blocks|key|915249|text|更容易将nm转换为字符，然后进行更改：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|915250|junk$nm+<-+as.character(junk$nm)
junk$nm[junk$nm+==+"B"]+<-+"b"|code-block|syntax|javascript|915251|编辑:如果您确实需要维护nm作为因子，请在末尾添加以下内容：|915252|junk$nm+<-+as.factor(junk$nm)|915253|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|$]]

Easier to convert nm to characters and then make the change:

<pre><code>junk$nm &lt;- as.character(junk$nm)
junk$nm[junk$nm == "B"] &lt;- "b"
</code></pre>

EDIT: And if indeed you need to maintain nm as factors, add this in the end:

<pre><code>junk$nm &lt;- as.factor(junk$nm)
</code></pre>

blocks|key|915416|text|替换值的另一种有用的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|915417|library(plyr)
junk$nm+<-+revalue(junk$nm,+c("B"="b"))|code-block|syntax|javascript|915418|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|$]]

another useful way to replace values 

<pre><code>library(plyr)
junk$nm &lt;- revalue(junk$nm, c("B"="b"))
</code></pre>

blocks|key|915383|text|在一个命令中完成此操作的最简单方法是使用which命令，并且也不需要通过执行以下操作将因子更改为字符：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|915384|junk$nm[which(junk$nm=="B")]<-"b"|code-block|syntax|javascript|915385|entityMap^0|K|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|$]]

The easiest way to do this in one command is to use <code>which</code> command and also need not to change the factors into character by doing this:

<pre><code>junk$nm[which(junk$nm=="B")]&lt;-"b"
</code></pre>

blocks|key|699229|text|您已经在nm中创建了一个因子变量，因此需要避免这样做，或者向因子属性添加一个额外的级别。您还应该避免在data.frame()的参数中使用<-|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|699230|选项1：|699231|junk+<-+data.frame(x+=+rep(LETTERS[1:4],+3),+y+=letters[1:12],+stringsAsFactors=FALSE)
junk$nm[junk$nm+==+"B"]+<-+"b"|code-block|syntax|javascript|699232|选项2：|699233|levels(junk$nm)+<-+c(levels(junk$nm),+"b")
junk$nm[junk$nm+==+"B"]+<-+"b"
junk|699234|entityMap^0|4|2|1X|2|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

You have created a factor variable in <code>nm</code> so you either need to avoid doing so or add an additional level to the factor attributes. You should also avoid using <code>&lt;-</code> in the arguments to data.frame()

Option 1:

<pre><code>junk &lt;- data.frame(x = rep(LETTERS[1:4], 3), y =letters[1:12], stringsAsFactors=FALSE)
junk$nm[junk$nm == "B"] &lt;- "b"
</code></pre>

Option 2:

<pre><code>levels(junk$nm) &lt;- c(levels(junk$nm), "b")
junk$nm[junk$nm == "B"] &lt;- "b"
junk
</code></pre>

blocks|key|915460|text|如果您正在使用字符变量(请注意，此处的stringsAsFactors为false+)，则可以使用replace：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|915461|junk+<-+data.frame(x+<-+rep(LETTERS[1:4],+3),+y+<-+letters[1:12],+stringsAsFactors+=+FALSE)
colnames(junk)+<-+c("nm",+"val")

junk$nm+<-+replace(junk$nm,+junk$nm+==+"B",+"b")
junk
#++++nm+val
#+1+++A+++a
#+2+++b+++b
#+3+++C+++c
#+4+++D+++d
#+...|code-block|syntax|javascript|915462|entityMap^0|J|G|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|$]]

If you are working with character variables (note that <code>stringsAsFactors</code> is false here) you can use replace: 

<pre><code>junk &lt;- data.frame(x &lt;- rep(LETTERS[1:4], 3), y &lt;- letters[1:12], stringsAsFactors = FALSE)
colnames(junk) &lt;- c("nm", "val")

junk$nm &lt;- replace(junk$nm, junk$nm == "B", "b")
junk
# nm val
# 1 A a
# 2 b b
# 3 C c
# 4 D d
# ...
</code></pre>

blocks|key|699416|text|您也可以使用ifelse，它非常容易理解|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|699417|junk$val+<-+ifelse(junk$nm+==+"B",+"b",+junk$val)|code-block|syntax|javascript|699418|如果您仍然想通过for+loop来做这件事，那么正确的方法就是|699419|for(i+in+1:nrow(junk)){
++if(junk[i,+"nm"]+==+"B"){
++++junk[i,+"val"]+<-+"b"
++}
}

junk
>+junk
+++nm+val
1+++A+++a
2+++B+++b
3+++C+++c
4+++D+++d
5+++A+++e
6+++B+++b
7+++C+++g
8+++D+++h
9+++A+++i
10++B+++b
11++C+++k
12++D+++l|699420|entityMap^0|6|6|0|0|8|8|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

You can use <code>ifelse</code> too, which is very simple to understand
<pre><code>junk$val &lt;- ifelse(junk$nm == &quot;B&quot;, &quot;b&quot;, junk$val)
</code></pre>
If you still want to do it through <code>for loop</code> the correct way of doing it
<pre><code>for(i in 1:nrow(junk)){
 if(junk[i, &quot;nm&quot;] == &quot;B&quot;){
 junk[i, &quot;val&quot;] &lt;- &quot;b&quot;
 }
}

junk
&gt; junk
 nm val
1 A a
2 B b
3 C c
4 D d
5 A e
6 B b
7 C g
8 D h
9 A i
10 B b
11 C k
12 D l
</code></pre>

blocks|key|915480|text|stata.replace<-function(data,replacevar,replacevalue,ifs)+{
++ifs=parse(text=ifs)
++yy=as.numeric(eval(ifs,data,parent.frame()))
++x=sum(yy)
++data=cbind(data,yy)
++data[yy==1,replacevar]=replacevalue
++message=noquote(paste0(x,+"+replacement+are+made"))
++print(message)
++return(data[,1:(ncol(data)-1)])
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|915481|使用下面的代码行调用此函数。|unstyled|915482|d=stata.replace(d,"under20",1,"age<20")|915483|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>stata.replace&lt;-function(data,replacevar,replacevalue,ifs) {
 ifs=parse(text=ifs)
 yy=as.numeric(eval(ifs,data,parent.frame()))
 x=sum(yy)
 data=cbind(data,yy)
 data[yy==1,replacevar]=replacevalue
 message=noquote(paste0(x, " replacement are made"))
 print(message)
 return(data[,1:(ncol(data)-1)])
}
</code></pre>

Call this function using below line.

<pre><code>d=stata.replace(d,"under20",1,"age&lt;20")
</code></pre>

blocks|key|699447|text|我也有同样的问题，你也可以对每一列做同样的事情，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|699448|+fix_junk+<-+function(x){
++++++#x+<-+as.character(x)
++++++x[x+==+"B"]+<-+"b"
++++++x
++++}
++++junk[]+<-+lapply(junk,+fix_junk);+junk+#+junk[]+to+get+a+data+frame+rather+than+a+list
++++junk[1:3]+<-+lapply(junk[1:3],+fix_junk);+junk|code-block|syntax|javascript|699449|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 got the same issue, you can also do the same thing for each columns,
<pre><code> fix_junk &lt;- function(x){
 #x &lt;- as.character(x)
 x[x == &quot;B&quot;] &lt;- &quot;b&quot;
 x
 }
 junk[] &lt;- lapply(junk, fix_junk); junk # junk[] to get a data frame rather than a list
 junk[1:3] &lt;- lapply(junk[1:3], fix_junk); junk
</code></pre>

In the R data frame coded for below, I would like to replace all of the times that <code>B</code> 
appears with <code>b</code>. 

<pre><code>junk &lt;- data.frame(x &lt;- rep(LETTERS[1:4], 3), y &lt;- letters[1:12])
colnames(junk) &lt;- c("nm", "val")
</code></pre>

this provides:

<pre><code> nm val
1 A a
2 B b
3 C c
4 D d
5 A e
6 B f
7 C g
8 D h
9 A i
10 B j
11 C k
12 D l
</code></pre>

My initial attempt was to use a <code>for</code> and <code>if</code> statements like so: 

<pre><code>for(i in junk$nm) if(i %in% "B") junk$nm &lt;- "b"
</code></pre>

but as I am sure you can see, this replaces ALL of the values of <code>junk$nm</code> with <code>b</code>. I can see why this is doing this but I can't seem to get it to replace only those cases of junk$nm where the original value was <code>B</code>.

NOTE: I managed to solve the problem with <code>gsub</code> but in the interest of learning R I still would like to know how to get my original approach to work (if it is possible)

Replace a value in a data frame based on a conditional (`if`) statement

在下面编码的R数据帧中，我想用b替换所有出现B的时间。junk <- data.frame(x <- rep(LETTERS[1:4], 3), y <- letters[1:12])colnames(junk) <- c("nm", "val")这提供了： nm val1 A a2 B b3 C c4 D d5 A e6 B f7 C

问根据条件(`if`)语句替换数据框中的值
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问根据条件(`if`)语句替换数据框中的值
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