blocks|key|771381|text|这里有一些简单的东西可以完成这项工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|771382|key+<-+c('AA','AC','AG')
val+<-+c('0101','0102','0103')

lapply(1:3,FUN+=+function(i){foo[foo+==+key[i]]+<<-+val[i]})
foo

+snp1+snp2+snp3
1+0101+0101+<NA>
2+0103+++AT+++GG
3+0101+0103+++GG
4+0101+0101+++GC|code-block|syntax|javascript|771383|在这种情况下，lapply将输出一个我们实际上并不关心的列表。如果你喜欢，你可以将结果赋给某个东西，然后直接丢弃它。我在这里迭代索引，但您也可以很容易地将键放在列表中并直接迭代它们。请注意在<<-中使用全局赋值。|offset|length|style|CODE|771384|我用mapply修补了一种方法，但我的第一次尝试不起作用，所以我换了。不过，我怀疑使用mapply的解决方案是可能的。|771385|entityMap^0|0|0|7|6|2N|3|0|2|6|17|6|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]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@$I|Y|J|Z|K|L]|$I|10|J|11|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|P|$]]

Here's something simple that will do the job:

<pre><code>key &lt;- c('AA','AC','AG')
val &lt;- c('0101','0102','0103')

lapply(1:3,FUN = function(i){foo[foo == key[i]] &lt;&lt;- val[i]})
foo

 snp1 snp2 snp3
1 0101 0101 &lt;NA&gt;
2 0103 AT GG
3 0101 0103 GG
4 0101 0101 GC
</code></pre>

<code>lapply</code> will output a list in this case that we don't actually care about. You could assign the result to something if you like and then just discard it. I'm iterating over the indices here, but you could just as easily place the key/vals in a list themselves and iterate over them directly. Note the use of global assignment with <code>&lt;&lt;-</code>.

I tinkered with a way to do this with <code>mapply</code> but my first attempt didn't work, so I switched. I suspect a solution with <code>mapply</code> is possible, though.

blocks|key|554914|text|这是一个快速的解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|554915|dict+=+list(AA+=+'0101',+AC+=+'0102',+AG+=+'0103')
foo2+=+foo
for+(i+in+1:3){foo2+<-+replace(foo2,+foo2+==+names(dict[i]),+dict[i])}|code-block|syntax|javascript|554916|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|$]]

Here is a quick solution

<pre><code>dict = list(AA = '0101', AC = '0102', AG = '0103')
foo2 = foo
for (i in 1:3){foo2 &lt;- replace(foo2, foo2 == names(dict[i]), dict[i])}
</code></pre>

blocks|key|555007|text|map+=+setNames(c("0101",+"0102",+"0103"),+c("AA",+"AC",+"AG"))
foo[]+<-+map[unlist(foo)]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|555008|假设map涵盖了foo中的所有情况。如果foo是一个(字符()的)矩阵，那么这会让人感觉不像是'hack‘，并且在空间和时间上都更有效，那么|unstyled|offset|length|style|CODE|555009|matrix(map[foo],+nrow=nrow(foo),+dimnames=dimnames(foo))|555010|当有数百万个SNP和数千个样本时，矩阵和数据帧变体都违反了R对向量大小的2%5E31-1限制。|555011|entityMap^0|0|2|3|8|3|K|3|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@$G|S|H|T|I|J]|$G|U|H|V|I|J]|$G|W|H|X|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|Z|8|@]|9|@]|A|$]]|$1|O|3|-4|5|F|7|10|8|@]|9|@]|A|$]]]|P|$]]

<pre><code>map = setNames(c("0101", "0102", "0103"), c("AA", "AC", "AG"))
foo[] &lt;- map[unlist(foo)]
</code></pre>

assuming that <code>map</code> covers all the cases in <code>foo</code>. This would feel less like a 'hack' and be more efficient in both space and time if <code>foo</code> were a matrix (of character()), then

<pre><code>matrix(map[foo], nrow=nrow(foo), dimnames=dimnames(foo))
</code></pre>

Both matrix and data frame variants run afoul of R's 2^31-1 limit on vector size when there are millions of SNPs and thousands of samples.

blocks|key|771562|text|如果你愿意使用包，plyr是一个非常流行的包，它有这个方便的mapvalues()函数，它可以做你想做的事情：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|771563|foo+<-+mapvalues(foo,+from=c("AA",+"AC",+"AG"),+to=c("0101",+"0102",+"0103"))|code-block|syntax|javascript|771564|请注意，它适用于所有类型的数据类型，而不仅仅是字符串。|771565|entityMap|0|LINK|mutability|MUTABLE|url|http://www.inside-r.org/packages/cran/plyr/docs/mapvalues^0|9|4|U|B|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

If you're open to using packages, <code>plyr</code> is a very popular one and has this handy <a href="http://www.inside-r.org/packages/cran/plyr/docs/mapvalues" rel="noreferrer">mapvalues()</a> function that will do just what you're looking for:

<pre><code>foo &lt;- mapvalues(foo, from=c("AA", "AC", "AG"), to=c("0101", "0102", "0103"))
</code></pre>

Note that it works for data types of all kinds, not just strings.

blocks|key|555105|text|使用了上面@Ramnath的答案，但让它从文件中读取(要替换什么和要替换什么)，并使用gsub而不是replace。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|555106|hrw+<-+read.csv("hgWords.txt",+header=T,+stringsAsFactor=FALSE,+encoding="UTF-8",+sep="\t")+

for+(i+in+nrow(hrw))+
{
document+<-+gsub(hrw$from[i],+hrw$to[i],+document,+ignore.case=TRUE)
}|code-block|syntax|javascript|555107|hgword.txt包含以下用分隔符分隔的选项卡|555108|"from"++"to"
"AA"++++"0101"
"AC"++++"0102"
"AG"++++"0103"+|555109|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|$]]

Used @Ramnath's answer above, but made it read (what to be replaced and what to be replaced with) from a file and use gsub rather than replace.

<pre><code>hrw &lt;- read.csv("hgWords.txt", header=T, stringsAsFactor=FALSE, encoding="UTF-8", sep="\t") 

for (i in nrow(hrw)) 
{
document &lt;- gsub(hrw$from[i], hrw$to[i], document, ignore.case=TRUE)
}
</code></pre>

hgword.txt contains the following tab separated 

<pre><code>"from" "to"
"AA" "0101"
"AC" "0102"
"AG" "0103" 
</code></pre>

blocks|key|771679|text|因为距离上次回答已经有几年了，今晚有一个关于这个话题的新问题出现了，一位版主关闭了它，我将在这里添加它。发帖者有一个包含0、1和2的大型数据帧，并希望将它们分别更改为AA、AB和BB。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|771680|使用plyr|offset|length|style|CODE|771681|>+df+<-+data.frame(matrix(sample(c(NA,+c("0","1","2")),+100,+replace+=+TRUE),+10))
>+df
+++++X1+++X2+++X3+X4+++X5+++X6+++X7+++X8+++X9++X10
1+++++1++++2+<NA>++2++++1++++2++++0++++2++++0++++2
2+++++0++++2++++1++1++++2++++1++++1++++0++++0++++1
3+++++1++++0++++2++2++++1++++0+<NA>++++0++++1+<NA>
4+++++1++++2+<NA>++2++++2++++2++++1++++1++++0++++1
...+to+10th+row

>+df[]+<-+lapply(df,+as.character)|code-block|syntax|javascript|771682|使用revalue在数据框上创建一个函数来替换多个术语：|771683|>+library(plyr)
>+apply(df,+2,+function(x)+{x+<-+revalue(x,+c("0"="AA","1"="AB","2"="BB"));+x})
++++++X1+++X2+++X3+++X4+++X5+++X6+++X7+++X8+++X9+++X10+
+[1,]+"AB"+"BB"+NA+++"BB"+"AB"+"BB"+"AA"+"BB"+"AA"+"BB"
+[2,]+"AA"+"BB"+"AB"+"AB"+"BB"+"AB"+"AB"+"AA"+"AA"+"AB"
+[3,]+"AB"+"AA"+"BB"+"BB"+"AB"+"AA"+NA+++"AA"+"AB"+NA++
+[4,]+"AB"+"BB"+NA+++"BB"+"BB"+"BB"+"AB"+"AB"+"AA"+"AB"
...+and+so+on|771684|entityMap^0|0|2|4|0|0|2|7|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|W|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|X|8|@$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|O|3|P|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

Since it's been a few years since the last answer, and a new question came up tonight on this topic and a moderator closed it, I'll add it here. The poster has a large data frame containing 0, 1, and 2, and wants to change them to AA, AB, and BB respectively. 

