blocks|key|1767336|text|您可以为此使用sapply：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1767337|ID+<-+c(1,2,3,4,5,6,7,8,9)
Basket_List+<-+list(integer(0),c(123,987),c(123,123),456,
++++++++++++++++++++c(456,123),456,c(123,987),c(987,123),987)
d.f+<-+data.frame(ID)

sel+<-+sapply(+Basket_List,+function(bl,searchItem)+{
++any(searchItem+%25in%25+bl)
},+searchItem=c(123)+)

>+sel
[1]+FALSE++TRUE++TRUE+FALSE++TRUE+FALSE++TRUE++TRUE+FALSE

>+d.f[sel,,drop=FALSE]
++ID
2++2
3++3
5++5
7++7
8++8|code-block|syntax|javascript|1767338|请小心你的术语。data.frame不是矩阵。这是一种名单。|1767339|就速度而言，sapply不是最快的，但是选择将非常快，因为它是矢量化的。如果你需要更多的速度，data.table时间。|1767340|entityMap^0|7|6|0|0|0|6|6|1B|A|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]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

<p>You can use <code>sapply</code> for this:</p>

<pre><code>ID &lt;- c(1,2,3,4,5,6,7,8,9)
Basket_List &lt;- list(integer(0),c(123,987),c(123,123),456,
                    c(456,123),456,c(123,987),c(987,123),987)
d.f &lt;- data.frame(ID)

sel &lt;- sapply( Basket_List, function(bl,searchItem) {
  any(searchItem %in% bl)
}, searchItem=c(123) )

&gt; sel
[1] FALSE  TRUE  TRUE FALSE  TRUE FALSE  TRUE  TRUE FALSE

&gt; d.f[sel,,drop=FALSE]
  ID
2  2
3  3
5  5
7  7
8  8
</code></pre>

<p>Please be careful with your terminology.  A data.frame is not a matrix.  It's a type of list.</p>

<p>Speed-wise, <code>sapply</code> is not the fastest, but the selection will be very fast since it is vectorized.  If you need more speed, <code>data.table</code> time.</p>


blocks|key|2805237|text|@AriB的一种类似方法是使用any操作符，跨行运行apply，如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2805238|d.f[+apply(+d.f+,+1+,+function(x)+any(unlist(x)+%25in%25+123)+)+,+]
#++ID+Basket_List
#2++2++++123,+987
#3++3++++123,+123
#5++5++++456,+123
#7++7++++123,+987
#8++8++++987,+123|code-block|syntax|javascript|2805239|在第二次设置数据时，我想它会非常快，因为您可以简单地像这样的子集：|2805240|df[+df$Basket+%25in%25+123+,+]
#+++ID+Basket
#NA+NA+++++NA
#2+++2++++123
#4+++3++++123
#5+++3++++123
#8+++5++++123
#10++7++++123
#13++8++++123|2805241|如果您只想要包含Basket值的行的第一个实例，那么随后可以在惟一in中使用match，因为match返回它的第二个参数的第一个匹配项：|2805242|df2+<-+df[+df$Basket+%25in%25+123+,+]
df2[+match(+unique(df2$ID)+,+df2$ID),]
#+++ID+Basket
#NA+NA+++++NA
#2+++2++++123
#4+++3++++123
#8+++5++++123
#10++7++++123
#13++8++++123|2805243|数据的第二次设置将比我认为的第一次要快得多。实际上，让我们在100万行表上使用它做一个粗略的基准测试：|2805244|DF+<-+data.frame(+ID+=+sample(ID+,+1e6+,+repl=TRUE)+,+Basket+=+sample(Basket+,+1e6+,+repl+=+TRUE)+)
df<-DF

system.time({
++df2+<-+df[+df$Basket+%25in%25+123+,+]
++df2[+match(+unique(df2$ID)+,+df2$ID),]
})
#+++user++system+elapsed+
#+++0.16++++0.00++++0.16+

nrow(df)
#[1]+1000000
nrow(df2)
#[1]+428187|2805245|entityMap^0|F|3|Q|5|0|0|0|0|8|6|12|5|1A|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1E|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|X|$]]

<p>A similar approach to @AriB is to use the <code>any</code> operator, <code>apply</code> ing across rows, like so:</p>

<pre><code>d.f[ apply( d.f , 1 , function(x) any(unlist(x) %in% 123) ) , ]
#  ID Basket_List
#2  2    123, 987
#3  3    123, 123
#5  5    456, 123
#7  7    123, 987
#8  8    987, 123
</code></pre>

<p>With the second set up of your data I imagine that it would be very fast, because you could simply subset like so:</p>

<pre><code>df[ df$Basket %in% 123 , ]
#   ID Basket
#NA NA     NA
#2   2    123
#4   3    123
#5   3    123
#8   5    123
#10  7    123
#13  8    123
</code></pre>

<p>And if you only want the first instance of a row that contains <code>Basket</code> value you can subsequently use <code>match</code> with the unique IDs, as <code>match</code> returns first match of it's first argument in it's second:</p>

<pre><code>df2 &lt;- df[ df$Basket %in% 123 , ]
df2[ match( unique(df2$ID) , df2$ID),]
#   ID Basket
#NA NA     NA
#2   2    123
#4   3    123
#8   5    123
#10  7    123
#13  8    123
</code></pre>

<p>The second setup of your data will be far faster than the first I think. In fact, let's do a rough benchmark with it on a 1 million row table:</p>

<pre><code>DF &lt;- data.frame( ID = sample(ID , 1e6 , repl=TRUE) , Basket = sample(Basket , 1e6 , repl = TRUE) )
df&lt;-DF

system.time({
  df2 &lt;- df[ df$Basket %in% 123 , ]
  df2[ match( unique(df2$ID) , df2$ID),]
})
#   user  system elapsed 
#   0.16    0.00    0.16 

nrow(df)
#[1] 1000000
nrow(df2)
#[1] 428187
</code></pre>


blocks|key|2805279|text|使用purrr+&+dplyr库(和magrittr管道操作符)的一个稍微容易读的解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2805280|library(dplyr)
library(purrr)++++

d.f+%25>%25+filter(map_lgl(Basket_List,contains,as.integer(123)))|code-block|syntax|javascript|2805281|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|$]]

<p>A slightly more readable solution using the purrr &amp; dplyr libraries (and the magrittr pipe operator) would be:</p>

<pre><code>library(dplyr)
library(purrr)    

d.f %&gt;% filter(map_lgl(Basket_List,contains,as.integer(123)))
</code></pre>


<h3>Set-Up</h3>

<p>I have a list matrix, where one of the "columns" is a list (I realize it's an odd dataset to work with, but I find it useful for other operations). Each entry of the list is either; (1) empty (integer(0)), (2) an integer, or (3) a vector of integers. </p>

<p>E.g. the R object "d.f", With d.f$ID an index vector, and d.f$Basket_List the list. </p>

<pre><code>ID &lt;- c(1,2,3,4,5,6,7,8,9)
Basket_List &lt;- list(integer(0),c(123,987),c(123,123),456,
                    c(456,123),456,c(123,987),c(987,123),987)
d.f &lt;- data.frame(ID)
d.f$Basket_List &lt;- Basket_List
</code></pre>

<h2>My Question</h2>

<h3>Issue 1</h3>

<p>I'd like to create a new dataset that's a subset of the initial, based on whether or not "Basket_List" contains certain value(s). E.g. a subset of all the rows in d.f such that Bask_list has "123" or "123" &amp; "987" -- or other more complicated conditions. </p>

<p>I've tried every variation of the following, but to no avail. </p>

<pre><code>d.f2 &lt;- subset(d.f, 123 %in% Basket_List)
d.f2 &lt;- subset(d.f, 123 == any(Basket_List))
d.f2 &lt;- d.f[which(123 %in% d.f$Basket_List,]
# should return the subset, with rows 2,3,5,7 &amp; 8
</code></pre>

<h3>Issue 2</h3>

<p>My other issue is that'd I'll be running this operation over many millions of rows (it's transaction data), so I'd like to optimize it as much as possible for speed (I have a complicated for loop now, but it takes too much time).  </p>

<hr>

<h3>Alternative Set-Up of Data</h3>

<p>If you think it might be useful, the data might also be set-up as the following: </p>

<pre><code>ID &lt;- c(1,2,2,3,3,4,5,5,6,7,7,8,8,9)
Basket &lt;- c(NA,123,987,123,123,456,456,123,456,123,987,987,123,987)
alt.d.f &lt;- data.frame(ID,Basket)
</code></pre>


Subsetting Data Frame Based on Contents of a "Column" List

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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架设我有一个列表矩阵，其中一个“列”是一个列表(我意识到它是一个奇怪的数据集，但我发现它对于其他操作很有用)。列表的每个条目都是；(1)空(整数(0))，(2)整数，或(3)整数向量。例如R对象"d.f"，其中d.f$ID是索引向量，d.f$Basket_List是列表。ID <- c(1,2,3,4,5,6,7,8,...

问根据“列”列表的内容设置数据框架
EN

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问根据“列”列表的内容设置数据框架EN