R：在R中重新排序多个文件路径

发布于 2025-01-28 07:38:34 字数 591 浏览 2 评论 0原文

我有100个文件，每个文件命名为“ ABC -Day -1（至100）.csv”。

当我将它们读成R时，它是这样的：Day1，Day10，Day100等（请参见图1）。我知道R是这样做的，因为它是按字符对其进行排序，而不是按数字对其进行排序。有没有办法可以在数值正确的顺序（Day1，Day2，Day3，...）中重新排序路径，而无需我实际上必须手动更改我的原始文件名？

这是我到目前为止所拥有的：

filenames <- list.files(path="../STEP_ONE/Test_raw",
                    pattern="ADD_Day+.*sav",
                    full.names = TRUE)  # Reads in path of the 100 files

原文

I have 100 files, each named "ABC - Day - 1(to 100).csv".

When I read them into R, it is ordered like this: Day1, Day10, Day100, etc. (see figure 1). I know R does this because it is sorting it by character, not by number. Is there a way that I could reorder the path in numerically correct order (Day1, Day2, Day3, ...) without me actually having to manually change my raw file names?

Here is what I have so far:

filenames <- list.files(path="../STEP_ONE/Test_raw",
                    pattern="ADD_Day+.*sav",
                    full.names = TRUE)  # Reads in path of the 100 files

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流年已逝 2025-02-04 07:38:34

让我们假设您有一个带有文件名的向量v（根据您所说的___ day __. sav）。您可以减去一天的数量，并使用以下代码重新排序名称：

# Load library
  library(stringr)
# Matrix with your files' names and the day
  tab <- as.data.frame(str_match(v, "Day\\s*(.*?)\\s*.sav"))
  # Column names
    colnames(tab) <- c("file.name", "day")
  # Day as numeric
    tab$day <- as.numeric(tab$day)
# Reorder `tab` according to $day
  tab <- tab[order(tab$day),]

Let’s suppose you have a vector v with the names of your file (according to what you said, ___Day__.sav). You can subtract the number of the day and reorder the names with the following code:

# Load library
  library(stringr)
# Matrix with your files' names and the day
  tab <- as.data.frame(str_match(v, "Day\\s*(.*?)\\s*.sav"))
  # Column names
    colnames(tab) <- c("file.name", "day")
  # Day as numeric
    tab$day <- as.numeric(tab$day)
# Reorder `tab` according to $day
  tab <- tab[order(tab$day),]

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