如何使用R以排序顺序读取目录中的文件？

Question

下面给出的代码运行良好,并读取我的目录中的文件并提取值:

X <- c(75:85) ; Y <- c(208:215) 
extract <- vector()
files <- list.files("C:\\New folder (10)", "*.img",full.names=TRUE)

}

我试图通过使用sprintf指定,但我得到了错误.任何帮助请:

for (i in c(1:365)) {
   fileName <- sprintf("C:New folder (10)/Climate_Rad_%d.img", i)
}

Answer 1

为什么不使用 list.files() 检索所有文件（具有特定模式），然后对其进行排序。然后，您从排序向量中检索文件，这会以正确的排序顺序提供给您。这样做的优点是，当 1:365 序列中缺少数字时，它也能起作用

就像是：

myFiles <- list.files(pattern = "^Climate_Rad_") #all files starting with Climate_
myFiles <- sort(myFiles)
# then read them in, for instance through
for (fileNames in myFiles) {READ.IN.AND.DO.YOUR.MAGIC.ON.THEM}

Answer 2

啊,好吧,我明白了,问题在于排序.排序顺序按字母顺序排列.Climate_Rad_1按字母顺序排在Climate_Rad_10之后,而不是Climate_Rad_2顺序不是"随机"的,它按字母顺序正确.

但是,您希望Climate_Rad_2在Climate_Rad_10之前处理,而不是之后.有几种方法可以解决这个问题.首先,您应该注意Climate_Rad_002在Climate_Rad_010之前按字母顺序排列,因此如果您在生成文件时添加前导零,这将使以后按数字顺序处理文件变得容易.

或者,假设您在创建文件时无法添加零.然后至少有两种方法可以按顺序访问文件.通过之后向文件名添加零,或者只是对文件名的数字部分进行排序.

让我告诉你后者.

myFiles <- paste("Climate_Rad_", c(1:15, 95:110), ".img", sep = "") # create some test names, you get the actual myFiles through a call to list.files()

myFiles.sorted <- sort(myFiles) # this gives the alphabetic sorting, not what you want

> myFiles.sorted
 [1] "Climate_Rad_1.img"   "Climate_Rad_10.img"  "Climate_Rad_100.img"
 [4] "Climate_Rad_101.img" "Climate_Rad_102.img" "Climate_Rad_103.img"
 [7] "Climate_Rad_104.img" "Climate_Rad_105.img" "Climate_Rad_106.img"
[10] "Climate_Rad_107.img" "Climate_Rad_108.img" "Climate_Rad_109.img"
[13] "Climate_Rad_11.img"  "Climate_Rad_110.img" "Climate_Rad_12.img" 
[16] "Climate_Rad_13.img"  "Climate_Rad_14.img"  "Climate_Rad_15.img" 
[19] "Climate_Rad_2.img"   "Climate_Rad_3.img"   "Climate_Rad_4.img"  
[22] "Climate_Rad_5.img"   "Climate_Rad_6.img"   "Climate_Rad_7.img"  
[25] "Climate_Rad_8.img"   "Climate_Rad_9.img"   "Climate_Rad_95.img" 
[28] "Climate_Rad_96.img"  "Climate_Rad_97.img"  "Climate_Rad_98.img" 
[31] "Climate_Rad_99.img" 

# split between the part that comes before the numerics and the "1.img" etc.--adjust appropriately
split <- strsplit(myFiles.sorted, "Climate_Rad_") 
# strip the "1.img" etc such that only the numeric part is left
# turn the characters in numeric
split <- as.numeric(sapply(split, function(x) x <- sub(".img", "", x[2])))
# not you can sort, by using order, that gives the original filenames, ordered on the numeric part of the filename
myFiles.correct.order <- myFiles.sorted[order(split)]

 [1] "Climate_Rad_1.img"   "Climate_Rad_2.img"   "Climate_Rad_3.img"  
 [4] "Climate_Rad_4.img"   "Climate_Rad_5.img"   "Climate_Rad_6.img"  
 [7] "Climate_Rad_7.img"   "Climate_Rad_8.img"   "Climate_Rad_9.img"  
[10] "Climate_Rad_10.img"  "Climate_Rad_11.img"  "Climate_Rad_12.img" 
[13] "Climate_Rad_13.img"  "Climate_Rad_14.img"  "Climate_Rad_15.img"   
[16] "Climate_Rad_95.img"  "Climate_Rad_96.img"  "Climate_Rad_97.img" 
[19] "Climate_Rad_98.img"  "Climate_Rad_99.img"  "Climate_Rad_100.img"
[22] "Climate_Rad_101.img" "Climate_Rad_102.img" "Climate_Rad_103.img"
[25] "Climate_Rad_104.img" "Climate_Rad_105.img" "Climate_Rad_106.img"
[28] "Climate_Rad_107.img" "Climate_Rad_108.img" "Climate_Rad_109.img"
[31] "Climate_Rad_110.img"

这将按照您要查找的顺序为您提供文件.现在根据它来拉取文件,例如

for (fileNames in myFiles.correct.order) {READ.IN.AND.DO.YOUR.THING}

应该这样做.确保根据文件名调整"Climate_Rad_"和".img"(您可能还需要在"Climate_Rad_"之前添加路径,以使其类似于"C:/ filefolder/Climate_Rad_",如果是需要).