blocks|key|4227539|text|可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4227540|bind_rows(list(tibble1=tibble1,+tibble2=tibble2),+.id='source')
#+A+tibble:+3+x+3
++source++++++a+++++b
++<chr>+++<dbl>+<dbl>
1+tibble1+++++1+++++2
2+tibble1+++++3+++++4
3+tibble2+++++5+++++6|code-block|syntax|javascript|4227541|4227542|如果您没有输入名称：|4227543|bind_rows_with_source+<-+function(...,+.id+=+'source'){
++bind_rows(setNames(list(...),+as.character(substitute(...()))),+.id+=+.id)
}

bind_rows_with_source(tibble1,tibble2)
#+A+tibble:+3+x+3
++source++++++a+++++b
++<chr>+++<dbl>+<dbl>
1+tibble1+++++1+++++2
2+tibble1+++++3+++++4
3+tibble2+++++5+++++6|4227544|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|O|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|P|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|Q|8|@]|9|@]|A|$]]|$1|J|3|K|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|L|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|M|$]]

This could be done as:
<pre><code>bind_rows(list(tibble1=tibble1, tibble2=tibble2), .id='source')
# A tibble: 3 x 3
 source a b
 &lt;chr&gt; &lt;dbl&gt; &lt;dbl&gt;
1 tibble1 1 2
2 tibble1 3 4
3 tibble2 5 6
</code></pre>
<hr />
If you refer not inputing names:
<pre><code>bind_rows_with_source &lt;- function(..., .id = 'source'){
 bind_rows(setNames(list(...), as.character(substitute(...()))), .id = .id)
}

bind_rows_with_source(tibble1,tibble2)
# A tibble: 3 x 3
 source a b
 &lt;chr&gt; &lt;dbl&gt; &lt;dbl&gt;
1 tibble1 1 2
2 tibble1 3 4
3 tibble2 5 6
</code></pre>

blocks|key|951024|text|如果您真的希望您的函数签名与...一起使用，则可以使用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|951025|bind_rows_with_source+<-+function(...){
++tibbleNames+<-+as.character(unlist(as.list(match.call())[-1]))
++tibbleList+<-+setNames(lapply(tibbleNames,get),tibbleNames)
++sourceCol+<-+rep(tibbleNames,times=sapply(tibbleList,NROW))
++out+<-+do.call("rbind",tibbleList)
++out$source+<-+sourceCol
++return(out)
}|code-block|syntax|javascript|951026|或者如果您可以使用dplyr|951027|bind_rows_with_source+<-+function(...){
++tibbleNames+<-+as.character(unlist(as.list(match.call())[-1]))
++tibbleList+<-+setNames(lapply(tibbleNames,get),tibbleNames)
++dplyr::bind_rows(tibbleList,+.id='source')
}|951028|entityMap^0|E|3|0|0|9|5|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|$]]

If you really want your function-signature to be with <code>...</code>, you can use
<pre><code>bind_rows_with_source &lt;- function(...){
 tibbleNames &lt;- as.character(unlist(as.list(match.call())[-1]))
 tibbleList &lt;- setNames(lapply(tibbleNames,get),tibbleNames)
 sourceCol &lt;- rep(tibbleNames,times=sapply(tibbleList,NROW))
 out &lt;- do.call(&quot;rbind&quot;,tibbleList)
 out$source &lt;- sourceCol
 return(out)
}
</code></pre>
or if you can use <code>dplyr</code>
<pre><code>bind_rows_with_source &lt;- function(...){
 tibbleNames &lt;- as.character(unlist(as.list(match.call())[-1]))
 tibbleList &lt;- setNames(lapply(tibbleNames,get),tibbleNames)
 dplyr::bind_rows(tibbleList, .id='source')
}
</code></pre>

