Question

我有以下代码 -

zz3 <- 'data,key
"VA1,VA2,20140524,,0,0,5969,20140523134902,S7,S1147,140,20140523134902,m/t",4503632376496128
"VA2,VA3,20140711,,0,0,8824,20140601095714,S1,S6402,175,20140601095839,m/t",4503643113914368
"VA1,VA3,20140710,,0,0,11678,20140604085203,S1,S1430,250,20140604085329,m/t",4503666467799040
"VA2,VA1,20140724,,0,0,7109,20140523133835,S7,S793,130,20140523133835,m/t",4503679218483200
"VA3,VA1,20140925,,0,0,10592,20140604092548,S7,S109,395,20140604092714,m/t",4503694653521920'

columnClasses <- c("or"="factor", "d"="factor", "ddate"="factor", "rdate"="factor", "changes"="integer", "class"="factor", "price"="integer", "fdate"="factor", "company"="factor", "number"="factor", "dur"="integer", "added"="factor", "source"="factor", "key"="NULL") # skip last column "key"
data <- fread(zz3, header = FALSE, sep = ",", skip = 1, na.strings = c(""), colClasses = columnClasses)

但它返回错误 -

Error in fread(zz3, header = FALSE, sep = ",", skip = 1, na.strings = c(""),  : 
  Column name 'or' in colClasses[[1]] not found

我希望colClasses在header = FALSE时分配列名，但看起来并非如此。

我该如何解决这个问题？类似的read.csv代码效果很好。

Answer 1

确实不是这样。

colClasses可让您使用fread定义列类型。假设您有|分割的文件，其中包含一个名为＆＃39; key＆＃39;的列。如果你想让它成为一个角色，你将运行命令：fread(filePath, sep='|', colClasses=c(key='character'))。

如果文件中没有名称，则可以使用setnames在读取后为data.table分配列名。

Answer 2

您应该将其分为列名和列类

设置列名称应该在单独的步骤中完成。

column_names <-c("or", "d", "ddate", "rdate", "changes", "class", "price", "fdate", "company", "number", "dur", "added", "source", "key") 
column_classes <- c("factor", "factor", "factor", "factor", "integer", "factor", "integer", "factor", "factor", "factor", "integer", "factor", "factor", "NULL") 

data <- fread(zz3, header = FALSE, sep = ",", skip = 1, na.strings = c(""), colClasses = column_classes)
setnames(data, column_names)

如何在R中使用fread分配列名？

2 个答案: