blocks|key|1821760|text|一种“简单”的方法是在导入文本数据时不将字符串设置为因子。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1821761|注意，read.{table,csv,...}函数接受一个stringsAsFactors参数，该参数在缺省情况下设置为TRUE。您可以在导入和rbind-ing数据时将其设置为FALSE。|offset|length|style|CODE|1821762|如果您希望将列设置为结尾的一个因子，也可以这样做。|1821763|例如：|1821764|alltime+<-+read.table("alltime.txt",+stringsAsFactors=FALSE)
all2008+<-+read.table("all2008.txt",+stringsAsFactors=FALSE)
alltime+<-+rbind(alltime,+all2008)
#+If+you+want+the+doctor+column+to+be+a+factor,+make+it+so:
alltime$doctor+<-+as.factor(alltime$doctor)|code-block|syntax|javascript|1821765|entityMap^0|0|3|K|T|G|1O|4|20|5|2H|5|0|0|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]|$D|W|E|X|F|G]|$D|Y|E|Z|F|G]|$D|10|E|11|F|G]|$D|12|E|13|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|14|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|15|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|16|8|@]|9|@]|A|$O|P]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$]]

An "easy" way is to simply not have your strings set as factors when importing text data.

Note that the <code>read.{table,csv,...}</code> functions take a <code>stringsAsFactors</code> parameter, which is by default set to <code>TRUE</code>. You can set this to <code>FALSE</code> while you're importing and <code>rbind</code>-ing your data.

If you'd like to set the column to be a factor at the end, you can do that too.

For example:

<pre><code>alltime &lt;- read.table("alltime.txt", stringsAsFactors=FALSE)
all2008 &lt;- read.table("all2008.txt", stringsAsFactors=FALSE)
alltime &lt;- rbind(alltime, all2008)
# If you want the doctor column to be a factor, make it so:
alltime$doctor &lt;- as.factor(alltime$doctor)
</code></pre>

blocks|key|1821775|text|正如前一个答案中所建议的，将列作为字符读取，并在rbind之后将其转换为因子。SQLFetch+(我假设是RODBC)也有stringsAsFactors或as.is参数来控制字符的转换。对于read.table，允许的值是as，例如as.is=TRUE或某个列号。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1821776|entityMap^0|O|5|13|8|1P|G|26|5|2O|A|39|A|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|G|$]]

As suggested in the previous answer, read the columns as character and do the conversion to factors after <code>rbind</code>.
<code>SQLFetch</code> (I assume RODBC) has also the <code>stringsAsFactors</code> or the <code>as.is</code> argument to control the conversion of characters.
Allowed values are as for <code>read.table</code>, e.g., <code>as.is=TRUE</code> or some column number.

blocks|key|2035408|text|我在类型不匹配方面也遇到了同样的问题，尤其是在因子方面。我必须将两个原本兼容的数据集粘合在一起。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2035409|我的解决方案是将两个数据帧中的因子转换为“字符”。然后它就像一个护身符一样工作:-)|2035410|++++convert.factors.to.strings.in.dataframe+<-+function(dataframe)
++++{
++++++++class.data++<-+sapply(dataframe,+class)
++++++++factor.vars+<-+class.data[class.data+==+"factor"]
++++++++for+(colname+in+names(factor.vars))
++++++++{
++++++++++++dataframe[,colname]+<-+as.character(dataframe[,colname])
++++++++}
++++++++return+(dataframe)
++++}|code-block|syntax|javascript|2035411|如果您想要查看两个数据帧运行中的类型(更改变量名称)：|2035412|++++cbind("orig"=sapply(allSurveyData,+class),+
++++++++++"merge"+=+sapply(curSurveyDataMerge,+class),
++++++++++"eq"=sapply(allSurveyData,+class)+==+sapply(curSurveyDataMerge,+class)
++++)|2035413|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I had the same problem with type mismatches, especially with factors. I had to glue together two otherwise compatible datasets.

My solution is to convert factors in both dataframes to "character". Then it works like a charm :-)

<pre><code> convert.factors.to.strings.in.dataframe &lt;- function(dataframe)
 {
 class.data &lt;- sapply(dataframe, class)
 factor.vars &lt;- class.data[class.data == "factor"]
 for (colname in names(factor.vars))
 {
 dataframe[,colname] &lt;- as.character(dataframe[,colname])
 }
 return (dataframe)
 }
</code></pre>

If you want to see the types in your two dataframes run (change var names):

<pre><code> cbind("orig"=sapply(allSurveyData, class), 
 "merge" = sapply(curSurveyDataMerge, class),
 "eq"=sapply(allSurveyData, class) == sapply(curSurveyDataMerge, class)
 )
</code></pre>

blocks|key|1821798|text|1)创建stringsAsFactor设置为FALSE的数据框。这应该可以解决因数问题|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1821799|2)之后不要使用rbind+-如果数据框为空，它会弄乱列名。只需这样做：|1821800|df[nrow(df)%2B1,]+<-+c("d","gsgsgd",4)|code-block|syntax|javascript|1821801|/|1821802|>+df+<-+data.frame(a+=+character(0),+b=character(0),+c=numeric(0))

