Question

我想根据正则表达式模式从几百个txt文件中提取值，重新排列它们并将它们写入数据框。

文件的开头如下：http://pastebin.com/embed_js.php?i=vdbXfDhC

并以此结尾：http://pastebin.com/embed_js.php?i=hse7SDJd

我之前有过类似的问题（Rearranging the structure of many txt files and then merging them in one data frame） rawr为我提供了这段代码：

(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
})

但是，这段代码使用的是txt文件，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（1/745 ......等）所示，所以我可以不要使用相同的循环。

有人可以帮我调整上面发布的代码来提取

1）MeshSize：XXX [ha] 2）rel。碎片化：XXX 3）MeshSize comp.time [sec]：XXX

对于每个图像（例如，============== 703/745 ==============表示图像名称进入的新图像路径）

与我的其他问题类似，我需要重新排列数据，以便图像名称（以.tif：20130815 225017 957 000000 0892 0464结尾的路径中的字符串）是行名称（我不需要.tif结束）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

编辑为rawr的出色解决方案

如果您希望将整个事物循环并保存为CSV，则可以这样做：

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

我认为这个更直截了当。（或者我可能比其他问题更难了）

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

递归地从文本文件中提取值并循环其中的更多值以及重新排列行和列

1 个答案: