Question

可能重复：
  R sorts a vector on its own accord - bad boy!
  How can I read the files in a directory in sorted order using R?

下面给出的代码效果很好。但问题是，当我输入dir1来查看结果时，我发现R将文件命令为：

[1] "data1.flt"   "data10.flt"  "data100.flt" "data101.flt"
[5] "data102.flt" "data103.flt" "data104.flt" "data105.flt"
[9] "data106.flt" "data107.flt" "data108.flt" "data109.flt"
[13] "data11.flt"  "data110.flt" "data111.flt" "data112.flt"
[17] "data113.flt" "data114.flt" "data115.flt" "data116.flt"
.
.
to
.
.
[357] "data91.flt"  "data92.flt"  "data93.flt"  "data94.flt"
[361] "data95.flt"  "data96.flt"  "data97.flt"  "data98.flt"
[365] "data99.flt"

会导致错误的结果。如何告诉R按顺序从1到365开始读取（我甚至使用了sort（dir1）但没有对它们进行排序）。类似的东西：

[1] "data1.flt"   "data2.flt"  "data3.flt" "data4.flt"

不喜欢：

[1] "data1.flt"   "data10.flt"  "data100.flt" "data101.flt"

以下是代码：

dir1 <- list.files("C:\\Users", "*.flt", full.names = TRUE)
results <- list()
for (.files in seq_along(dir1)){
    file2 <- readBin(dir2[.files], double(), size = 4, n = w * 67420, signed = TRUE)
    results[[length(results) + 1L]] <- file1[file1 != -9999]*10
}
for (i in seq_along(results)){
    fileName <- sprintf("C:\\New folder (2)\\NewFile%03d.bin", i)
    writeBin(as.integer(results[[i]]), fileName, size = 2)
}

Answer 1

因为它不解释数字 - 而是排序字符串

Answer 2

如果您知道文件名称的结构，可以使用它而不是直接从dir抓取它们。

for(i in seq(365)){
    filename <- paste0("data", i, ".flt")
    # Do some stuff with filename
}

R 2.15中引入了

paste0，因此对于旧版本，您需要：

for(i in seq(365)){
    filename <- paste("data", i, ".flt", sep = "")
    # Do some stuff with filename
}

进一步编辑因为你似乎很困惑。 filename通过循环每次迭代都会更新。您可以通过每次打印文件名来查看。

for(i in seq(365)){
    filename <- paste("data", i, ".flt", sep = "")
    print(filename)
    # Do some stuff with filename
}

为什么R订单文件为1 10 100而非1 2 3？

2 个答案: