blocks|key|1032838|text|另一种方法是使用apply进行转换|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1032839|bob2+<-+apply(bob,2,as.character)|code-block|syntax|javascript|1032840|和一个更好的(前一个是‘matrix’类)|1032841|bob2+<-+as.data.frame(as.matrix(bob),stringsAsFactors=F)|1032842|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|$]]

Another way is to convert it using apply

<pre><code>bob2 &lt;- apply(bob,2,as.character)
</code></pre>

And a better one (the previous is of class 'matrix')

<pre><code>bob2 &lt;- as.data.frame(as.matrix(bob),stringsAsFactors=F)
</code></pre>

blocks|key|1032919|text|只是跟着马特和德克。如果要在不更改全局选项的情况下重新创建现有数据框，可以使用apply语句重新创建它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1032920|bob+<-+data.frame(lapply(bob,+as.character),+stringsAsFactors=FALSE)|code-block|syntax|javascript|1032921|这会将所有变量转换为类"character"，如果您只想转换因子，请参见Marek's+solution+below。|offset|length|1032922|正如@hadley指出的那样，下面的内容更简洁。|1032923|bob[]+<-+lapply(bob,+as.character)|1032924|在这两种情况下，lapply都输出一个列表；但是，由于R的神奇属性，在第二种情况下使用[]保留了bob对象的data.frame类，从而消除了使用带有参数data.frame的as.data.frame转换回stringsAsFactors+=+FALSE的需要。|style|CODE|1032925|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/2853231/180892^0|0|0|10|M|0|0|0|0|8|6|17|2|1C|3|2G|D|2W|O|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@$I|13|J|14|1|15]]|A|$]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|18|8|@$I|19|J|1A|Q|R]|$I|1B|J|1C|Q|R]|$I|1D|J|1E|Q|R]|$I|1F|J|1G|Q|R]|$I|1H|J|1I|Q|R]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

Just following on Matt and Dirk. If you want to recreate your existing data frame without changing the global option, you can recreate it with an apply statement:

<pre><code>bob &lt;- data.frame(lapply(bob, as.character), stringsAsFactors=FALSE)
</code></pre>

This will convert all variables to class "character", if you want to only convert factors, see <a href="https://stackoverflow.com/a/2853231/180892">Marek's solution below</a>.

As @hadley points out, the following is more concise. 

<pre><code>bob[] &lt;- lapply(bob, as.character)
</code></pre>

In both cases, <code>lapply</code> outputs a list; however, owing to the magical properties of R, the use of <code>[]</code> in the second case keeps the data.frame class of the <code>bob</code> object, thereby eliminating the need to convert back to a data.frame using <code>as.data.frame</code> with the argument <code>stringsAsFactors = FALSE</code>.

blocks|key|1248007|text|要仅替换因子，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1248008|i+<-+sapply(bob,+is.factor)
bob[i]+<-+lapply(bob[i],+as.character)|code-block|syntax|javascript|1248009|在dplyr+in+version+0.5.0+new+function+mutate_if+was+introduced包中|offset|length|style|CODE|1248010|library(dplyr)
bob+%25>%25+mutate_if(is.factor,+as.character)+->+bob|1248011|...and+in+version+1.0.0+was+replaced+by+across|1248012|library(dplyr)
bob+%25>%25+mutate(across(where(is.factor),+as.character))+->+bob|1248013|Package+purrr+from+RStudio提供了另一种选择：|1248014|library(purrr)
bob+%25>%25+modify_if(is.factor,+as.character)+->+bob|1248015|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/hadley/dplyr/blob/master/NEWS.md#dplyr-050|1|https://github.com/tidyverse/dplyr/blob/master/NEWS.md#across|2|http://blog.rstudio.org/2015/09/29/purrr-0-1-0/^0|0|0|11|9|7|1I|0|0|0|14|6|7|13|1|0|0|0|Q|2|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1A|8|@$I|1B|J|1C|K|L]]|9|@$I|1D|J|1E|1|1F]]|A|$]]|$1|M|3|N|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|1H|8|@$I|1I|J|1J|K|L]]|9|@$I|1K|J|1L|1|1M]]|A|$]]|$1|Q|3|R|5|D|7|1N|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1O|8|@]|9|@$I|1P|J|1Q|1|1R]]|A|$]]|$1|U|3|V|5|D|7|1S|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]|14|$5|Z|10|11|A|$12|15]]|16|$5|Z|10|11|A|$12|17]]]]

To replace only factors:
<pre><code>i &lt;- sapply(bob, is.factor)
bob[i] &lt;- lapply(bob[i], as.character)
</code></pre>
In package dplyr <a href="https://github.com/hadley/dplyr/blob/master/NEWS.md#dplyr-050" rel="noreferrer">in version 0.5.0 new function <code>mutate_if</code> was introduced</a>:
<pre><code>library(dplyr)
bob %&gt;% mutate_if(is.factor, as.character) -&gt; bob
</code></pre>
...and <a href="https://github.com/tidyverse/dplyr/blob/master/NEWS.md#across" rel="noreferrer">in version 1.0.0 was replaced by <code>across</code></a>:
<pre><code>library(dplyr)
bob %&gt;% mutate(across(where(is.factor), as.character)) -&gt; bob
</code></pre>
<a href="http://blog.rstudio.org/2015/09/29/purrr-0-1-0/" rel="noreferrer">Package purrr from RStudio</a> gives another alternative:
<pre><code>library(purrr)
bob %&gt;% modify_if(is.factor, as.character) -&gt; bob
</code></pre>

blocks|key|1033015|text|我通常会把这个函数作为我所有项目的一部分。快速而简单。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1033016|unfactorize+<-+function(df){
++for(i+in+which(sapply(df,+class)+==+"factor"))+df[[i]]+=+as.character(df[[i]])
++return(df)
}|code-block|syntax|javascript|1033017|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|$]]

I typically make this function apart of all my projects. Quick and easy.

<pre><code>unfactorize &lt;- function(df){
 for(i in which(sapply(df, class) == "factor")) df[[i]] = as.character(df[[i]])
 return(df)
}
</code></pre>

blocks|key|1248132|text|如果您了解因子是如何存储的，就可以避免使用基于应用的函数来实现这一点。这并不意味着应用解决方案不能很好地工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1248133|因子被构造为与“级别”列表相关联的数字索引。如果您将一个因子转换为数字，就可以看到这一点。所以：|1248134|>+fact+<-+as.factor(c("a","b","a","d")
>+fact
[1]+a+b+a+d
Levels:+a+b+d

>+as.numeric(fact)
[1]+1+2+1+3|code-block|syntax|javascript|1248135|最后一行中返回的数字对应于因子的级别。|1248136|>+levels(fact)
[1]+"a"+"b"+"d"|1248137|请注意，levels()返回一个字符数组。您可以使用这一事实轻松而简洁地将因子转换为字符串或数字，如下所示：|offset|length|style|CODE|1248138|>+fact_character+<-+levels(fact)[as.numeric(fact)]
>+fact_character
[1]+"a"+"b"+"a"+"d"|1248139|这也适用于数值，前提是您将表达式包装在as.numeric()中。|1248140|>+num_fact+<-+factor(c(1,2,3,6,5,4))
>+num_fact
[1]+1+2+3+6+5+4
Levels:+1+2+3+4+5+6
>+num_num+<-+as.numeric(levels(num_fact)[as.numeric(num_fact)])
>+num_num
[1]+1+2+3+6+5+4|1248141|entityMap^0|0|0|0|0|0|4|8|0|0|J|C|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|15|8|@$O|16|P|17|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|F|7|18|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|19|8|@$O|1A|P|1B|Q|R]]|9|@]|A|$]]|$1|W|3|X|5|F|7|1C|8|@]|9|@]|A|$G|H]]|$1|Y|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|Z|$]]

If you understand how factors are stored, you can avoid using apply-based functions to accomplish this. Which isn't at all to imply that the apply solutions don't work well.

Factors are structured as numeric indices tied to a list of 'levels'. This can be seen if you convert a factor to numeric. So:

<pre><code>&gt; fact &lt;- as.factor(c("a","b","a","d")
&gt; fact
[1] a b a d
Levels: a b d

&gt; as.numeric(fact)
[1] 1 2 1 3
</code></pre>

The numbers returned in the last line correspond to the levels of the factor.

<pre><code>&gt; levels(fact)
[1] "a" "b" "d"
</code></pre>

Notice that <code>levels()</code> returns an array of characters. You can use this fact to easily and compactly convert factors to strings or numerics like this:

<pre><code>&gt; fact_character &lt;- levels(fact)[as.numeric(fact)]
&gt; fact_character
[1] "a" "b" "a" "d"
</code></pre>

This also works for numeric values, provided you wrap your expression in <code>as.numeric()</code>.

<pre><code>&gt; num_fact &lt;- factor(c(1,2,3,6,5,4))
&gt; num_fact
[1] 1 2 3 6 5 4
Levels: 1 2 3 4 5 6
&gt; num_num &lt;- as.numeric(levels(num_fact)[as.numeric(num_fact)])
&gt; num_num
[1] 1 2 3 6 5 4
</code></pre>

blocks|key|1033112|text|这对我来说很管用--我终于想出了一个单行|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1033113|df+<-+as.data.frame(lapply(df,function+(y)+if(class(y)=="factor"+)+as.character(y)+else+y),stringsAsFactors=F)|code-block|syntax|javascript|1033114|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|$]]

This works for me - I finally figured a one liner

<pre><code>df &lt;- as.data.frame(lapply(df,function (y) if(class(y)=="factor" ) as.character(y) else y),stringsAsFactors=F)
</code></pre>

blocks|key|1033162|text|如果您在data.frame上使用data.table包进行操作，那么问题就不存在了。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1033163|​|1033164|library(data.table)
dt+=+data.table(col1+=+c("a","b","c"),+col2+=+1:3)
sapply(dt,+class)
#+++++++col1++++++++col2+
#"character"+++"integer"+|code-block|syntax|javascript|1033165|如果数据集中已经有因子列，并且要将它们转换为字符，则可以执行以下操作。|1033166|library(data.table)
dt+=+data.table(col1+=+factor(c("a","b","c")),+col2+=+1:3)
sapply(dt,+class)
#+++++col1++++++col2+
#+"factor"+"integer"+
upd.cols+=+sapply(dt,+is.factor)
dt[,+names(dt)[upd.cols]+:=+lapply(.SD,+as.character),+.SDcols+=+upd.cols]
sapply(dt,+class)
#+++++++col1++++++++col2+
#"character"+++"integer"+|1033167|entityMap^0|H|A|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

If you would use <code>data.table</code> package for the operations on data.frame then the problem is not present. 



<pre><code>library(data.table)
dt = data.table(col1 = c("a","b","c"), col2 = 1:3)
sapply(dt, class)
# col1 col2 
#"character" "integer" 
</code></pre>

If you have a factor columns in you dataset already and you want to convert them to character you can do the following. 

<pre><code>library(data.table)
dt = data.table(col1 = factor(c("a","b","c")), col2 = 1:3)
sapply(dt, class)
# col1 col2 
# "factor" "integer" 
upd.cols = sapply(dt, is.factor)
dt[, names(dt)[upd.cols] := lapply(.SD, as.character), .SDcols = upd.cols]
sapply(dt, class)
# col1 col2 
#"character" "integer" 
</code></pre>

blocks|key|1248303|text|在数据帧的开头包含stringsAsFactors+=+FALSE以忽略所有误解。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1248304|entityMap^0|9|O|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

At the beginning of your data frame include <code>stringsAsFactors = FALSE</code> to ignore all misunderstandings.

blocks|key|1033242|text|此函数可以实现此功能|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1033243|df+<-+stacomirtools::killfactor(df)|code-block|syntax|javascript|1033244|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|$]]

This function does the trick

<pre><code>df &lt;- stacomirtools::killfactor(df)
</code></pre>

blocks|key|1248376|text|您应该在hablar中使用convert，它提供了与tidyverse管道兼容的可读性语法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1248377|library(dplyr)
library(hablar)

df+<-+tibble(a+=+factor(c(1,+2,+3,+4)),
+++++++++++++b+=+factor(c(5,+6,+7,+8)))

df+%25>%25+convert(chr(a:b))|code-block|syntax|javascript|1248378|这为您提供了：|1248379|++a+++++b++++
++<chr>+<chr>
1+1+++++5++++
2+2+++++6++++
3+3+++++7++++
4+4+++++8+++|1248380|entityMap^0|4|6|D|7|Q|9|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]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

You should use <code>convert</code> in <code>hablar</code> which gives readable syntax compatible with <code>tidyverse</code> pipes:

<pre><code>library(dplyr)
library(hablar)

df &lt;- tibble(a = factor(c(1, 2, 3, 4)),
 b = factor(c(5, 6, 7, 8)))

df %&gt;% convert(chr(a:b))
</code></pre>

which gives you:

<pre><code> a b 
 &lt;chr&gt; &lt;chr&gt;
1 1 5 
2 2 6 
3 3 7 
4 4 8 
</code></pre>

blocks|key|1033306|text|也许是更新的选择？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1033307|library("tidyverse")

bob+<-+bob+%25>%25+group_by_if(is.factor,+as.character)|code-block|syntax|javascript|1033308|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|$]]

Maybe a newer option?

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

bob &lt;- bob %&gt;% group_by_if(is.factor, as.character)
</code></pre>

blocks|key|1033394|text|这样可以将所有字符转换为字符，然后将数字转换为数字：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1033395|makenumcols<-function(df){
++df<-as.data.frame(df)
++df[]+<-+lapply(df,+as.character)
++cond+<-+apply(df,+2,+function(x)+{
++++x+<-+x[!is.na(x)]
++++all(suppressWarnings(!is.na(as.numeric(x))))
++})
++numeric_cols+<-+names(df)[cond]
++df[,numeric_cols]+<-+sapply(df[,numeric_cols],+as.numeric)
++return(df)
}|code-block|syntax|javascript|1033396|改编自Get+column+types+of+excel+sheet+automatically|offset|length|1033397|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/42735346/get-column-types-of-excel-sheet-automatically^0|0|0|3|19|0|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

This works transforming all to character and then the numeric to numeric:

<pre><code>makenumcols&lt;-function(df){
 df&lt;-as.data.frame(df)
 df[] &lt;- lapply(df, as.character)
 cond &lt;- apply(df, 2, function(x) {
 x &lt;- x[!is.na(x)]
 all(suppressWarnings(!is.na(as.numeric(x))))
 })
 numeric_cols &lt;- names(df)[cond]
 df[,numeric_cols] &lt;- sapply(df[,numeric_cols], as.numeric)
 return(df)
}
</code></pre>

Adapted from: <a href="https://stackoverflow.com/questions/42735346/get-column-types-of-excel-sheet-automatically">Get column types of excel sheet automatically</a>

blocks|key|1248532|text|加载dplyr-package后使用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1248533|bob=bob%25>%25mutate_at("phenotype",+as.character)|code-block|syntax|javascript|1248534|如果您只想专门更改phenotype-column。|1248535|entityMap^0|2|5|0|0|9|9|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|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

With the <code>dplyr</code>-package loaded use

<pre><code>bob=bob%&gt;%mutate_at("phenotype", as.character)
</code></pre>

if you only want to change the <code>phenotype</code>-column specifically.

blocks|key|1033524|text|在dplyr+version+1.0.0中引入了新的函数"across“。新函数将取代作用域变量(_if、_at、_all)。这是官方的documentation|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1033525|library(dplyr)
bob+<-+bob+%25>%25+
+++++++mutate(across(where(is.factor),+as.character))|code-block|syntax|javascript|1033526|entityMap|0|LINK|mutability|MUTABLE|url|https://dplyr.tidyverse.org/reference/across.html|1|https://dplyr.tidyverse.org/reference/mutate_all.html^0|7|D|0|1W|D|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]]]]

New function &quot;across&quot; was introduced in dplyr <a href="https://dplyr.tidyverse.org/reference/across.html" rel="nofollow noreferrer">version 1.0.0</a>. The new function will supersede scoped variables (_if, _at, _all). Here's the official <a href="https://dplyr.tidyverse.org/reference/mutate_all.html" rel="nofollow noreferrer">documentation</a>
<pre><code>library(dplyr)
bob &lt;- bob %&gt;% 
 mutate(across(where(is.factor), as.character))
</code></pre>

I have a data frame. Let's call him <code>bob</code>:

<pre><code>&gt; head(bob)
 phenotype exclusion
GSM399350 3- 4- 8- 25- 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-
GSM399351 3- 4- 8- 25- 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-
GSM399352 3- 4- 8- 25- 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-
GSM399353 3- 4- 8- 25+ 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-
GSM399354 3- 4- 8- 25+ 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-
GSM399355 3- 4- 8- 25+ 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-
</code></pre>

I'd like to concatenate the rows of this data frame (this will be another question). But look:

<pre><code>&gt; class(bob$phenotype)
[1] "factor"
</code></pre>

<code>Bob</code>'s columns are factors. So, for example:

<pre><code>&gt; as.character(head(bob))
[1] "c(3, 3, 3, 6, 6, 6)" "c(3, 3, 3, 3, 3, 3)" 
[3] "c(29, 29, 29, 30, 30, 30)"
</code></pre>

I don't begin to understand this, but I guess these are indices into the levels of the factors of the columns (of the court of king caractacus) of <code>bob</code>? Not what I need.

Strangely I can go through the columns of <code>bob</code> by hand, and do

<pre><code>bob$phenotype &lt;- as.character(bob$phenotype)
</code></pre>

which works fine. And, after some typing, I can get a data.frame whose columns are characters rather than factors. So my question is: how can I do this automatically? How do I convert a data.frame with factor columns into a data.frame with character columns without having to manually go through each column? 

Bonus question: why does the manual approach work?

Convert data.frame columns from factors to characters

我有个数据框。我们叫他bob吧> head(bob)                 phenotype                         exclusionGSM399350 3- 4- 8- 25- 44+ 11b- 11c- 19- NK1.1- Gr1- TER119-GSM399351 3- 4...

问将data.frame列从因子转换为字符
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