blocks|key|4453215|text|您可以计算为每个MACHINE_NUMBER找到的不同MACHINE_KIND的数量|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4453216|library(data.table)
df[,n+:=+uniqueN("MACHINE_KIND"),+by+=+"MACHINE_NUMBER"][n>=2]|code-block|syntax|javascript|4453217|如果您只对A和B类别感兴趣，请先过滤：|4453218|library(data.table)
df["MACHINE_KIND"+%25in%25+c("A","B"),n+:=+uniqueN("MACHINE_KIND"),+by+=+"MACHINE_NUMBER"][n>=2]|4453219|entityMap^0|8|E|R|C|0|0|5|1|7|1|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

You can count the number of different <code>MACHINE_KIND</code> you find for each <code>MACHINE_NUMBER</code>

<pre class="lang-r prettyprint-override"><code>library(data.table)
df[,n := uniqueN("MACHINE_KIND"), by = "MACHINE_NUMBER"][n&gt;=2]
</code></pre>

If you are only interested in <code>A</code> and <code>B</code> categories, filter first:

<pre class="lang-r prettyprint-override"><code>library(data.table)
df["MACHINE_KIND" %in% c("A","B"),n := uniqueN("MACHINE_KIND"), by = "MACHINE_NUMBER"][n&gt;=2]
</code></pre>

blocks|key|1664807|text|Base+R一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1664808|unique(Reduce(function(x,+y){setdiff(x,+y)},+split(df$MACHINE_NUMBER,+df$MACHINE_NUMBER)))|code-block|syntax|javascript|1664809|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|$]]

Base R one-liner:

<pre><code>unique(Reduce(function(x, y){setdiff(x, y)}, split(df$MACHINE_NUMBER, df$MACHINE_NUMBER)))
</code></pre>

blocks|key|2136315|text|您可以使用dplyr包和stringr包来检查MACHINE_NUMBER是否具有这两种(或更多)类型。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2136316|library(tidyverse)
library(stringr)

a+<-+tibble::tribble(
++++~MACHINE_NUMBER,+~MACHINE_KIND,
++++12345L,+++++++++++"A",
++++12346L,+++++++++++"A",
++++12347L,+++++++++++"A",
++++12345L,+++++++++++"B",
++++12348L,+++++++++++"B"
)+

a+%25>%25+
++++group_by(MACHINE_NUMBER)+%25>%25+
++++summarize(both+=+if_else(all(str_detect(MACHINE_KIND,+unique(a$MACHINE_KIND))),+"yes",+"no"))
#>+#+A+tibble:+4+x+2
#>+++MACHINE_NUMBER+both+
#>+*++++++++++<int>+<chr>
#>+1++++++++++12345+yes++
#>+2++++++++++12346+no+++
#>+3++++++++++12347+no+++
#>+4++++++++++12348+no|code-block|syntax|javascript|2136317|然后，您可以根据需要过滤它们，以防您想要对数据进行更多分析。|2136318|entityMap^0|5|5|C|7|N|E|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]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

You could use the <code>dplyr</code> package and the <code>stringr</code> package to check if the <code>MACHINE_NUMBER</code> has both (or more) kinds.

<pre><code>library(tidyverse)
library(stringr)

a &lt;- tibble::tribble(
 ~MACHINE_NUMBER, ~MACHINE_KIND,
 12345L, "A",
 12346L, "A",
 12347L, "A",
 12345L, "B",
 12348L, "B"
) 

a %&gt;% 
 group_by(MACHINE_NUMBER) %&gt;% 
 summarize(both = if_else(all(str_detect(MACHINE_KIND, unique(a$MACHINE_KIND))), "yes", "no"))
#&gt; # A tibble: 4 x 2
#&gt; MACHINE_NUMBER both 
#&gt; * &lt;int&gt; &lt;chr&gt;
#&gt; 1 12345 yes 
#&gt; 2 12346 no 
#&gt; 3 12347 no 
#&gt; 4 12348 no
</code></pre>

Then you could filter them if you want, just in case you want to do more analyses with your data.

