blocks|key|1180976|text|plyr包中的rbind.fill可能就是您要查找的内容。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1180977|entityMap^0|0|4|7|A|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

<code>rbind.fill</code> from the package <code>plyr</code> might be what you are looking for.

blocks|key|1181291|text|一个较新的解决方案是使用dplyr的bind_rows函数，我认为它比smartbind更有效。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1181292|df1+<-+data.frame(a+=+c(1:5),+b+=+c(6:10))
df2+<-+data.frame(a+=+c(11:15),+b+=+c(16:20),+c+=+LETTERS[1:5])
dplyr::bind_rows(df1,+df2)
++++a++b++++c
1+++1++6+<NA>
2+++2++7+<NA>
3+++3++8+<NA>
4+++4++9+<NA>
5+++5+10+<NA>
6++11+16++++A
7++12+17++++B
8++13+18++++C
9++14+19++++D
10+15+20++++E|code-block|syntax|javascript|1181293|entityMap^0|C|5|I|9|Z|9|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

A more recent solution is to use <code>dplyr</code>'s <code>bind_rows</code> function which I assume is more efficient than <code>smartbind</code>.

<pre><code>df1 &lt;- data.frame(a = c(1:5), b = c(6:10))
df2 &lt;- data.frame(a = c(11:15), b = c(16:20), c = LETTERS[1:5])
dplyr::bind_rows(df1, df2)
 a b c
1 1 6 &lt;NA&gt;
2 2 7 &lt;NA&gt;
3 3 8 &lt;NA&gt;
4 4 9 &lt;NA&gt;
5 5 10 &lt;NA&gt;
6 11 16 A
7 12 17 B
8 13 18 C
9 14 19 D
10 15 20 E
</code></pre>

blocks|key|966544|text|您可以使用gtools包中的smartbind。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|966545|示例：|966546|library(gtools)
df1+<-+data.frame(a+=+c(1:5),+b+=+c(6:10))
df2+<-+data.frame(a+=+c(11:15),+b+=+c(16:20),+c+=+LETTERS[1:5])
smartbind(df1,+df2)
#+result
+++++a++b++++c
1.1++1++6+<NA>
1.2++2++7+<NA>
1.3++3++8+<NA>
1.4++4++9+<NA>
1.5++5+10+<NA>
2.1+11+16++++A
2.2+12+17++++B
2.3+13+18++++C
2.4+14+19++++D
2.5+15+20++++E|code-block|syntax|javascript|966547|entityMap^0|5|6|E|9|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

You can use <code>smartbind</code> from the <code>gtools</code> package.

Example:

<pre><code>library(gtools)
df1 &lt;- data.frame(a = c(1:5), b = c(6:10))
df2 &lt;- data.frame(a = c(11:15), b = c(16:20), c = LETTERS[1:5])
smartbind(df1, df2)
# result
 a b c
1.1 1 6 &lt;NA&gt;
1.2 2 7 &lt;NA&gt;
1.3 3 8 &lt;NA&gt;
1.4 4 9 &lt;NA&gt;
1.5 5 10 &lt;NA&gt;
2.1 11 16 A
2.2 12 17 B
2.3 13 18 C
2.4 14 19 D
2.5 15 20 E
</code></pre>

blocks|key|1180911|text|您也可以只提取常见的列名。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1180912|>+cols+<-+intersect(colnames(df1),+colnames(df2))
>+rbind(df1[,cols],+df2[,cols])|code-block|syntax|javascript|1180913|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 could also just pull out the common column names. 

<pre><code>&gt; cols &lt;- intersect(colnames(df1), colnames(df2))
&gt; rbind(df1[,cols], df2[,cols])
</code></pre>

blocks|key|1181131|text|我写了一个函数来做这件事，因为我喜欢我的代码来告诉我是否出了什么问题。此函数将显式地告诉您哪些列名不匹配，以及是否存在类型不匹配。无论如何，它都会尽最大努力结合data.frames。限制是一次只能组合两个data.frames。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1181132|###+combines+data+frames+(like+rbind)+but+by+matching+column+names
#+columns+without+matches+in+the+other+data+frame+are+still+combined
#+but+with+NA+in+the+rows+corresponding+to+the+data+frame+without
#+the+variable
#+A+warning+is+issued+if+there+is+a+type+mismatch+between+columns+of
#+the+same+name+and+an+attempt+is+made+to+combine+the+columns
combineByName+<-+function(A,B)+{
++++a.names+<-+names(A)
++++b.names+<-+names(B)
++++all.names+<-+union(a.names,b.names)
++++print(paste("Number+of+columns:",length(all.names)))
++++a.type+<-+NULL
++++for+(i+in+1:ncol(A))+{
++++++++a.type[i]+<-+typeof(A[,i])
++++}
++++b.type+<-+NULL
++++for+(i+in+1:ncol(B))+{
++++++++b.type[i]+<-+typeof(B[,i])
++++}
++++a_b.names+<-+names(A)[!names(A)%25in%25names(B)]
++++b_a.names+<-+names(B)[!names(B)%25in%25names(A)]
++++if+(length(a_b.names)>0+%7C+length(b_a.names)>0){
++++++++print("Columns+in+data+frame+A+but+not+in+data+frame+B:")
++++++++print(a_b.names)
++++++++print("Columns+in+data+frame+B+but+not+in+data+frame+A:")
++++++++print(b_a.names)
++++}+else+if(a.names==b.names+&+a.type==b.type){
++++++++C+<-+rbind(A,B)
++++++++return(C)
++++}
++++C+<-+list()
++++for(i+in+1:length(all.names))+{
++++++++l.a+<-+all.names[i]%25in%25a.names
++++++++pos.a+<-+match(all.names[i],a.names)
++++++++typ.a+<-+a.type[pos.a]
++++++++l.b+<-+all.names[i]%25in%25b.names
++++++++pos.b+<-+match(all.names[i],b.names)
++++++++typ.b+<-+b.type[pos.b]
++++++++if(l.a+&+l.b)+{
++++++++++++if(typ.a==typ.b)+{
++++++++++++++++vec+<-+c(A[,pos.a],B[,pos.b])
++++++++++++}+else+{
++++++++++++++++warning(c("Type+mismatch+in+variable+named:+",all.names[i],"\n"))
++++++++++++++++vec+<-+try(c(A[,pos.a],B[,pos.b]))
++++++++++++}
++++++++}+else+if+(l.a)+{
++++++++++++vec+<-+c(A[,pos.a],rep(NA,nrow(B)))
++++++++}+else+{
++++++++++++vec+<-+c(rep(NA,nrow(A)),B[,pos.b])
++++++++}
++++++++C[[i]]+<-+vec
++++}
++++names(C)+<-+all.names
++++C+<-+as.data.frame(C)
++++return(C)
}|code-block|syntax|javascript|1181133|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 wrote a function to do this because I like my code to tell me if something is wrong. This function will explicitly tell you which column names don't match and if you have a type mismatch. Then it will do its best to combine the data.frames anyway. The limitation is that you can only combine two data.frames at a time.

<pre><code>### combines data frames (like rbind) but by matching column names
# columns without matches in the other data frame are still combined
# but with NA in the rows corresponding to the data frame without
# the variable
# A warning is issued if there is a type mismatch between columns of
# the same name and an attempt is made to combine the columns
combineByName &lt;- function(A,B) {
 a.names &lt;- names(A)
 b.names &lt;- names(B)
 all.names &lt;- union(a.names,b.names)
 print(paste("Number of columns:",length(all.names)))
 a.type &lt;- NULL
 for (i in 1:ncol(A)) {
 a.type[i] &lt;- typeof(A[,i])
 }
 b.type &lt;- NULL
 for (i in 1:ncol(B)) {
 b.type[i] &lt;- typeof(B[,i])
 }
 a_b.names &lt;- names(A)[!names(A)%in%names(B)]
 b_a.names &lt;- names(B)[!names(B)%in%names(A)]
 if (length(a_b.names)&gt;0 | length(b_a.names)&gt;0){
 print("Columns in data frame A but not in data frame B:")
 print(a_b.names)
 print("Columns in data frame B but not in data frame A:")
 print(b_a.names)
 } else if(a.names==b.names &amp; a.type==b.type){
 C &lt;- rbind(A,B)
 return(C)
 }
 C &lt;- list()
 for(i in 1:length(all.names)) {
 l.a &lt;- all.names[i]%in%a.names
 pos.a &lt;- match(all.names[i],a.names)
 typ.a &lt;- a.type[pos.a]
 l.b &lt;- all.names[i]%in%b.names
 pos.b &lt;- match(all.names[i],b.names)
 typ.b &lt;- b.type[pos.b]
 if(l.a &amp; l.b) {
 if(typ.a==typ.b) {
 vec &lt;- c(A[,pos.a],B[,pos.b])
 } else {
 warning(c("Type mismatch in variable named: ",all.names[i],"\n"))
 vec &lt;- try(c(A[,pos.a],B[,pos.b]))
 }
 } else if (l.a) {
 vec &lt;- c(A[,pos.a],rep(NA,nrow(B)))
 } else {
 vec &lt;- c(rep(NA,nrow(A)),B[,pos.b])
 }
 C[[i]] &lt;- vec
 }
 names(C) &lt;- all.names
 C &lt;- as.data.frame(C)
 return(C)
}
</code></pre>

blocks|key|966794|text|也许我完全误解了您的问题，但“我希望在绑定后保留不匹配的列”使我认为您正在寻找类似于SQL查询的left+join或right+join。R具有merge函数，它允许您指定left、right或inner，类似于SQL表的连接。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|966795|这里已经有一个关于这个主题的很好的问题和答案：How+to+join+(merge)+data+frames+(inner,+outer,+left,+right)?|966796|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/1299871/how-to-join-data-frames-in-r-inner-outer-left-right^0|1C|9|1M|A|20|5|0|N|1O|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|H|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Maybe I completely misread your question, but the "I am hoping to retain the columns that do not match after the bind" makes me think you are looking for a <code>left join</code> or <code>right join</code> similar to an SQL query. R has the <code>merge</code> function that lets you specify left, right, or inner joins similar to joining tables in SQL.

There is already a great question and answer on this topic here: <a href="https://stackoverflow.com/questions/1299871/how-to-join-data-frames-in-r-inner-outer-left-right">How to join (merge) data frames (inner, outer, left, right)?</a>

blocks|key|1181190|text|gtools/smartbind不喜欢处理日期，可能是因为它是as.vectoring。所以这是我的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1181191|sbind+=+function(x,+y,+fill=NA)+{
++++sbind.fill+=+function(d,+cols){+
++++++++for(c+in+cols)
++++++++++++d[[c]]+=+fill
++++++++d
++++}

++++x+=+sbind.fill(x,+setdiff(names(y),names(x)))
++++y+=+sbind.fill(y,+setdiff(names(x),names(y)))

++++rbind(x,+y)
}|code-block|syntax|javascript|1181192|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|$]]

gtools/smartbind didnt like working with Dates, probably because it was as.vectoring. So here's my solution...

<pre><code>sbind = function(x, y, fill=NA) {
 sbind.fill = function(d, cols){ 
 for(c in cols)
 d[[c]] = fill
 d
 }

 x = sbind.fill(x, setdiff(names(y),names(x)))
 y = sbind.fill(y, setdiff(names(x),names(y)))

 rbind(x, y)
}
</code></pre>

blocks|key|967122|text|只是为了文档。您可以尝试使用以下形式的Stack库及其函数Stack：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|967123|Stack(df_1,+df_2)|code-block|syntax|javascript|967124|我的印象也是，对于大型数据集，它比其他方法更快。|967125|entityMap^0|J|5|T|5|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

Just for the documentation. You can try the <code>Stack</code> library and its function <code>Stack</code> in the following form:

<pre><code>Stack(df_1, df_2)
</code></pre>

I have also the impression that it is faster than other methods for large data sets.

blocks|key|1181147|text|rbind.ordered=function(x,y){

++diffCol+=+setdiff(colnames(x),colnames(y))
++if+(length(diffCol)>0){
++++cols=colnames(y)
++++for+(i+in+1:length(diffCol))+y=cbind(y,NA)
++++colnames(y)=c(cols,diffCol)
++}

++diffCol+=+setdiff(colnames(y),colnames(x))
++if+(length(diffCol)>0){
++++cols=colnames(x)
++++for+(i+in+1:length(diffCol))+x=cbind(x,NA)
++++colnames(x)=c(cols,diffCol)
++}
++return(rbind(x,+y[,+colnames(x)]))
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1181148|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>rbind.ordered=function(x,y){

 diffCol = setdiff(colnames(x),colnames(y))
 if (length(diffCol)&gt;0){
 cols=colnames(y)
 for (i in 1:length(diffCol)) y=cbind(y,NA)
 colnames(y)=c(cols,diffCol)
 }

 diffCol = setdiff(colnames(y),colnames(x))
 if (length(diffCol)&gt;0){
 cols=colnames(x)
 for (i in 1:length(diffCol)) x=cbind(x,NA)
 colnames(x)=c(cols,diffCol)
 }
 return(rbind(x, y[, colnames(x)]))
}
</code></pre>

Is it possible to row bind two data frames that don't have the same set of columns? I am hoping to retain the columns that do not match after the bind.

Combine two data frames by rows (rbind) when they have different sets of columns

有没有可能行绑定两个不具有相同列集的数据帧？我希望在绑定后保留不匹配的列。

问当两个数据框具有不同的列集时，按行组合两个数据框(rbind
EN

回答 9

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问当两个数据框具有不同的列集时，按行组合两个数据框(rbindEN

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问当两个数据框具有不同的列集时，按行组合两个数据框(rbind
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