字段名称 [英] split character columns and get names of field in string

问题描述

我需要将包含信息的列拆分为几列.
我会使用 tstrsplit ，但是相同的信息在行之间的顺序并不相同，因此我需要在变量中提取新列的名称.重要信息:可能有很多信息(字段变成新变量)，我不知道所有这些信息，因此，我不需要逐字段"解决方案.

I need to split a column that contains information into several columns.
I'd use tstrsplit but the same kind of information is not in the same order among the rows and I need to extract the name of the new column within the variable. Important to know: there can be many pieces of information (fields to become new variables) and I don't know all of them, so I don't want a "field by field" solution.

下面是我所拥有的一个例子:

Below is an example of what I have:

library(data.table)

myDT <- structure(list(chr = c("chr1", "chr2", "chr4"), pos = c(123L,
                  435L, 120L), info = c("type=3;end=4", "end=6", "end=5;pos=TRUE;type=2"
                  )), class = c("data.table", "data.frame"), row.names = c(NA,-3L))

#    chr pos                  info
#1: chr1 123          type=3;end=4
#2: chr2 435                 end=6
#3: chr4 120 end=5;pos=TRUE;type=2

我想得到:

#    chr pos end  pos type
#1: chr1 123   4 <NA>    3
#2: chr2 435   6 <NA> <NA>
#3: chr4 120   5 TRUE    2

最简单的方法将不胜感激！(注意:我不愿意采用dplyr/tidyr方式)

A most straightforward way to get that would be much appreciated! (Note: I'm not willing to go with a dplyr/tidyr way)

推荐答案

使用 regex 和 stringi 软件包:

setDT(myDT) # After creating data.table from structure()

library(stringi)

fields <- unique(unlist(stri_extract_all(regex = "[a-z]+(?==)", myDT$info)))
patterns <- sprintf("(?<=%s=)[^;]+", fields)
myDT[, (fields) := lapply(patterns, function(x) stri_extract(regex = x, info))]
myDT[, !"info"]

    chr  pos type end
1: chr1 <NA>    3   4
2: chr2 <NA> <NA>   6
3: chr4 TRUE    2   5

要获取正确的类型，请使用似乎(?) type.convert():

To get the correct type it seems (?) type.convert() can be used:

myDT[, (fields) := lapply(patterns, function(x) type.convert(stri_extract(regex = x, info), as.is = TRUE))]

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