blocks|key|4453301|text|#+example+dataset
dt+=+read.table(text+=+"
MACHINE_NUMBER+%7C+MACHINE_KIND
12345+%7C+A
12346+%7C+A
12347+%7C+A
12345+%7C+B
12348+%7C+B
",+header=T,+sep="%7C")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4453302|使用基数R的一种选择|unstyled|4453303|#+create+table+based+on+unique+rows+of+your+dataframe
tbl+=+table(unique(dt)$MACHINE_NUMBER)

#+get+those+that+have+2+rows+(i.e.+belong+in+both+groups)
names(tbl)[tbl+==+2]

#+[1]+"12345"|4453304|另一种选择是使用dplyr|offset|length|style|CODE|4453305|library(dplyr)

dt+%25>%25
++group_by(MACHINE_NUMBER)+%25>%25
++summarise(Counts+=+n_distinct(MACHINE_KIND))+%25>%25
++filter(Counts+==+2)

#+#+A+tibble:+1+x+2
#++++++MACHINE_NUMBER+Counts
#++++++++++++++<dbl>++<int>
#+++1++++++++++12345++++++2|4453306|请注意，如果您有多个组，则可以使用>=+2而不是==+2。|4453307|entityMap^0|0|0|0|8|5|0|0|H|4|O|4|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|X|8|@$K|Y|L|Z|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|Q|3|R|5|F|7|11|8|@$K|12|L|13|M|N]|$K|14|L|15|M|N]]|9|@]|A|$]]|$1|S|3|-4|5|F|7|16|8|@]|9|@]|A|$]]]|T|$]]

<pre><code># example dataset
dt = read.table(text = "
MACHINE_NUMBER | MACHINE_KIND
12345 | A
12346 | A
12347 | A
12345 | B
12348 | B
", header=T, sep="|")
</code></pre>

One option using base R

<pre><code># create table based on unique rows of your dataframe
tbl = table(unique(dt)$MACHINE_NUMBER)

# get those that have 2 rows (i.e. belong in both groups)
names(tbl)[tbl == 2]

# [1] "12345"
</code></pre>

Another option using <code>dplyr</code>:

<pre><code>library(dplyr)

dt %&gt;%
 group_by(MACHINE_NUMBER) %&gt;%
 summarise(Counts = n_distinct(MACHINE_KIND)) %&gt;%
 filter(Counts == 2)

# # A tibble: 1 x 2
# MACHINE_NUMBER Counts
# &lt;dbl&gt; &lt;int&gt;
# 1 12345 2
</code></pre>

Note that if you have multiple groups you can use <code>&gt;= 2</code>, instead of <code>== 2</code>.

I have a data set <code>age_structure</code> something like this:

<pre><code>MACHINE_NUMBER | MACHINE_KIND
 12345 | A
 12346 | A
 12347 | A
 12345 | B
 12348 | B
</code></pre>

I need to somehow determine which MACHINE_NUMBER are in both MACHINE_KIND subsets. In this case, only 12345 should arise. So far I have tried subsetting the data frame like this, but because the MACHINE_KIND column is not in the subset, R throws an error:

<pre><code>common_numbers &lt;- Reduce(intersect, list(subset(age_structure$MACHINE_NUMBER, MACHINE_KIND == "A"), subset(age_structure$MACHINE_NUMBER, MACHINE_KIND == "B")))

object 'MACHINE_KIND' not found
</code></pre>

Defining the <code>age_structure</code> name as part of the condition runs the code without error, but the resulting <code>common_numbers</code> is empty.

<pre><code>common_numbers &lt;- Reduce(intersect, list(subset(age_structure$MACHINE_NUMBER, age_structure$MACHINE_KIND == "A"), subset(age_structure$MACHINE_NUMBER, age_structure$MACHINE_KIND == "B")))
</code></pre>

MACHINE_NUMBER and MACHINE_KIND are both factors. What is the advisable thing to do?

R: finding intersections in a subset by different column

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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我有一个类似下面的数据集age_structure：MACHINE_NUMBER | MACHINE_KIND         12345 | A         12346 | A         12347 | A         12345 | B         12348 | B我需要以某种方式确定两个MAC...

问R:按不同列查找子集中的交集
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问R:按不同列查找子集中的交集EN