blocks|key|919952|text|在将因子更改为数字时，必须小心。下面这行代码将一组列从factor更改为numeric。我在这里假设要更改为数字的列分别为1、3、4和5。您可以相应地更改它|type|unstyled|depth|inlineStyleRanges|entityRanges|data|919953|cols+=+c(1,+3,+4,+5);++++
df[,cols]+=+apply(df[,cols],+2,+function(x)+as.numeric(as.character(x)));|code-block|syntax|javascript|919954|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|$]]

You have to be careful while changing factors to numeric. Here is a line of code that would change a set of columns from factor to numeric. I am assuming here that the columns to be changed to numeric are 1, 3, 4 and 5 respectively. You could change it accordingly

<pre><code>cols = c(1, 3, 4, 5); 
df[,cols] = apply(df[,cols], 2, function(x) as.numeric(as.character(x)));
</code></pre>

blocks|key|1134525|text|根据Ramnath的回答，您正在经历的行为是由于as.numeric(x)在R级别返回因子x的内部数字表示。如果您希望保留作为因子级别的数字(而不是其内部表示)，则需要首先按照Ramnath的示例通过as.character()转换为字符。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1134526|您的for循环与apply调用一样合理，并且可能比代码的意图更具可读性。只需更改这一行：|1134527|stats[,i]+<-+as.numeric(stats[,i])|code-block|syntax|javascript|1134528|阅读|1134529|stats[,i]+<-+as.numeric(as.character(stats[,i]))|1134530|这是R常见问题中的FAQ+7.10。|1134531|HTH|1134532|entityMap|0|LINK|mutability|MUTABLE|url|http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-do-I-convert-factors-to-numeric_003f^0|O|D|19|1|2S|E|0|2|3|8|5|0|0|0|0|9|8|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1E|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1F|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|1G|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1H|8|@]|D|@$9|1I|A|1J|1|1K]]|E|$]]|$1|S|3|T|5|6|7|1L|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|1M|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]]]

Further to Ramnath's answer, the behaviour you are experiencing is that due to <code>as.numeric(x)</code> returning the internal, numeric representation of the factor <code>x</code> at the R level. If you want to preserve the numbers that are the levels of the factor (rather than their internal representation), you need to convert to character via <code>as.character()</code> first as per Ramnath's example.

Your <code>for</code> loop is just as reasonable as an <code>apply</code> call and might be slightly more readable as to what the intention of the code is. Just change this line:

<pre><code>stats[,i] &lt;- as.numeric(stats[,i])
</code></pre>

to read

<pre><code>stats[,i] &lt;- as.numeric(as.character(stats[,i]))
</code></pre>

This is <a href="http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-do-I-convert-factors-to-numeric_003f" rel="noreferrer">FAQ 7.10</a> in the R FAQ.

HTH

blocks|key|920080|text|这可以在一行中完成，不需要循环，无论是for循环还是apply循环。改用unlist()：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|920081|#+testdata
Df+<-+data.frame(
++x+=+as.factor(sample(1:5,30,r=TRUE)),
++y+=+as.factor(sample(1:5,30,r=TRUE)),
++z+=+as.factor(sample(1:5,30,r=TRUE)),
++w+=+as.factor(sample(1:5,30,r=TRUE))
)
##

Df[,c("y","w")]+<-+as.numeric(as.character(unlist(Df[,c("y","w")])))

str(Df)|code-block|syntax|javascript|920082|编辑:对于您的代码，这将变成：|920083|id+<-+c(1,3:ncol(stats)))+
stats[,id]+<-+as.numeric(as.character(unlist(stats[,id])))|920084|显然，如果您有一个单列数据框，并且您不希望R的自动降维将其转换为向量，则必须添加drop=FALSE参数。|offset|length|style|CODE|920085|entityMap^0|0|0|0|0|14|A|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

This can be done in one line, there's no need for a loop, be it a for-loop or an apply. Use unlist() instead :

<pre><code># testdata
Df &lt;- data.frame(
 x = as.factor(sample(1:5,30,r=TRUE)),
 y = as.factor(sample(1:5,30,r=TRUE)),
 z = as.factor(sample(1:5,30,r=TRUE)),
 w = as.factor(sample(1:5,30,r=TRUE))
)
##

Df[,c("y","w")] &lt;- as.numeric(as.character(unlist(Df[,c("y","w")])))

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

Edit : for your code, this becomes :

<pre><code>id &lt;- c(1,3:ncol(stats))) 
stats[,id] &lt;- as.numeric(as.character(unlist(stats[,id])))
</code></pre>

Obviously, if you have a one-column data frame and you don't want the automatic dimension reduction of R to convert it to a vector, you'll have to add the <code>drop=FALSE</code> argument.

blocks|key|1134620|text|lapply很大程度上是为此而设计的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1134621|unfactorize<-c("colA","colB")
df[,unfactorize]<-lapply(unfactorize,+function(x)+as.numeric(as.character(df[,x])))|code-block|syntax|javascript|1134622|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|$]]

lapply is pretty much designed for this

<pre><code>unfactorize&lt;-c("colA","colB")
df[,unfactorize]&lt;-lapply(unfactorize, function(x) as.numeric(as.character(df[,x])))
</code></pre>

blocks|key|920138|text|我在使用apply()调用将所有列转换为数字时遇到问题：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|920139|apply(data,+2,+as.numeric)|code-block|syntax|javascript|920140|问题被证明是因为一些字符串中有一个逗号--例如，"1024.63“而不是”1024.63“--而R不喜欢这种格式化数字的方式。所以我删除了它们，然后运行as.numeric()|920141|data+=+as.data.frame(apply(data,+2,+function(x)+{
++y+=+str_replace_all(x,+",",+"")+#remove+commas
++return(as.numeric(y))+#then+convert
}))|920142|请注意，这需要加载stringr包。|920143|entityMap^0|4|7|0|0|24|C|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

I had problems converting all columns to numeric with an <code>apply()</code> call:

<pre><code>apply(data, 2, as.numeric)
</code></pre>

The problem turns out to be because some of the strings had a comma in them -- e.g. "1,024.63" instead of "1024.63" -- and R does not like this way of formatting numbers. So I removed them and then ran <code>as.numeric()</code>:

<pre><code>data = as.data.frame(apply(data, 2, function(x) {
 y = str_replace_all(x, ",", "") #remove commas
 return(as.numeric(y)) #then convert
}))
</code></pre>

Note that this requires the stringr package to be loaded.

blocks|key|920177|text|我想指出的是，如果您在任何列中都有NAs，则简单地使用下标将不起作用。如果因子中有NAs，则必须使用Ramnath提供的应用脚本。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|920178|例如。|920179|Df+<-+data.frame(
++x+=+c(NA,as.factor(sample(1:5,30,r=T))),
++y+=+c(NA,as.factor(sample(1:5,30,r=T))),
++z+=+c(NA,as.factor(sample(1:5,30,r=T))),
++w+=+c(NA,as.factor(sample(1:5,30,r=T)))
)

Df[,c(1:4)]+<-+as.numeric(as.character(Df[,c(1:4)]))|code-block|syntax|javascript|920180|返回以下内容：|920181|Warning+message:
NAs+introduced+by+coercion+

++++>+head(Df)
+++++++x++y++z++w
++++1+NA+NA+NA+NA
++++2+NA+NA+NA+NA
++++3+NA+NA+NA+NA
++++4+NA+NA+NA+NA
++++5+NA+NA+NA+NA
++++6+NA+NA+NA+NA|920182|但是：|920183|Df[,c(1:4)]=+apply(Df[,c(1:4)],+2,+function(x)+as.numeric(as.character(x)))|920184|返回：|920185|>+head(Df)
+++x++y++z++w
1+NA+NA+NA+NA
2++2++3++4++1
3++1++5++3++4
4++2++3++4++1
5++5++3++5++5
6++4++2++4++4|920186|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

I would like to point out that if you have NAs in any column, simply using subscripts will not work. If there are NAs in the factor, you must use the apply script provided by Ramnath.

E.g.

<pre><code>Df &lt;- data.frame(
 x = c(NA,as.factor(sample(1:5,30,r=T))),
 y = c(NA,as.factor(sample(1:5,30,r=T))),
 z = c(NA,as.factor(sample(1:5,30,r=T))),
 w = c(NA,as.factor(sample(1:5,30,r=T)))
)

Df[,c(1:4)] &lt;- as.numeric(as.character(Df[,c(1:4)]))
</code></pre>

Returns the following:

<pre><code>Warning message:
NAs introduced by coercion 

 &gt; head(Df)
 x y z w
 1 NA NA NA NA
 2 NA NA NA NA
 3 NA NA NA NA
 4 NA NA NA NA
 5 NA NA NA NA
 6 NA NA NA NA
</code></pre>

But:

<pre><code>Df[,c(1:4)]= apply(Df[,c(1:4)], 2, function(x) as.numeric(as.character(x)))
</code></pre>

Returns:

<pre><code>&gt; head(Df)
 x y z w
1 NA NA NA NA
2 2 3 4 1
3 1 5 3 4
4 2 3 4 1
5 5 3 5 5
6 4 2 4 4
</code></pre>

blocks|key|920203|text|这就是对我有效的方法。apply()函数尝试将df转换为矩阵，并返回NA's。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|920204|numeric.df+<-+as.data.frame(sapply(df,+2,+as.numeric))|920205|entityMap^0|B|7|0|0|1I|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

That's what's worked for me. The <code>apply()</code> function tries to coerce df to matrix and it returns NA's.

<code>numeric.df &lt;- as.data.frame(sapply(df, 2, as.numeric))</code>

blocks|key|1134774|text|我在其他几个重复的线程上发现了这个函数，并且发现它是解决这个问题的一种优雅而通用的方法。这个帖子在大多数关于这个话题的搜索中都是最先出现的，所以我在这里分享它，以节省人们的时间。我不接受这一点，所以请查看原始帖子here和here的详细信息。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1134775|df+<-+data.frame(x+=+1:10,
+++++++++++++++++y+=+rep(1:2,+5),
+++++++++++++++++k+=+rnorm(10,+5,2),
+++++++++++++++++z+=+rep(c(2010,+2012,+2011,+2010,+1999),+2),
+++++++++++++++++j+=+c(rep(c("a",+"b",+"c"),+3),+"d"))

convert.magic+<-+function(obj,+type){
++FUN1+<-+switch(type,
+++++++++++++++++character+=+as.character,
+++++++++++++++++numeric+=+as.numeric,
+++++++++++++++++factor+=+as.factor)
++out+<-+lapply(obj,+FUN1)
++as.data.frame(out)
}

str(df)
str(convert.magic(df,+"character"))
str(convert.magic(df,+"factor"))
df[,+c("x",+"y")]+<-+convert.magic(df[,+c("x",+"y")],+"factor")|code-block|syntax|javascript|1134776|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/11261399/function-for-converting-dataframe-column-type|1|https://stackoverflow.com/questions/7680959/convert-type-of-multiple-columns-of-a-dataframe-at-once^0|2Y|4|0|33|4|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]]]

I found this function on a couple other duplicate threads and have found it an elegant and general way to solve this problem. This thread shows up first on most searches on this topic, so I am sharing it here to save folks some time. I take no credit for this just so see the original posts <a href="https://stackoverflow.com/questions/11261399/function-for-converting-dataframe-column-type">here</a> and <a href="https://stackoverflow.com/questions/7680959/convert-type-of-multiple-columns-of-a-dataframe-at-once">here</a> for details.

<pre><code>df &lt;- data.frame(x = 1:10,
 y = rep(1:2, 5),
 k = rnorm(10, 5,2),
 z = rep(c(2010, 2012, 2011, 2010, 1999), 2),
 j = c(rep(c("a", "b", "c"), 3), "d"))

convert.magic &lt;- function(obj, type){
 FUN1 &lt;- switch(type,
 character = as.character,
 numeric = as.numeric,
 factor = as.factor)
 out &lt;- lapply(obj, FUN1)
 as.data.frame(out)
}

str(df)
str(convert.magic(df, "character"))
str(convert.magic(df, "factor"))
df[, c("x", "y")] &lt;- convert.magic(df[, c("x", "y")], "factor")
</code></pre>

blocks|key|1134819|text|你可以使用"varhandle“包中的unfactor()函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1134820|library("varhandle")

my_iris+<-+data.frame(Sepal.Length+=+factor(iris$Sepal.Length),
++++++++++++++++++++++sample_id+=+factor(1:nrow(iris)))

my_iris+<-+unfactor(my_iris)|code-block|syntax|javascript|1134821|entityMap^0|J|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

you can use <code>unfactor()</code> function from "varhandle" package form CRAN:

<pre><code>library("varhandle")

my_iris &lt;- data.frame(Sepal.Length = factor(iris$Sepal.Length),
 sample_id = factor(1:nrow(iris)))

my_iris &lt;- unfactor(my_iris)
</code></pre>

blocks|key|920326|text|我喜欢这段代码，因为它非常方便：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|920327|++data[]+<-+lapply(data,+function(x)+type.convert(as.character(x),+as.is+=+TRUE))+#change+all+vars+to+their+best+fitting+data+type|code-block|syntax|javascript|920328|这并不完全是所要求的(转换为数字)，但在许多情况下甚至更合适。|920329|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I like this code because it's pretty handy:

<pre><code> data[] &lt;- lapply(data, function(x) type.convert(as.character(x), as.is = TRUE)) #change all vars to their best fitting data type
</code></pre>

It is not exactly what was asked for (convert to numeric), but in many cases even more appropriate.

blocks|key|920378|text|根据@SDahm的回答，这对于我的tibble来说是一个“最佳”的解决方案|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|920379|data+%25<>%25+lapply(type.convert)+%25>%25+as.data.table()|code-block|syntax|javascript|920380|这需要dplyr和magrittr。|920381|entityMap^0|H|6|0|0|3|5|9|8|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

Based on @SDahm's answer, this was an "optimal" solution for my <code>tibble</code>:

<pre><code>data %&lt;&gt;% lapply(type.convert) %&gt;% as.data.table()
</code></pre>

This requires <code>dplyr</code> and <code>magrittr</code>.

blocks|key|1134958|text|以下是一些dplyr选项：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1134959|#+by+column+type:
df+%25>%25+
++mutate_if(is.factor,+~as.numeric(as.character(.)))

#+by+specific+columns:
df+%25>%25+
++mutate_at(vars(x,+y,+z),+~as.numeric(as.character(.)))+

#+all+columns:
df+%25>%25+
++mutate_all(~as.numeric(as.character(.)))+|code-block|syntax|javascript|1134960|entityMap^0|5|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Here are some <code>dplyr</code> options:

<pre><code># by column type:
df %&gt;% 
 mutate_if(is.factor, ~as.numeric(as.character(.)))

# by specific columns:
df %&gt;% 
 mutate_at(vars(x, y, z), ~as.numeric(as.character(.))) 

# all columns:
df %&gt;% 
 mutate_all(~as.numeric(as.character(.))) 
</code></pre>

blocks|key|920480|text|df$colname+<-+as.numeric(df$colname)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|920481|我尝试用这种方法更改一种列类型，如果您不打算更改所有列类型，我认为它比许多其他版本更好|920482|df$colname+<-+as.character(df$colname)|920483|反之亦然。|920484|entityMap^0|0|10|0|0|0|12|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|U|8|@]|D|@]|E|$]]|$1|L|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|M|$]]

<code>df$colname &lt;- as.numeric(df$colname)</code>

I tried this way for changing one column type and I think it is better than many other versions, if you are not going to change all column types

<code>df$colname &lt;- as.character(df$colname)</code>

for the vice versa.

What is the quickest/best way to change a large number of columns to numeric from factor? 

I used the following code but it appears to have re-ordered my data.

<pre><code>&gt; head(stats[,1:2])
 rk team
1 1 Washington Capitals*
2 2 San Jose Sharks*
3 3 Chicago Blackhawks*
4 4 Phoenix Coyotes*
5 5 New Jersey Devils*
6 6 Vancouver Canucks*

for(i in c(1,3:ncol(stats))) {
 stats[,i] &lt;- as.numeric(stats[,i])
}

&gt; head(stats[,1:2])
 rk team
1 2 Washington Capitals*
2 13 San Jose Sharks*
3 24 Chicago Blackhawks*
4 26 Phoenix Coyotes*
5 27 New Jersey Devils*
6 28 Vancouver Canucks*
</code></pre>

What is the best way, short of naming every column as in:

<pre><code>df$colname &lt;- as.numeric(ds$colname)
</code></pre>

Change the class from factor to numeric of many columns in a data frame

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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将大量列从factor更改为数字的最快/最佳方法是什么？我使用了以下代码，但它似乎对我的数据进行了重新排序。> head(stats[,1:2])  rk                 team1  1 Washington Capitals*2  2     San Jose Sharks*3  3  Chicag...

问将数据框中许多列的类从因子更改为数字
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