从文本文件中递归提取值并遍历更多值并重新排列行和列

Question

我想根据正则表达式模式从数百个 txt 文件中提取值，重新排列它们并将它们写入数据框。

文件的开头是这样的：http://pastebin.com/embed_js.php?i=vdbXfDhC

然后这样结束：http://pastebin.com/embed_js.php?i=hse7SDJd

我之前有过类似的问题 () rawr 向我提供此代码的地方：

(lf <- list.files('~/desktop', pattern = '^image\d+.txt', full.names = TRUE))
# [1] "/Users/rawr/desktop/image001.txt" "/Users/rawr/desktop/image002.txt"
# [3] "/Users/rawr/desktop/image003.txt"

res <- lapply(lf, function(xx) {
  rl <- readLines(con <- file(xx), warn = FALSE)
  close(con)
  img_name <- gsub('.*file:\s+(.*).tif', '\1', rl[1])
  rl <- rl[-(1:grep('==', rl))]
  rl <- gsub('^\s+', '', rl)
  mat <- do.call('rbind', strsplit(rl, '\s{2, }'))
  dat <- as.data.frame(mat, stringsAsFactors = FALSE)
  tmp <- `colnames<-`(do.call('rbind', strsplit(dat$V2, '[-\/\s]+', perl = TRUE)),
                      c('Foreground','Data pixels'))
  dat <- cbind(dat[, -2], tmp, image_name = img_name)
  dat[] <- lapply(dat, as.character)
  dat[dat == ''] <- NA
  names(dat)[1:2] <- c('MSPA-class','Frequency')

  zzz <- reshape(dat, direction = 'wide', idvar = 'image_name', timevar = 'MSPA-class')
  names(zzz)[-1] <- gsub('(.*)\.(.*) (?:.*)', '\2_\1', names(zzz)[-1], perl = TRUE)
  zzz
})

然而，这段代码使用的是 txt 文件，其中每个文件只有一个分析步骤，现在我在一个 .log 文件中有很多分析，如 pastebin 示例所示（1/745 ...等）所以我不能使用相同的循环。

有人可以帮我修改上面发布的代码以提取

1) 网目尺寸：XXX [ha] 2）相对。碎片：XXX 3) 网格大小 comp.time [秒]: XXX

对于每个图像（例如，============== 703/745 ============== 表示具有图像名称的新图像进来的路径）

与我的其他问题类似，我需要重新排列数据，以便图像名称（路径中以 .tif 结尾的字符串：20130815 225017 957 000000 0892 0464）是行名称（我不需要.tif 结尾) 和 1) MeshSize[ha], 2) rel。碎片 3) MeshSize comp.time [sec] 是列。

image name    1) mesh size 2).....  3)......
   row1              xx        xx      xx
   row2

编辑到rawr的绝妙解决方案

如果你想把整个事情循环并保存为 CSV，你可以这样做：

lf =  list.files(path="xx", pattern = '^batch_mesh8\d.log', full.names = TRUE)

mesh2<-NULL

for (i in lf)
{

#rawr's code here:

#final lines of code:
mesh1<-cbind(data.frame('image_name' = img_names), mat)
mesh2 <- rbind(mesh2, mesh1)
}
write.csv(mesh2, file = "all_mesh_th8.csv")

Answer 1

我觉得这个更直接。（或者也许我让另一个问题比原来更难了）

path <- '~/desktop/log.log'
x <- readLines(con <- file(path))
close(con)

# m <- gregexpr('(\d+/\d+)', x, perl = TRUE)
# img_names <- head(unlist(regmatches(x, m)), -1)
# completed <- tail(img_names, 1)

y <- x[grepl('File', x)]
img_names <- basename(gsub('File: ', '', gsub('\\+','/', y), perl = TRUE))
img_names <- gsub('\.([[:alnum:]]+)$','', img_names)

(x <- x[grepl('\d+\.\d+', x)])

# [1] "MeshSize: 0.30289606 [ha]; rel. fragmentation: 83.1300" 
# [2] "MeshSize comp.time [sec]:    0.00099992752"             
# [3] "MeshSize: 0.39157622 [ha]; rel. fragmentation: 81.4600" 
# [4] "MeshSize comp.time [sec]:    0.00099992752"             
# [5] "MeshSize: 0.45971902 [ha]; rel. fragmentation: 76.8700" 
# [6] "MeshSize comp.time [sec]:       0.00000000"             
# [7] "MeshSize: 0.032965344 [ha]; rel. fragmentation: 94.5500"
# [8] "MeshSize comp.time [sec]:       0.00000000"             
# [9] "MeshSize: 0.034653125 [ha]; rel. fragmentation: 93.6300"
# [10] "MeshSize comp.time [sec]:       0.00000000"             
# [11] "MeshSize: 0.74313322 [ha]; rel. fragmentation: 90.2700" 
# [12] "MeshSize comp.time [sec]:    0.00099992752"             
# [13] "MeshSize: 0.48677515 [ha]; rel. fragmentation: 85.5700" 
# [14] "MeshSize comp.time [sec]:    0.00099992752" 

nums <- unlist(regmatches(x, gregexpr('\d+\.\d+', x, perl = TRUE)))
mat <- matrix(nums, ncol = 3, byrow = TRUE, 
              dimnames = list(NULL, c('Mesh size','rel frag','comp time')))
cbind(data.frame('image_name' = img_names), mat)

#                               image_name   Mesh size rel frag     comp time
# 1 20130815 143656  507  000000 0952 0536  0.30289606  83.1300 0.00099992752
# 2 20130815 143657  673  000002 0244 0284  0.39157622  81.4600 0.00099992752
# 3 20130815 143657  706  000000 0764 0304  0.45971902  76.8700    0.00000000
# 4 20130815 143658  806  000000 0776 0672 0.032965344  94.5500    0.00000000
# 5 20130815 143700  005  000000 0232 0116 0.034653125  93.6300    0.00000000
# 6 20130815 225020  589  000000 0188 0564  0.74313322  90.2700 0.00099992752
# 7 20130815 225033  917  000000 0288 0804  0.48677515  85.5700 0.00099992752

从文本文件中递归提取值并遍历更多值并重新排列行和列

Recursively extracting values from within a text file and looping over more of them plus rearranging rows and columns

regex

r

function

lapply

dataframe