Question

我有一个类似这样的数据

1.1.1.1     Alcohol dehydrogenase.
1.1.1.2     Alcohol dehydrogenase (NADP(+)).
1.1.1.3     Homoserine dehydrogenase.
1.1.1.4     (R,R)-butanediol dehydrogenase.
1.1.1.5     Transferred entry: 1.1.1.303 and 1.1.1.304.
1.1.1.6     Glycerol dehydrogenase.
1.1.1.7     Propanediol-phosphate dehydrogenase.
1.1.1.8     Glycerol-3-phosphate dehydrogenase (NAD(+)).
1.1.1.9     D-xylulose reductase.
1.1.1.10    L-xylulose reductase.

我用read.table这样加载它

df <- read.table("path to data", header=F, fill=T)

我得到以下数据

df <- structure(list(V1 = structure(c(1L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 
10L, 2L), .Label = c("1.1.1.1", "1.1.1.10", "1.1.1.2", "1.1.1.3", 
"1.1.1.4", "1.1.1.5", "1.1.1.6", "1.1.1.7", "1.1.1.8", "1.1.1.9"
), class = "factor"), V2 = structure(c(2L, 2L, 6L, 1L, 9L, 4L, 
8L, 5L, 3L, 7L), .Label = c("(R,R)-butanediol dehydrogenase.", 
"Alcohol", "D-xylulose", "Glycerol", "Glycerol-3-phosphate", 
"Homoserine", "L-xylulose", "Propanediol-phosphate", "Transferred"
), class = "factor"), V3 = structure(c(3L, 2L, 3L, 1L, 4L, 3L, 
3L, 2L, 5L, 5L), .Label = c("", "dehydrogenase", "dehydrogenase.", 
"entry:", "reductase."), class = "factor"), V4 = structure(c(1L, 
3L, 1L, 1L, 4L, 1L, 1L, 2L, 1L, 1L), .Label = c("", "(NAD(+)).", 
"(NADP(+)).", "1.1.1.303"), class = "factor"), V5 = structure(c(1L, 
1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L), .Label = c("", "and"), class = "factor"), 
    V6 = structure(c(1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L), .Label = c("", 
    "1.1.1.304."), class = "factor")), class = "data.frame", row.names = c(NA, 
-10L))

我使用fill = T是因为如果不这样做，它将给我错误

df <- read.table("path/example.txt", header=F, fill=F)
Error in scan(file = file, what = what, sep = sep, quote = quote, dec = dec,  : 
  line 1 did not have 6 elements

有没有一种方法可以加载数据并具有两列？或将数据汇总在一起，以便在R中有两列？

请注意，我可以使用read.delim做到这一点，但是它将使我正在使用的另一个代码出现问题

我的愿望输出就像

Answer 1

不好看，但这是可能的解决方法：

# seperator is multi-space, but not possible in R
file_data <- "1.1.1.1     Alcohol dehydrogenase.
1.1.1.2     Alcohol dehydrogenase (NADP(+)).
1.1.1.3     Homoserine dehydrogenase.
1.1.1.4     (R,R)-butanediol dehydrogenase.
1.1.1.5     Transferred entry: 1.1.1.303 and 1.1.1.304.
1.1.1.6     Glycerol dehydrogenase.
1.1.1.7     Propanediol-phosphate dehydrogenase.
1.1.1.8     Glycerol-3-phosphate dehydrogenase (NAD(+)).
1.1.1.9     D-xylulose reductase.
1.1.1.10    L-xylulose reductase."

# change sep from 4 spaces to \t, which is identifiable. 
# replace textConnection(file_data) with your data file name
read_text <- readLines(textConnection(file_data ))
altered_text <- gsub("    ", "\t", read_text)

# parsing from altered text
df <- read.delim(textConnection(altered_text), header=FALSE, sep="\t", fill=TRUE)
df

问题是您的分隔符超过一个字符（http://r.789695.n4.nabble.com/multiple-separators-in-sep-argument-for-read-table-td856567.html）。

替代方法是更改预加载日期，以在列之间使用通用分隔符。否则，请按原样读取数据，然后添加数据步骤以将第2列向前连接到1列，例如使用paste。

Answer 2

使用基础R，可以将Reduce()与paste()一起使用，然后使用trimws()修剪空白以产生另一个data.frame：

df2 <- data.frame(V1 = df[1], V2 = trimws(Reduce(paste, df[-1])))

> df2
         V1                                           V2
1   1.1.1.1                       Alcohol dehydrogenase.
2   1.1.1.2             Alcohol dehydrogenase (NADP(+)).
3   1.1.1.3                    Homoserine dehydrogenase.
4   1.1.1.4              (R,R)-butanediol dehydrogenase.
5   1.1.1.5  Transferred entry: 1.1.1.303 and 1.1.1.304.
6   1.1.1.6                      Glycerol dehydrogenase.
7   1.1.1.7         Propanediol-phosphate dehydrogenase.
8   1.1.1.8 Glycerol-3-phosphate dehydrogenase (NAD(+)).
9   1.1.1.9                        D-xylulose reductase.
10 1.1.1.10                        L-xylulose reductase.

如何将数据合并为一个？

2 个答案: