entityMap|blocks|key|9cd2s|text|这不是很有效，但它可以完成工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|fllso|blue+<-c("Ti","Ka","Vv","Cn")
Turq+<-c("Cc","Wx","Oq","Yt")
Indigo+<-c("Gb","Ap")

#assumes+no+duplicates+in+the+original+vectors.
df<-as.data.frame(cbind(blue,+Turq,+Indigo))

library(tidyr)
df<-unique(gather(df))|code-block|syntax|javascript|dsl29|这将生成一些警告，但会提供您请求的结果。cbind开始重复向量，这样所有向量都有相同的长度。|a449b^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|K|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|L|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|M|9|@]|A|@]|B|$]]|$2|J|4|-4|6|7|8|N|9|@]|A|@]|B|$]]]]

This is not very efficient, but it gets the job done.

<pre><code>blue &lt;-c("Ti","Ka","Vv","Cn")
Turq &lt;-c("Cc","Wx","Oq","Yt")
Indigo &lt;-c("Gb","Ap")

#assumes no duplicates in the original vectors.
df&lt;-as.data.frame(cbind(blue, Turq, Indigo))

library(tidyr)
df&lt;-unique(gather(df))
</code></pre>

This will generate a few warnings, but provide your requested result. cbind with start repeating the vectors so that all of the vectors are the same length.

entityMap|blocks|key|1n0ps|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7h0on|df<-data.frame(blue,Turq,Indigo)
library(reshape)
unique(melt(cbind(1:nrow(df),df),id=1)[,-1])

+++variable+value
1++++++blue++++Ti
2++++++blue++++Ka
3++++++blue++++Vv
4++++++blue++++Cn
5++++++Turq++++Cc
6++++++Turq++++Wx
7++++++Turq++++Oq
8++++++Turq++++Yt
9++++Indigo++++Gb
10+++Indigo++++Ap|code-block|syntax|javascript|9mlfl^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

Try:

<pre><code>df&lt;-data.frame(blue,Turq,Indigo)
library(reshape)
unique(melt(cbind(1:nrow(df),df),id=1)[,-1])

 variable value
1 blue Ti
2 blue Ka
3 blue Vv
4 blue Cn
5 Turq Cc
6 Turq Wx
7 Turq Oq
8 Turq Yt
9 Indigo Gb
10 Indigo Ap
</code></pre>

entityMap|blocks|key|6apff|text|您可以将向量放在命名列表中，然后使用stack|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|cgha0|lst+<-+list(blue+=+blue,+Turq+=+Turq,+Indigo+=+Indigo)
lst
#+$blue
#+[1]+"Ti"+"Ka"+"Vv"+"Cn"

#+$Turq
#+[1]+"Cc"+"Wx"+"Oq"+"Yt"

#+$Indigo
#+[1]+"Gb"+"Ap"

stack(lst)
#+++values++++ind
#1++++++Ti+++blue
#2++++++Ka+++blue
#3++++++Vv+++blue
#4++++++Cn+++blue
#5++++++Cc+++Turq
#6++++++Wx+++Turq
#7++++++Oq+++Turq
#8++++++Yt+++Turq
#9++++++Gb+Indigo
#10+++++Ap+Indigo|code-block|syntax|javascript|1oevn^0|I|5|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@$A|N|B|O|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|P|9|@]|E|@]|F|$J|K]]|$2|L|4|-4|6|7|8|Q|9|@]|E|@]|F|$]]]]

You can put the vectors in a named list and then use <code>stack</code>:

<pre><code>lst &lt;- list(blue = blue, Turq = Turq, Indigo = Indigo)
lst
# $blue
# [1] "Ti" "Ka" "Vv" "Cn"

# $Turq
# [1] "Cc" "Wx" "Oq" "Yt"

# $Indigo
# [1] "Gb" "Ap"

stack(lst)
# values ind
#1 Ti blue
#2 Ka blue
#3 Vv blue
#4 Cn blue
#5 Cc Turq
#6 Wx Turq
#7 Oq Turq
#8 Yt Turq
#9 Gb Indigo
#10 Ap Indigo
</code></pre>

entityMap|blocks|key|23v34|text|我们可以在将向量放入list之后使用melt。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|ark10|library(reshape2)
melt(mget(c("blue",+"Turq",+"Indigo")))[2:1]
#+++++++L1+value
#1++++blue++++Ti
#2++++blue++++Ka
#3++++blue++++Vv
#4++++blue++++Cn
#5++++Turq++++Cc
#6++++Turq++++Wx
#7++++Turq++++Oq
#8++++Turq++++Yt
#9++Indigo++++Gb
#10+Indigo++++Ap|code-block|syntax|javascript|c7ol9^0|A|4|I|4|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@$A|N|B|O|C|D]|$A|P|B|Q|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|R|9|@]|E|@]|F|$J|K]]|$2|L|4|-4|6|7|8|S|9|@]|E|@]|F|$]]]]

We can use <code>melt</code> after placing the vectors in a <code>list</code>.

<pre><code>library(reshape2)
melt(mget(c("blue", "Turq", "Indigo")))[2:1]
# L1 value
#1 blue Ti
#2 blue Ka
#3 blue Vv
#4 blue Cn
#5 Turq Cc
#6 Turq Wx
#7 Turq Oq
#8 Turq Yt
#9 Indigo Gb
#10 Indigo Ap
</code></pre>

I have three character vectors:

<pre><code>blue &lt;-c("Ti","Ka","Vv","Cn")
Turq &lt;-c("Cc","Wx","Oq","Yt")
Indigo &lt;-c("Gb","Ap")
</code></pre>

and I want to combine them into one dataframe so that they look like this:

<pre><code>blue Ti
blue Ka
blue Vv
blue Cn
Turq Cc
Turq Wx
Turq Oq
Turq Yt
Indigo Gb
Indigo Ap
</code></pre>

I cannot seem to find a way

R combine character vectors into dataframe, with vector name as row id

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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我有三个字符向量： blue <-c("Ti","Ka","Vv","Cn")Turq <-c("Cc","Wx","Oq","Yt")Indigo <-c("Gb","Ap") 我想把它们组合成一个数据帧，这样它们看起来就像这样： blue Tiblue Kablue Vvblue CnTurq ...

问R将字符向量组合为dataframe，并将向量名称作为行id
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问R将字符向量组合为dataframe，并将向量名称作为行idEN