>+df[nrow(df)%2B1,]+<-+c("d","gsgsgd",4)

Warnmeldungen:
1:+In+`[<-.factor`(`*tmp*`,+iseq,+value+=+"d")+:
++invalid+factor+level,+NAs+generated
2:+In+`[<-.factor`(`*tmp*`,+iseq,+value+=+"gsgsgd")+:
++invalid+factor+level,+NAs+generated

>+df+<-+data.frame(a+=+character(0),+b=character(0),+c=numeric(0),+stringsAsFactors=F)

>+df[nrow(df)%2B1,]+<-+c("d","gsgsgd",4)

>+df
++a++++++b+c
1+d+gsgsgd+4|1821803|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

1) create the data frame with stringsAsFactor set to FALSE. This should resolve the factor-issue

2) afterwards don't use rbind - it messes up the column names if the data frame is empty. simply do it this way:

<pre><code>df[nrow(df)+1,] &lt;- c("d","gsgsgd",4)
</code></pre>

/

<pre><code>&gt; df &lt;- data.frame(a = character(0), b=character(0), c=numeric(0))

&gt; df[nrow(df)+1,] &lt;- c("d","gsgsgd",4)

Warnmeldungen:
1: In `[&lt;-.factor`(`*tmp*`, iseq, value = "d") :
 invalid factor level, NAs generated
2: In `[&lt;-.factor`(`*tmp*`, iseq, value = "gsgsgd") :
 invalid factor level, NAs generated

&gt; df &lt;- data.frame(a = character(0), b=character(0), c=numeric(0), stringsAsFactors=F)

&gt; df[nrow(df)+1,] &lt;- c("d","gsgsgd",4)

&gt; df
 a b c
1 d gsgsgd 4
</code></pre>

blocks|key|1821831|text|这是一个函数，获取2个数据帧的公共行名并执行rbind，其中我们基本上找到了作为因子的字段，然后添加新的因子，然后执行rbind。这应该会照顾到任何因素问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1821832|rbindCommonCols<-function(x，y){|1821833|commonColNames+=+intersect(colnames(x),+colnames(y))
x+=+x[,commonColNames]
y+=+y[,commonColNames]

colClassesX+=+sapply(x,+class)
colClassesY+=+sapply(y,+class)
classMatch+=+paste(+colClassesX,+colClassesY,+sep+=+"-"+)
factorColIdx+=+grep("factor",+classMatch)

for(n+in+factorColIdx){+
++++x[,n]+=+as.factor(x[,n])
++++y[,n]+=+as.factor(y[,n])
}

for(n+in+factorColIdx){+
++++x[,n]+=+factor(x[,n],+levels+=+unique(c(+levels(x[,n]),+levels(y[,n])+)))
++++y[,n]+=+factor(y[,n],+levels+=+unique(c(+levels(y[,n]),+levels(x[,n])+)))++
}+

res+=+rbind(x,y)
res|code-block|syntax|javascript|1821834|}|1821835|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

here's a function to take the common row names of 2 data frames and do an rbind where we basically find the fields that are factors, add the new factors then do the rbind. This should take care of any factor issues:

rbindCommonCols&lt;-function(x, y){

<pre><code>commonColNames = intersect(colnames(x), colnames(y))
x = x[,commonColNames]
y = y[,commonColNames]

colClassesX = sapply(x, class)
colClassesY = sapply(y, class)
classMatch = paste( colClassesX, colClassesY, sep = "-" )
factorColIdx = grep("factor", classMatch)

for(n in factorColIdx){ 
 x[,n] = as.factor(x[,n])
 y[,n] = as.factor(y[,n])
}

for(n in factorColIdx){ 
 x[,n] = factor(x[,n], levels = unique(c( levels(x[,n]), levels(y[,n]) )))
 y[,n] = factor(y[,n], levels = unique(c( levels(y[,n]), levels(x[,n]) ))) 
} 

res = rbind(x,y)
res
</code></pre>

}

I have a dataframe (14.5K rows by 15 columns) containing billing data from 2001 to 2007.

I append new 2008 data to it with: <code>alltime &lt;- rbind(alltime,all2008)</code>

Unfortunately that generates a warning:

<pre><code>&gt; Warning message:
In `[&lt;-.factor`(`*tmp*`, ri, value = c(NA, NA, NA, NA, NA, NA, NA, :
 invalid factor level, NAs generated
</code></pre>

My guess is that there are some new patients whose names were not in the previous dataframe and therefore it would not know what level to give those. Similarly new unseen names in the 'referring doctor' column.

What's the solution?

Unseen factor levels when appending new records with unseen string values to a dataframe, cause Warning and result in NA

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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我有一个数据帧(14.5K行乘15列)，其中包含2001到2007年的账单数据。我将新的2008年的数据添加到其中：alltime <- rbind(alltime,all2008)不幸的是，这会生成一个警告：> Warning message:In `[<-.factor`(`*tmp*`, ri, value = ...

问看不见的因子级别将具有看不见的字符串值的新记录附加到数据帧时，会导致警告并导致不可见
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问看不见的因子级别将具有看不见的字符串值的新记录附加到数据帧时，会导致警告并导致不可见
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