blocks|key|4227612|text|我们可以使用lazyeval包:一种使用公式进行非标准评估的替代方法。提供LISP样式‘准引号’的完整实现，使与其他代码一起生成代码更容易。https://cran.r-project.org/web/packages/lazyeval/lazyeval.pdf|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4227613|library(lazyeval)
my_function+<-+function(df)+{
++df+<-+df+%25>%25+mutate(ref+=+expr_label(df))
++return(df)
}

a+<-+my_function(tibble1)
b+<-+my_function(tibble2)
bind_rows(a,+b)|code-block|syntax|javascript|4227614|输出：|4227615|++++++a+++++b+ref++++++
++<dbl>+<dbl>+<chr>++++
1+++++1+++++2+`tibble1`
2+++++3+++++4+`tibble1`
3+++++5+++++6+`tibble2`|4227616|entityMap|0|LINK|mutability|MUTABLE|url|https://cran.r-project.org/web/packages/lazyeval/lazyeval.pdf^0|6|8|1Y|1P|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

We could use <code>lazyeval</code> package: An alternative approach to non-standard evaluation using formulas. Provides a full implementation of LISP style 'quasiquotation',making it easier to generate code with other code.
<a href="https://cran.r-project.org/web/packages/lazyeval/lazyeval.pdf" rel="nofollow noreferrer">https://cran.r-project.org/web/packages/lazyeval/lazyeval.pdf</a>
<pre><code>library(lazyeval)
my_function &lt;- function(df) {
 df &lt;- df %&gt;% mutate(ref = expr_label(df))
 return(df)
}

a &lt;- my_function(tibble1)
b &lt;- my_function(tibble2)
bind_rows(a, b)
</code></pre>
Output:
<pre><code> a b ref 
 &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt; 
1 1 2 `tibble1`
2 3 4 `tibble1`
3 5 6 `tibble2`
</code></pre>

blocks|key|951046|text|另一个选择是rbindlist|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|951047|library(data.table)
rbindlist(list(tibble1,+tibble2),+idcol+=+'source')|code-block|syntax|javascript|951048|entityMap^0|6|9|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|$]]

Another option is <code>rbindlist</code>
<pre><code>library(data.table)
rbindlist(list(tibble1, tibble2), idcol = 'source')
</code></pre>

I want to be able to take data that's in the same format but from different sources and concatenate the rows, but to keep track of the source of the data I'd like to also introduce a source column.
This seems routine enough that I thought I'd create a utility function to do it, but I'm having trouble getting it to work.
Here's what I tried:
<pre><code>library(tidyverse)

tibble1 = tribble(
 ~a, ~b,
 1,2,
 3,4
)

tibble2 = tribble(
 ~a, ~b,
 5,6
)

bind_rows_with_source &lt;- function(...){
 out = tibble()
 for (newtibb in list(...)){
 out &lt;- bind_rows(out, newtibb %&gt;% mutate(source = deparse(substitute(newtibb))))
 }
 return(out)
}

bind_rows_with_source(tibble1,tibble2)
#source column just contains the string 'newtibb' on all rows
#I want it to contain tibble1 for the first two rows and tibble2 for the third:
#~a, ~b, ~source
# 1, 2, tibble1
# 3, 4, tibble1
# 5, 6, tibble2
</code></pre>
Is there already a function that could achieve this?
Is there a better approach than the utility function I tried to create?
Is there a way to correct my approach?
Sincere thanks for reading my question

Function to concatenate rows of data with an additional source column in R (and tidyverse)

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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我希望能够获取格式相同但来自不同来源的数据，并将行连接起来，但是要跟踪数据的来源，我还想介绍一个源列。这似乎是例行公事，所以我想我应该创建一个实用程序函数来完成它，但是我很难让它开始工作。以下是我尝试过的：library(tidyverse)tibble1 = tribble(  ~a, ~b,  1,2,  3,4)...

问函数将数据行与R(和tidyverse)中的附加源列连接起来。
EN

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问函数将数据行与R(和tidyverse)中的附加源列连接起来。EN