在r中，如何根据文件名将固定宽度平面文件的X数读取到Y数据帧中？

Question

因此，我有80个文件以文件名格式出现：

P.A3588.ACO.CCLF0.D00001.TO30000
P.A3588.ACO.CCLF0.D30001.TO60000
...
P.A3588.ACO.CCLF1.D30001.TO60000    
P.A3588.ACO.CCLF1.D30001.TO60000
...
P.A3588.ACO.CCLF9.D30001.TO60000    
P.A3588.ACO.CCLF9.D30001.TO60000

有80个固定宽度的文本文件:10个CCLF数字(CCLF0、CCLF1、.、CCLF9)中每一个都有8个部分。我希望能够按CCLF编号分组，应用列宽向量，添加列名，并绑定CCLF部件的行。

下面是我已经尝试过的。它不起作用，但给了我一个想法，我正在尝试。

filenames <- list.files(dataPath)
names <- substr(filenames,13,17)

CCLF1_width <- c(13,6,11,2,10,10,1,1,7,7,2,17,1,2,2,4,1,10,10,10,10,10,2,10,10,10,11,2,2,1,1,1)
CCLF2_width <- c(13,10,11,2,10,10,4,10,5,11,6,10,10,24,17,2,2,2,2,2)
CCLF3_width <- c(13,11,2,2,7,10,11,6,10,10,1)
CCLF4_width <- c(13,11,2,1,2,7,11,6,10,10,7,1)
CCLF5_width <- c(13,10,11,2,10,10,3,2,2,1,2,10,10,5,15,1,7,10,10,2,2,2,10,10,40,11,17,24,2,2,2,2,2,2,7,7,7,7,7,7,7,7,1)
CCLF6_width <- c(13,10,11,2,10,10,1,2,10,10,5,15,1,10,10,2,2,2,10,10,40,11,17,2)
CCLF7_width <- c(13,11,11,2,10,2,20,1,1,24,9,2,20,13,2,10,10,12,9)
CCLF8_width <- c(11,2,3,5,10,1,1,3,2,2,10,10,10,30,15,40,1,1)
CCLF9_width <- c(11,11,10,10,12)
CCLF0_width <- c(11,11)

for (i in length(filenames)){
  assign(paste0(substr(filenames,13,17)),
  read_fwf(grepl("CCLF1",filenames),
  paste0(i,"_width")))
  }

Answer 1

考虑通过两个等长列表进行映射，并使用Map (包装器到mapply)迭代元素：

cclf_files <- paste0("CCLF", seq(0:9))

cclf_widths <- list(
  CCLF0_width = c(11,11)
  CCLF1_width = c(13,6,11,2,10,10,1,1,7,7,2,17,1,2,2,4,1,10,10,10,10,10,2,10,10,10,11,2,2,1,1,1)
  CCLF2_width = c(13,10,11,2,10,10,4,10,5,11,6,10,10,24,17,2,2,2,2,2)
  CCLF3_width = c(13,11,2,2,7,10,11,6,10,10,1)
  CCLF4_width = c(13,11,2,1,2,7,11,6,10,10,7,1)
  CCLF5_width = c(13,10,11,2,10,10,3,2,2,1,2,10,10,5,15,1,7,10,10,2,2,2,10,10,40,11,17,24,2,2,2,2,2,2,7,7,7,7,7,7,7,7,1)
  CCLF6_width = c(13,10,11,2,10,10,1,2,10,10,5,15,1,10,10,2,2,2,10,10,40,11,17,2)
  CCLF7_width = c(13,11,11,2,10,2,20,1,1,24,9,2,20,13,2,10,10,12,9)
  CCLF8_width = c(11,2,3,5,10,1,1,3,2,2,10,10,10,30,15,40,1,1)
  CCLF9_width = c(11,11,10,10,12)
)

proc_files <- function(f, w) {
    # RETRIEVE FILES WITH CURRENT CCF# IN NAME
    files <- list.files(path = "/path/to/cclf/files", pattern = f)
    print(files)

    # BUILD A LIST OF DFs FROM ALL FILES WITH CURRENT CCF#_width
    df_list <- lapply(files, function(x) {
                   tryCatch(read.fwf(x, widths=w), error = function(e) return(NA))
               })

    # ROW BIND ALL DFs TO FINAL FOR RETURN
    df <- do.call(rbind, df_list)
}

# BUILD A NAMED LIST OF DATA FRAMES FOR EACH CCLF#
df_list <- setNames(Map(proc_files, cclf_files, cclf_widths), cclf_files)

df_list$CCLF0
df_list$CCLF1
df_list$CCLF2
...

Answer 2

这个问题可以用for循环和函数来解决。我编写了函数combine_fwf，将所有固定宽度的文件与给定的CCLF编号组合成一个数据帧。list.files行查找当前工作目录中的所有文件，其中包含您的数字之一(0-9)。然后使用read.fwf读取文件名中的数据，对CCLF数字进行cbind，并将其重新绑定到整个结果。

combine_fwf = function(CCLF_num, colwidths) {
  filenames = list.files(pattern = paste0("CCLF", CCLF_num))
  df_list = vector("list", length(filenames))
  for (i in 1:length(filenames)) {
   df_list[[i]] = cbind.data.frame(CCLF_num, read.fwf(filenames[[i]], 
                                   colwidths)) 
  }
  return(do.call(rbind, df_list))
}

combine_fwf(2, c(12,6))
>>  CCLF_num           V1     V2
1        2 adsfasdfadsf 123123
2        2 lkjhlkjhlkjh  98098
3        2 adsfasdfadsf 123123
4        2 lkjhlkjhlkjh  98098
5        2 adsfasdfadsf 123123
6        2 lkjhlkjhlkjh  98098
> combine_fwf(1, c(12,6))
  CCLF_num           V1     V2
1        1 adsfasdfadsf 123123
2        1 lkjhlkjhlkjh  98098
3        1 adsfasdfadsf 123123
4        1 lkjhlkjhlkjh  98098
5        1 adsfasdfadsf 123123
6        1 lkjhlkjhlkjh  98098