的test.txt

Question

我写了一个代码来对我的大数据集进行排序。我测试了一个文件的代码（例如“K_7-1_0H.TXT”），它完全正常，没有任何警告和错误消息。我得到了我想要的正确.csv文件。但是当我运行循环时，出现错误消息：

Error in `[.default`(data, 1, ) : incorrect number of dimensions

这是我的代码：

kcl <- list.files(pattern = "K_.*\\.TXT",recursive = TRUE)
c <- c(0, 0.001,0.01,0.05,0.25,0.5,1.0,1.5,2.0,2.5,3.0,3.4)
Kconc. <- rep(c,each=8)
for (i in kcl) {
  data <- read.csv(i,header=FALSE,sep = ",")
  data <- data[-c(1:14),] #delete noise info
  colnames(data) <- as.character(unlist(data[1,])) #add column name
  data <- data[-1,]
  data <- data[,-1]
  #change cell number and sort
  data$`Well Label` <- as.character(data$`Well Label`)
  data[1,1] <- "T01"
  data[12,1] <- "T02"
  ...
  data[89,1] <- "T09"
  data <- arrange(data, `Well Label`) #sort according to table number
  data <- data[-1,] #delete noise info
  data <- cbind(data,Kconc.)
  j <- sub("\\.[[:alnum:]]+$","",i) #grep the isolate name without the extention
  write.csv(data, paste0(j,".csv"))
}

这里是列表内容“

> kcl
 [1] "K_10-1_0.TXT" "K_10-3_0.TXT"
 [9] "K_10-3_6.TXT" "K_10-3_7.TXT" "K_11-1_8.TXT"
[17] "K_11-2_0.TXT" "K_11-3_8.TXT"
[25] "K_7-1_0H.TXT" "K_7-3_82.TXT" "K_8-1_0H.TXT" "K_8-1_60.TXT" "K_8-1_72.TXT" "K_8-1_84.TXT"
[49] "K_9-1_0Z.TXT" "K_9-1_60.TXT" "K_9-1_72.TXT" "K_9-1_84.TXT" "K_9-2-84.TXT"

当我检查我的文件时，创建了像“K_10 * .csv”和“K_11 * .csv”这样的文件，我得到了我想要的东西，但文件如“K_7 * .TXT”，“K_8 * .TXT”和“K_9 * .TXT”根本不起作用，这意味着我甚至没有为这些文件创建.csv。

我真的不明白错误信息以及代码仅适用于某些文件的原因。有人能帮助我吗？

编辑输入并期待输出
输入是.txt文件，如下所示：

[Assay],C:\REVEL\650-S.ASY
"Assay title",Untitled Assay
"Read Time",11.04.17,13:04:00
"Operator",
"Comments",
"Kit Lot Number",,
"Wells",A1 - H12
OD RESULTS
"Units",OD

[Results],Results are sorted on Sample ID,in ascending order

"Sample ID","Well Label","OD Results"
"T1","T1",0.045
"T10","T10",0.044
"T11","T11",0.045
"T2","T2",0.045

预期产出：

    Well Label  OD Results  Hconc.
2   T01 0.189   0
3   T02 0.11    0
4   T03 0.151   0
5   T04 0.053   0

Answer 1

试试这个例子：

library(dplyr)

# skip info rows
df1 <- read.csv("test.txt", skip = 12, stringsAsFactors = FALSE)
Kconc <- c(0, 0.001,0.01,0.05,0.25,0.5,1.0,1.5,2.0,2.5,3.0,3.4)

# Prerix with zero, e.g.: T1 to T01, then sort
res <- 
  df1 %>% 
  transmute(
    `Well Label` = if_else(nchar(df1$Sample.ID) == 2,
                           paste0(substr(df1$Sample.ID, 1, 1),
                                  0,
                                  substr(df1$Sample.ID, 2, 3)),
                           df1$Sample.ID),
    `OD Results` = OD.Results) %>% 
  arrange(`Well Label`)
res
#   Well Label OD Results
# 1        T01      0.045
# 2        T02      0.045
# 3        T10      0.044
# 4        T11      0.045

然后cbind Kconc，不再需要rep回收。在此示例中，我们只有4行，因此要获得此示例的正确结果，我们需要使用res <- cbind(res, Kconc[1:4])。

res <- cbind(res, Kconc)

另外，请阅读gtools的自然顺序：

  df1[ gtools::mixedorder(df1$Sample.ID), ]
#   Sample.ID Well.Label OD.Results
# 1        T1         T1      0.045
# 4        T2         T2      0.045
# 2       T10        T10      0.044
# 3       T11        T11      0.045

的test.txt

[Assay],C:\REVEL\650-S.ASY
"Assay title",Untitled Assay
"Read Time",11.04.17,13:04:00
"Operator",
"Comments",
"Kit Lot Number",,
"Wells",A1 - H12
OD RESULTS
"Units",OD

[Results],Results are sorted on Sample ID,in ascending order

"Sample ID","Well Label","OD Results"
"T1","T1",0.045
"T10","T10",0.044
"T11","T11",0.045
"T2","T2",0.045

错误：维数不正确

1 个答案:

的test.txt