blocks|key|248675|text|对于单个列：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|248676|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+...|code-block|syntax|javascript|248677|248678|使用lapply和as.character|offset|length|style|CODE|248679|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"+...|248680|entityMap^0|0|0|0|2|6|9|C|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|T|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|U|8|@$J|V|K|W|L|M]|$J|X|K|Y|L|M]]|9|@]|A|$]]|$1|N|3|O|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|P|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|Q|$]]

For a single column:

<pre><code>dtnew &lt;- 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 ...
</code></pre>

<hr>

Using <code>lapply</code> and <code>as.character</code>:

<pre><code>dtnew &lt;- 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" ...
</code></pre>

blocks|key|3145102|text|这是一种很糟糕的方式！我只留下这个答案，以防它解决其他奇怪的问题。这些更好的方法可能部分是较新的data.table版本的结果……因此，以这种艰难的方式进行记录是值得的。另外，这是eval+substitute语法的一个很好的语法示例。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3145103|library(data.table)
dt+<-+data.table(ID+=+c(rep("A",+5),+rep("B",5)),+
+++++++++++++++++fac1+=+c(1:5,+1:5),+
+++++++++++++++++fac2+=+c(1:5,+1:5)+*+2,+
+++++++++++++++++val1+=+rnorm(10),
+++++++++++++++++val2+=+rnorm(10))

names_factors+=+c('fac1',+'fac2')
names_values+=+c('val1',+'val2')

for+(col+in+names_factors){
++e+=+substitute(X+:=+as.factor(X),+list(X+=+as.symbol(col)))
++dt[+,+eval(e)]
}
for+(col+in+names_values){
++e+=+substitute(X+:=+as.numeric(X),+list(X+=+as.symbol(col)))
++dt[+,+eval(e)]
}

str(dt)|code-block|syntax|javascript|3145104|这给了你|3145105|Classes+‘data.table’+and+'data.frame':++10+obs.+of++5+variables:
+$+ID++:+chr++"A"+"A"+"A"+"A"+...
+$+fac1:+Factor+w/+5+levels+"1","2","3","4",..:+1+2+3+4+5+1+2+3+4+5
+$+fac2:+Factor+w/+5+levels+"2","4","6","8",..:+1+2+3+4+5+1+2+3+4+5
+$+val1:+num++0.0459+2.0113+0.5186+-0.8348+-0.2185+...
+$+val2:+num++-0.0688+0.6544+0.267+-0.1322+-0.4893+...
+-+attr(*,+".internal.selfref")=<externalptr>+|3145106|entityMap^0|2I|4|2N|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

This is a BAD way to do it! I'm only leaving this answer in case it solves other weird problems. These better methods are the probably partly the result of newer data.table versions... so it's worth while to document this hard way. Plus, this is a nice syntax example for <code>eval</code> <code>substitute</code> syntax.

<pre><code>library(data.table)
dt &lt;- data.table(ID = c(rep("A", 5), rep("B",5)), 
 fac1 = c(1:5, 1:5), 
 fac2 = c(1:5, 1:5) * 2, 
 val1 = rnorm(10),
 val2 = rnorm(10))

names_factors = c('fac1', 'fac2')
names_values = c('val1', 'val2')

for (col in names_factors){
 e = substitute(X := as.factor(X), list(X = as.symbol(col)))
 dt[ , eval(e)]
}
for (col in names_values){
 e = substitute(X := as.numeric(X), list(X = as.symbol(col)))
 dt[ , eval(e)]
}

str(dt)
</code></pre>

which gives you

<pre><code>Classes ‘data.table’ and 'data.frame': 10 obs. of 5 variables:
 $ ID : chr "A" "A" "A" "A" ...
 $ fac1: Factor w/ 5 levels "1","2","3","4",..: 1 2 3 4 5 1 2 3 4 5
 $ fac2: Factor w/ 5 levels "2","4","6","8",..: 1 2 3 4 5 1 2 3 4 5
 $ val1: num 0.0459 2.0113 0.5186 -0.8348 -0.2185 ...
 $ val2: num -0.0688 0.6544 0.267 -0.1322 -0.4893 ...
 - attr(*, ".internal.selfref")=&lt;externalptr&gt; 
</code></pre>

blocks|key|253709|text|我尝试了几种方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|253710|#+BY+{dplyr}
data.table(ID++++++=+c(rep("A",+5),+rep("B",5)),+
+++++++++++Quarter+=+c(1:5,+1:5),+
+++++++++++value+++=+rnorm(10))+->+df1
df1+%25<>%25+dplyr::mutate(ID++++++=+as.factor(ID),
+++++++++++++++++++++++Quarter+=+as.character(Quarter))
#+check+classes
dplyr::glimpse(df1)
#+Observations:+10
#+Variables:+3
#+$+ID++++++(fctr)+A,+A,+A,+A,+A,+B,+B,+B,+B,+B
#+$+Quarter+(chr)+"1",+"2",+"3",+"4",+"5",+"1",+"2",+"3",+"4",+"5"
#+$+value+++(dbl)+-0.07676732,+0.25376110,+2.47192852,+0.84929175,+-0.13567312,++-0.94224435,+0.80213218,+-0.89652819...|code-block|syntax|javascript|253711|，或者其他|253712|#+from+list+to+data.table+using+data.table::setDT
list(ID++++++=+as.factor(c(rep("A",+5),+rep("B",5))),+
+++++Quarter+=+as.character(c(1:5,+1:5)),+
+++++value+++=+rnorm(10))+%25>%25+setDT(list.df)+->+df2
class(df2)
#+[1]+"data.table"+"data.frame"|253713|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I tried several approaches.

<pre class="lang-r prettyprint-override"><code># BY {dplyr}
data.table(ID = c(rep("A", 5), rep("B",5)), 
 Quarter = c(1:5, 1:5), 
 value = rnorm(10)) -&gt; df1
df1 %&lt;&gt;% dplyr::mutate(ID = as.factor(ID),
 Quarter = as.character(Quarter))
# check classes
dplyr::glimpse(df1)
# Observations: 10
# Variables: 3
# $ ID (fctr) A, A, A, A, A, B, B, B, B, B
# $ Quarter (chr) "1", "2", "3", "4", "5", "1", "2", "3", "4", "5"
# $ value (dbl) -0.07676732, 0.25376110, 2.47192852, 0.84929175, -0.13567312, -0.94224435, 0.80213218, -0.89652819...
</code></pre>

, or otherwise

<pre class="lang-r prettyprint-override"><code># from list to data.table using data.table::setDT
list(ID = as.factor(c(rep("A", 5), rep("B",5))), 
 Quarter = as.character(c(1:5, 1:5)), 
 value = rnorm(10)) %&gt;% setDT(list.df) -&gt; df2
class(df2)
# [1] "data.table" "data.frame"
</code></pre>

blocks|key|248700|text|尝尝这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|248701|DT+<-+data.table(X1+=+c("a",+"b"),+X2+=+c(1,2),+X3+=+c("hello",+"you"))
changeCols+<-+colnames(DT)[which(as.vector(DT[,lapply(.SD,+class)])+==+"character")]

DT[,(changeCols):=+lapply(.SD,+as.factor),+.SDcols+=+changeCols]|code-block|syntax|javascript|248702|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this

<pre><code>DT &lt;- data.table(X1 = c("a", "b"), X2 = c(1,2), X3 = c("hello", "you"))
changeCols &lt;- colnames(DT)[which(as.vector(DT[,lapply(.SD, class)]) == "character")]

DT[,(changeCols):= lapply(.SD, as.factor), .SDcols = changeCols]
</code></pre>

blocks|key|253715|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|253716|dt+<-+data.table(A+=+c(1:5),+
+++++++++++++++++B=+c(11:15))

x+<-+ncol(dt)

for(i+in+1:x)+
{
+++++dt[[i]]+<-+as.character(dt[[i]])
}|code-block|syntax|javascript|253717|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

try:

<pre><code>dt &lt;- data.table(A = c(1:5), 
 B= c(11:15))

x &lt;- ncol(dt)

for(i in 1:x) 
{
 dt[[i]] &lt;- as.character(dt[[i]])
}
</code></pre>

blocks|key|250392|text|我提供了一种更通用、更安全的方法来做这些事情，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|250393|".."+<-+function+(x)+
{
++stopifnot(inherits(x,+"character"))
++stopifnot(length(x)+==+1)
++get(x,+parent.frame(4))
}


set_colclass+<-+function(x,+class){
++stopifnot(all(class+%25in%25+c("integer",+"numeric",+"double","factor","character")))
++for(i+in+intersect(names(class),+names(x))){
++++f+<-+get(paste0("as.",+class[i]))
++++x[,+(..("i")):=..("f")(get(..("i")))]
++}
++invisible(x)
}|code-block|syntax|javascript|250394|函数..确保我们得到一个超出data.table作用域的变量；set_colclass将设置您的cols的类。你可以这样使用它：|offset|length|style|CODE|250395|dt+<-+data.table(i=1:3,f=3:1)
set_colclass(dt,+c(i="character"))
class(dt$i)|250396|entityMap^0|0|0|2|2|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I provide a more general and safer way to do this stuff,

<pre><code>".." &lt;- function (x) 
{
 stopifnot(inherits(x, "character"))
 stopifnot(length(x) == 1)
 get(x, parent.frame(4))
}


set_colclass &lt;- function(x, class){
 stopifnot(all(class %in% c("integer", "numeric", "double","factor","character")))
 for(i in intersect(names(class), names(x))){
 f &lt;- get(paste0("as.", class[i]))
 x[, (..("i")):=..("f")(get(..("i")))]
 }
 invisible(x)
}
</code></pre>

The function <code>..</code> makes sure we get a variable out of the scope of data.table; set_colclass will set the classes of your cols.
You can use it like this:

<pre><code>dt &lt;- data.table(i=1:3,f=3:1)
set_colclass(dt, c(i="character"))
class(dt$i)
</code></pre>

blocks|key|3145169|text|如果您在data.table中有一个列名列表，则需要更改do的类：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3145170|convert_to_character+<-+c("Quarter",+"value")

dt[,+convert_to_character]+<-+dt[,+lapply(.SD,+as.character),+.SDcols+=+convert_to_character]|code-block|syntax|javascript|3145171|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you have a list of column names in data.table, you want to change the class of do:

<pre><code>convert_to_character &lt;- c("Quarter", "value")

dt[, convert_to_character] &lt;- dt[, lapply(.SD, as.character), .SDcols = convert_to_character]
</code></pre>

blocks|key|253751|text|下面是@Nera建议先检查类的相同方法，但不是使用.SD，而是使用data.table和set的快速循环作为@Matt+Dowle解决方案，并添加了类检查。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|253752|for+(j+in+seq_len(ncol(DT))){
++if(class(DT[[j]])+==+'factor')
++++set(DT,+j+=+j,+value+=+as.character(DT[[j]]))
}|code-block|syntax|javascript|253753|entityMap^0|P|3|18|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Here is the same way as @Nera suggested to check the class first but instead of using <code>.SD</code> is to use the fast loop of data.table with <code>set</code> as @Matt Dowle solution with added class check.
<pre><code>for (j in seq_len(ncol(DT))){
 if(class(DT[[j]]) == 'factor')
 set(DT, j = j, value = as.character(DT[[j]]))
}
</code></pre>

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:

<pre><code>df &lt;- 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 &lt;- data.frame(lapply(df, as.character), stringsAsFactors=FALSE)
#Another way
df[, "value"] &lt;- as.numeric(df[, "value"])

library(data.table)
dt &lt;- data.table(ID=c(rep("A", 5), rep("B",5)), Quarter=c(1:5, 1:5), value=rnorm(10))
dt &lt;- 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] &lt;- as.character(dt[, "ID", with=FALSE]) 
#Produces error: Error in `[&lt;-.data.table`(`*tmp*`, , "ID", with = FALSE, value = "c(1, 1, 1, 1, 1, 2, 2, 2, 2, 2)") : 
#unused argument(s) (with = FALSE)
</code></pre>

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:

<pre><code>classes &lt;- as.character(sapply(df, class))
colClasses &lt;- which(classes=="factor")
df[, colClasses] &lt;- sapply(df[, colClasses], as.character)
</code></pre>

Can I do something similar with data.table?

Update 3: 

<blockquote>
 sessionInfo()
 R version 2.13.1 (2011-07-08)
 Platform: x86_64-pc-mingw32/x64 (64-bit)
</blockquote>

<pre><code>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
</code></pre>

Convert column classes in data.table

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我在使用data.table时遇到了一个问题:如何转换列类？下面是一个简单的例子:对于data.frame，我没有问题，对于data.table，我只是不知道如何进行转换：df <- data.frame(ID=c(rep("A", 5), rep("B",5)), Quarter=c(1:5, 1:5), value...

问转换data.table中的列类
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问转换data.table中的列类
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