Use <code>plyr</code>:

<pre><code>&gt; df &lt;- data.frame(matrix(sample(c(NA, c("0","1","2")), 100, replace = TRUE), 10))
&gt; df
 X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 1 2 &lt;NA&gt; 2 1 2 0 2 0 2
2 0 2 1 1 2 1 1 0 0 1
3 1 0 2 2 1 0 &lt;NA&gt; 0 1 &lt;NA&gt;
4 1 2 &lt;NA&gt; 2 2 2 1 1 0 1
... to 10th row

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

Create a function over the data frame using <code>revalue</code> to replace multiple terms:

<pre><code>&gt; library(plyr)
&gt; apply(df, 2, function(x) {x &lt;- revalue(x, c("0"="AA","1"="AB","2"="BB")); x})
 X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 
 [1,] "AB" "BB" NA "BB" "AB" "BB" "AA" "BB" "AA" "BB"
 [2,] "AA" "BB" "AB" "AB" "BB" "AB" "AB" "AA" "AA" "AB"
 [3,] "AB" "AA" "BB" "BB" "AB" "AA" NA "AA" "AB" NA 
 [4,] "AB" "BB" NA "BB" "BB" "BB" "AB" "AB" "AA" "AB"
... and so on
</code></pre>

blocks|key|555233|text|用字典替换字符串或字符串向量中的值的最具可读性的方法之一是stringr包中的stringr::str_replace_all。str_replace_all所需的模式可以是字典，例如，|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|555234|#+1.+Made+your+dictionnary
dictio_replace=+c("AA"=+"0101",+
++++++++++++++++++"AC"=+"0102",
++++++++++++++++++"AG"=+"0103")+#+short+example+of+dictionnary.

+#+2.+Replace+all+pattern,+according+to+the+dictionary-values+(only+a+single+vector+of+string,+or+a+single+string)
+foo$snp1+<-+stringr::str_replace_all(string+=+foo$snp1,
++++++++++++++++++++++++++++++++++++++pattern=+dictio_replace)++#+we+only+use+the+'pattern'+option+here:+'replacement'+is+useless+since+we+provide+a+dictionnary.|code-block|syntax|javascript|555235|对foo$snp2和foo$snp3重复步骤2。如果你有更多的向量要转换，最好使用另一个函数，以便在不重复的情况下替换数据帧中每个列/向量中的值。|555236|entityMap^0|T|7|13|O|1S|F|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$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]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

One of the most readable way to replace value in a string or a vector of string with a dictionary is <code>stringr::str_replace_all</code>, from the <code>stringr</code> package. The pattern needed by <code>str_replace_all</code> can be a dictionnary, e.g.,
<pre><code># 1. Made your dictionnary
dictio_replace= c(&quot;AA&quot;= &quot;0101&quot;, 
 &quot;AC&quot;= &quot;0102&quot;,
 &quot;AG&quot;= &quot;0103&quot;) # short example of dictionnary.

 # 2. Replace all pattern, according to the dictionary-values (only a single vector of string, or a single string)
 foo$snp1 &lt;- stringr::str_replace_all(string = foo$snp1,
 pattern= dictio_replace) # we only use the 'pattern' option here: 'replacement' is useless since we provide a dictionnary.
</code></pre>
Repeat step 2 with foo$snp2 &amp; foo$snp3. If you have more vectors to transform it's a good idea to use another func', in order to replace values in each of the columns/vector in the dataframe without repeating yourself.

blocks|key|555311|text|不是太原始，但应该提供一个直观的界面来完成替换Base+R中的多个值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|555312|#+Function+performing+a+mapping+replacement:
#+replaceMultipleValues+=>+function()+
replaceMultipleValues+<-+function(df,+mapFrom,+mapTo){
++#+Extract+the+values+in+the+data.frame:+
++#+dfVals+=>+named+character+vector
++dfVals+<-+unlist(df)
++
++#+Get+all+values+in+the+mapping+&+data+
++#+and+assign+a+name+to+them:+tmp1+=>+named+character+vector+
++tmp1+<-+c(
++++setNames(mapTo,+mapFrom),+
++++setNames(dfVals,+dfVals)
++)
++
++#+Extract+the+unique+values:+
++#+valueMap+=>+named+character+vector
++valueMap+<-+tmp1[!(duplicated(names(tmp1)))]
++
++#+Recode+the+values+in+data.frame:+res+=>+data.frame
++res+<-+data.frame(
++++++matrix(
++++++++valueMap[dfVals],+
++++++++nrow+=+nrow(df),
++++++++ncol+=+ncol(df),
++++++++dimnames+=+dimnames(df)
++++)
++)
++
++#+Explicitly+define+the+returned+object:+data.frame+=>+env
++return(res)
}

