如何读取选项卡分隔的文件到data.table使用fread？ [英] How to read tab separated file into data.table using fread?

问题描述

示例数据（ emp.data ）

Beth  4.00  0
Dan   3.75  0
Kathy 4.00  10
Mark  5.00  20
Mary  5.50  22
Susie 4.25  18

我可以使用 read.table 将其读入 data.frame $ c>，然后将其转换为 data.table ：

I can read it into a data.frame using read.table, then convert it to data.table:

library(data.table)
df <- read.table("emp.data", col.names = c("Name", "PayRate", "HoursWorked"))
DT <- as.data.table(df, key = HoursWorked)

计算工资：

DT[HoursWorked > 0, .(Name, Pay = PayRate * HoursWorked),]

    Name   Pay
1: Kathy  40.0
2:  Mark 100.0
3:  Mary 121.0
4: Susie  76.5

然而，我认为转换有一个额外的步骤。由于在 data.table 中有 fread（），为什么不直接使用？

That works fine; however, I consider there's an extra step in converting. Since there's fread() in data.table, why not use it directly?

readDT <- fread("emp.data", header=FALSE, sep="\t")

               V1
1:  Beth  4.00  0
2:  Dan   3.75  0
3: Kathy 4.00  10
4: Mark  5.00  20
5: Mary  5.50  22
6: Susie 4.25  18

 str(readDT)
Classes 'data.table' and 'data.frame':  6 obs. of  1 variable:
 $ V1: chr  "Beth  4.00  0" "Dan   3.75  0" "Kathy 4.00  10" "Mark  5.00  20" ...
 - attr(*, ".internal.selfref")=<externalptr> 

数据被识别为一列;显然这不行。

The data is recognized as one column; obviously this doesn't work.

问题

如何读取此数据正确使用 fread（）？ （如果可能，也设置列名。）

How to read this data using fread() properly? (If possible, set the column names as well.)

推荐答案

这最近在devel版本v1.9.5 （将很快在CRAN上以v1.9.6提供）：

This has been fixed recently in the devel version, v1.9.5 (will be soon available on CRAN as v1.9.6):

require(data.table) # v1.9.5+
fread("~/Downloads/tmp.txt")
#       V1   V2 V3
# 1:  Beth 4.00  0
# 2:   Dan 3.75  0
# 3: Kathy 4.00 10
# 4:  Mark 5.00 20
# 5:  Mary 5.50 22
# 6: Susie 4.25 18

有关详细信息，请参阅项目页面中的 README.md 。 fread 获得 strip.white 参数（在其他功能/错误修复之间） TRUE 。

See README.md in the project page for more info. fread gained strip.white argument (amidst other functionalities / bug fixes) which is by default TRUE.

更新： col.names 参数now：

fread("~/Downloads/tmp.txt", col.names = c("Name", "PayRate", "HoursWorked"))
#     Name PayRate HoursWorked
# 1:  Beth    4.00           0
# 2:   Dan    3.75           0
# 3: Kathy    4.00          10
# 4:  Mark    5.00          20
# 5:  Mary    5.50          22
# 6: Susie    4.25          18

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