转换data.table中的列类 [英] Convert column classes in data.table
问题描述
我使用data.table时遇到问题:如何转换列类?这里是一个简单的例子:with data.frame我没有转换它的问题,与data.table我不知道如何:
df < - data.frame(ID = c(rep(A,5),rep(B,5)),Quarter = c(1:5,1:5) value = rnorm(10))
#一种方式:http://stackoverflow.com/questions/2851015/r-convert-data-frame-columns-from-factors-tocharacters
df< ; - data.frame(lapply(df,as.character),stringsAsFactors = FALSE)
#Another way
df [,value]< - as.numeric(df [,value ])
library(data.table)
dt< - data.table(ID = c(rep(A,5),rep(B,5)) ,Quarter = c(1:5,1:5),value = rnorm(10))
dt #Error in rep(,ncol(xi)):无效的'times'参数
#产生错误,data.table没有选项stringsAsFactors?
dt [,ID,with = FALSE]< - as.character(dt [,ID,with = FALSE])
#Produces错误: data.table`(`* tmp *`,,ID,其中= FALSE,value =c(1,1,1,1,1,2,2,2,2,2)):
#unused argument(s)(with = FALSE)
由于Matthew的帖子更新:我以前使用过较旧的版本,但即使更新到1.6.6(我现在使用的版本)后,我仍然得到一个错误。 / p>
更新2:假设我想将类factor的每一列转换为一个字符列,但不提前知道哪个列是哪个类。使用data.frame,我可以执行以下操作:
classes< - as.character(sapply(df,class) )
colClasses< - which(classes ==factor)
df [,colClasses]< - sapply(df [,colClasses],as.character)
我可以使用data.table类似的东西吗?
更新3: / p> sessionInfo() 对于单个列: 使用 I have a problem using data.table: How do I convert column classes? Here is a simple example: With data.frame I don't have a problem converting it, with data.table I just don't know how: Do I miss something obvious here? Update due to Matthew's post: I used an older version before, but even after updating to 1.6.6 (the version I use now) I still get an error. Update 2: Let's say I want to convert every column of class "factor" to a "character" column, but don't know in advance which column is of which class. With a data.frame, I can do the following: Can I do something similar with data.table? Update 3: sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-pc-mingw32/x64 (64-bit)
For a single column:
Using
这篇关于转换data.table中的列类的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
R版本2.13.1(2011-07-08)
平台:x86_64-pc- mingw32 / x64(64位)
locale:
[1] C
附加的基本包:
[1] stats graphics grDevices utils数据集方法base
其他附加包:
[1] data.table_1.6.6
通过命名空间加载(未附加):
[1] tools_2.13.1
dtnew < dt [,Quarter:= as.character(Quarter)]
str(dtnew)
类'data.table'和'data.frame':10 obs。的3个变量:
$ ID:因子w / 2级别A,B:1 1 1 1 1 2 2 2 2 2
$季度:chr12 4 ... ...
$ value:num -0.838 0.146 -1.059 -1.197 0.282 ...
lapply
和 as.character
:
dtnew <-dt [,lapply(.SD,as.character),by = ID]
str dtnew)
类'data.table'和'data.frame':10 obs。的3个变量:
$ ID:因子w / 2级别A,B:1 1 1 1 1 2 2 2 2 2
$季度:chr12 4 ... ...
$ value:chr1.487145280568-0.8278452183588810.0289771827700021.35392750102305...
df <- data.frame(ID=c(rep("A", 5), rep("B",5)), Quarter=c(1:5, 1:5), value=rnorm(10))
#One way: http://stackoverflow.com/questions/2851015/r-convert-data-frame-columns-from-factors-to-characters
df <- data.frame(lapply(df, as.character), stringsAsFactors=FALSE)
#Another way
df[, "value"] <- as.numeric(df[, "value"])
library(data.table)
dt <- data.table(ID=c(rep("A", 5), rep("B",5)), Quarter=c(1:5, 1:5), value=rnorm(10))
dt <- data.table(lapply(dt, as.character), stringsAsFactors=FALSE)
#Error in rep("", ncol(xi)) : invalid 'times' argument
#Produces error, does data.table not have the option stringsAsFactors?
dt[, "ID", with=FALSE] <- as.character(dt[, "ID", with=FALSE])
#Produces error: Error in `[<-.data.table`(`*tmp*`, , "ID", with = FALSE, value = "c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2)") :
#unused argument(s) (with = FALSE)
classes <- as.character(sapply(df, class))
colClasses <- which(classes=="factor")
df[, colClasses] <- sapply(df[, colClasses], as.character)
locale:
[1] C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] data.table_1.6.6
loaded via a namespace (and not attached):
[1] tools_2.13.1
dtnew <- dt[, Quarter:=as.character(Quarter)]
str(dtnew)
Classes ‘data.table’ and 'data.frame': 10 obs. of 3 variables:
$ ID : Factor w/ 2 levels "A","B": 1 1 1 1 1 2 2 2 2 2
$ Quarter: chr "1" "2" "3" "4" ...
$ value : num -0.838 0.146 -1.059 -1.197 0.282 ...
lapply
and as.character
:dtnew <- dt[, lapply(.SD, as.character), by=ID]
str(dtnew)
Classes ‘data.table’ and 'data.frame': 10 obs. of 3 variables:
$ ID : Factor w/ 2 levels "A","B": 1 1 1 1 1 2 2 2 2 2
$ Quarter: chr "1" "2" "3" "4" ...
$ value : chr "1.487145280568" "-0.827845218358881" "0.028977182770002" "1.35392750102305" ...