GCalignR R预处理数据中的字符串分裂问题

Question

尝试获取多个.txt (GC)文件，取出两列数据并将它们分配给我最终可以传递给GCalignR的对象。是否有更好的方法为GCalignR处理这些数据？

引起问题的辅助功能：

“”“

# nonempty strings after splitting a row by one space
get_nonempty_splits = function(row) {
  s = strsplit(row, "[[:space:]]")
  s = s[[1]]
  l = c()
  for (i in 1:length(s)) {
    if (s[i] != "") {
      l = c(l, s[i])
    }
  }
  return(l)
}

# filenames have .txt, names do not
filenames_to_names = function(x) {
  l = c()
  for (i in 1:length(x)) {
    x1 = strsplit(x, "[.]")[[i]][1]
    l = c(l, x1)
  }
  return(l)
}

# get data row indices
get_data_row_inds = function(df) {
  ind_start = 0
  ind_end = 0
  for (i in 1:length(df)) {
    row = df[i]
    # find start
    if (grepl("----", row)) {
      stopifnot(ind_start == 0)  # assert ind_start not set
      ind_start = i+1
    }
    # find end
    if (i == length(df) && ind_end == 0) {
      ind_end = length(df)
    } else if (grepl("Totals", row) && grepl(":", row)) {
      stopifnot(ind_end == 0)  # assert ind_end not set
      ind_end = i-1
    }
  }
  stopifnot(ind_start != 0)
  stopifnot(ind_end != 0)
  return(ind_start:ind_end)
}

“”“

“”“

path_to_raw_data = "/Users/input"
path_to_processed_data = "."
# get paths for all .txt files in pathToRawData directory
paths = list.files(path_to_raw_data, full.name=TRUE, pattern = ".txt")
filenames = list.files(path_to_raw_data, full.name=FALSE, pattern = ".txt")
names = filenames_to_names(filenames)  # without .txt extension

# get data from text file
processed_data = list()
df_lengths = c()
for (i in 1:length(paths)) {  # i indexes the raw files
  path = paths[i]
  df = read.delim(path, fileEncoding= 'UTF-16LE', header=TRUE)
  df = df[[1]]
  inds = get_data_row_inds(df)
  df_lengths = c(df_lengths, length(inds))
  times = c()
  areas = c()
  for (j in inds) {  # j indexes the data rows of a raw file
    row = df[j]
    row = get_nonempty_splits(row)
    time = row_to_time(row)
    area = row_to_area(row)
    times = c(times, time)
    areas = c(areas, area)
  }
  pairs = data.frame(time = times, area = areas)
  processed_data[[i]] = pairs
}

“”“

获取此错误:str拆分中的错误(行，"[:space:]")：非字符参数

有什么解决办法的建议吗？是文件编码吗？数据=list()不返回任何.？

输入标题：“

dput(头(Df))结构(c(59L，53L，45L，48L，47L，52L)，.Label = c(“Inj体积:1l "，”* Report *“，”面积百分比报告“，“1 1.353 1 BB 2.85703 2.453 e-5？"，”2 1.952 1 BV 4411.39551 0.03787？"，“3 2.058 1 VV 4693.20215 0.04029？"，”4 2.089 1 VV 6614.89502 0.05679？"，“5 2.139 2 0.00000 0.00000 NG "，”6 2.452 2 0.00000 0.00000 1，3 DNB "，“7 3.149 2 0.00000 0.00000 2，4-DNT，“8 3.315 1 VV S 1.15784e7 99.39858？"，”9 3.347 1 VV S 5169.44629 0.04438？"，“# min %"，”10 3.372 1 VV S 2.09449e4 0.17981？"，“11 3.466 1 VV S 2535.17432 0.02176？"，“12 3.547 1 VB S 2.45685e4 0.21092？"，”13 3.602 1 BV T 451.00174 0.00387？"，“14 3.686 1 VV T 40.45324 0.00035？"，”15 3.734 1 VV T 13.40936 0.00012？"，“16 3.819 1 VB T 508.57788 0.00437？"，”17 4.119 1 BB 13.01144 0.00011？"，“18 4.856 2 0.00000 0.00000 TNT "，”19 4.975 2 0.00000 0.00000 TNB "，“20 5.549 2 0.00000 0.00000 4-上午-DNT "，”21 5.869 2 0.00000 0.00000 RDX "，“22 5.943 2 0.00000 0.00000 2-Am-DNT "，”23 6.516 2 0.00000 0.00000四元"，“24 11.716 1 BB 1.75858 1.510 e-5？"，”25 14.243 1 BB 2.55644 2.195 e-5？"，“26 16.654 1 BB 3.81723 3.277 e-5？"，”27 18.826 1 BB 2.58369 2.218 e-5？"，“28 20.800 1 BB 1.51171 1.298 e-5？"，”29 24.159 1 BB 1.78975 1.536 e-5？"，“30 24.269 1 BB 1.81180 1.555e-5？"，”31 25.053 1 BB 2.96617 2.546 e-5？"，“32 25.658 1 BB 6.15337 5.283 e-5？"，”33 25.809 1 BB 3.89435 3.343 e-5？"，“34 26.577 1 BB 4.02199 3.453 e-5？"，”35 26.885 1 BB 2.48416 2.133 e-5？"，“36 27.219 1 BB 14.88012 0.00013？"，”37 27.465 1 BB 3.59732 3.088 e-5？"，“38 29.377 1 BB 18.55422 0.00016？"，”39 32.554 1 BB 17.15620 0.00015？"，"----|-------|---|------|----------|--------|-------------------------"，"====================================================================="，"2警告或错误:"，"Acq。仪器:仪器1位置:11小瓶，"Acq“。接线员: HHV Seq。线: 2"，“杯”。数据修改:2019年3月12日(星期二)- 6:13:25 PM、“稀释: 1.0000”、“不对ISTDs使用乘数和稀释系数”、“注射日期:24-2月20日、14:37:34 Inj : 1”、“仪器1 2/24/2020 3:13:35 PM HHV”、“最后一次更改: 2/6/2020 12:59:45下午被HHV"，“方法: C:\Chem32\1\DATA\IPOULIN\VOC_TEST_1 2020-02-24 13-49-15\VOC_TEST_HV.M”、“方法信息: VOC”、“乘数: 1.0000”、“峰值RetTime Sig类型面积名称”、“示例名称: P1U1十六进制022420"，“序列文件: C:\Chem32\1\DATA\IPOULIN\VOC_TEST_1 2020-02-24 13-49-15\VOC_TEST_1.S"，”信号1: FID1 B "，“按保留时间排序"，”总计: 1.16485e7“，”警告:未找到已校准的化合物“，”警告:校准警告(见校准表列表)“，类=”因子“)

Answer 1

通过添加: df =read.delim(路径，fileEncoding= 'UTF-16LE'，header=TRUE，stringsAsFactors = FALSE)进行求解。"stringsAsFactors = FALSE“是解决方案。谢谢。