#+Recode+values+in+data.frame:+
#+res+=>+data.frame
res+<-+replaceMultipleValues(
++foo,+
++c("AA",+"AC",+"AG"),+
++c("0101",+"0102",+"0103")
)

#+Print+data.frame+to+console:+
#+data.frame+=>+stdout(console)
res|code-block|syntax|javascript|555313|数据：|555314|#+Import+data:+foo+=>+data.frame
foo+<-+data.frame(snp1+=+c("AA",+"AG",+"AA",+"AA"),+snp2+=+c("AA",+"AT",+"AG",+"AA"),+snp3+=+c(NA,+"GG",+"GG",+"GC"),+stringsAsFactors=FALSE)|555315|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|$]]

Not overly original, but should provide an intuitive interface to accomplish replacing multiple values in Base R:
<pre><code># Function performing a mapping replacement:
# replaceMultipleValues =&gt; function() 
replaceMultipleValues &lt;- function(df, mapFrom, mapTo){
 # Extract the values in the data.frame: 
 # dfVals =&gt; named character vector
 dfVals &lt;- unlist(df)
 
 # Get all values in the mapping &amp; data 
 # and assign a name to them: tmp1 =&gt; named character vector 
 tmp1 &lt;- c(
 setNames(mapTo, mapFrom), 
 setNames(dfVals, dfVals)
 )
 
 # Extract the unique values: 
 # valueMap =&gt; named character vector
 valueMap &lt;- tmp1[!(duplicated(names(tmp1)))]
 
 # Recode the values in data.frame: res =&gt; data.frame
 res &lt;- data.frame(
 matrix(
 valueMap[dfVals], 
 nrow = nrow(df),
 ncol = ncol(df),
 dimnames = dimnames(df)
 )
 )
 
 # Explicitly define the returned object: data.frame =&gt; env
 return(res)
}

# Recode values in data.frame: 
# res =&gt; data.frame
res &lt;- replaceMultipleValues(
 foo, 
 c(&quot;AA&quot;, &quot;AC&quot;, &quot;AG&quot;), 
 c(&quot;0101&quot;, &quot;0102&quot;, &quot;0103&quot;)
)

# Print data.frame to console: 
# data.frame =&gt; stdout(console)
res
</code></pre>
Data:
<pre><code># Import data: foo =&gt; data.frame
foo &lt;- data.frame(snp1 = c(&quot;AA&quot;, &quot;AG&quot;, &quot;AA&quot;, &quot;AA&quot;), snp2 = c(&quot;AA&quot;, &quot;AT&quot;, &quot;AG&quot;, &quot;AA&quot;), snp3 = c(NA, &quot;GG&quot;, &quot;GG&quot;, &quot;GC&quot;), stringsAsFactors=FALSE)
</code></pre>

I have a large data.frame of character data that I want to convert based on what is commonly called a dictionary in other languages.

Currently I am going about it like so:

<pre><code>foo &lt;- data.frame(snp1 = c("AA", "AG", "AA", "AA"), snp2 = c("AA", "AT", "AG", "AA"), snp3 = c(NA, "GG", "GG", "GC"), stringsAsFactors=FALSE)
foo &lt;- replace(foo, foo == "AA", "0101")
foo &lt;- replace(foo, foo == "AC", "0102")
foo &lt;- replace(foo, foo == "AG", "0103")
</code></pre>

This works fine, but it is obviously not pretty and seems silly to repeat the <code>replace</code> statement each time I want to replace one item in the data.frame.

Is there a better way to do this since I have a dictionary of approximately 25 key/value pairs?

Dictionary style replace multiple items

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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我有一个很大的字符数据data.frame，我想根据其他语言中通常称为字典的东西来转换这些字符数据。目前我是这样做的：foo <- data.frame(snp1 = c("AA", "AG", "AA", "AA"), snp2 = c("AA", "AT", "AG", "AA"), snp3 = c(NA, "G...

问字典样式替换多个项目
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