将具有可变列数的文本文件读取到列表 [英] Read a text file with variable number of columns to a list

问题描述

我有一个像这样的文件:

I have a file like this:

mylist.txt
234984  10354  41175 932711 426928
1693237   13462

此文件的每一行具有不同数量的元素，每行最少1个元素. 我想将其读入这样的列表中:

Each line of this file has different number of elements, minimum of 1 element per line. I would like to read it into a list like this:

> print(head(mylist,2))
[[1]]
[1] 234984  10354  41175 932711 426928

[[2]]
[1] 1693237   13462

推荐答案

假定空格是定界符:

fc <- file("mylist.txt")
mylist <- strsplit(readLines(fc), " ")
close(fc)

如果值由多个空格定界(一个/或不一致的方式)，则可以将定界符与正则表达式匹配:

If the values are delimited by several spaces (an/or in unconsistent way), you can match delimiter with regular expression:

mylist.txt
234984   10354   41175 932711      426928
1693237               13462

fc <- file("mylist.txt")
mylist <- strsplit(readLines(fc), " +")
close(fc)

编辑#2

由于strsplit返回字符串，因此您需要将数据转换为数字(这很简单):

And since strsplit returns strings, you need to convert your data to numeric (that's an easy one):

mylist <- lapply(mylist, as.numeric)

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