blocks|key|1048858|text|which(match(df$Y,df$X)!=1:nrow(df))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1048859|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>which(match(df$Y,df$X)!=1:nrow(df))
</code></pre>

blocks|key|514124|text|1|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|514125|对于每个Y，我们检查X中的任何值(同一行除外)是否匹配。|CODE|514126|sapply(1:NROW(df),+function(i)+df$Y[i]+%25in%25+df$X[-i])
#[1]+FALSE++TRUE+FALSE|code-block|syntax|javascript|514127|如果需要索引，请将整个过程包装在which中|514128|which(sapply(1:NROW(df),+function(i)+df$Y[i]+%25in%25+df$X[-i]))
#[1]+2|514129|2+(测试不佳)|514130|df+<-+data.frame(X=c('a','b','c'),+Y=c('a','c','d'),+stringsAsFactors+=+FALSE)
temp+=+outer(df$X,+df$Y,+"==")++#Check+equality+among+values+of+X+and+Y
diag(temp)+=+FALSE++++++++++++++#Set+diagonal+values+as+FALSE+(for+same+row)
colSums(temp)+>+0
#[1]+FALSE++TRUE+FALSE|514131|entityMap^0|0|1|0|4|1|A|1|0|0|G|5|0|0|0|1|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|10|8|@$9|11|A|12|B|H]|$9|13|A|14|B|H]]|D|@]|E|$]]|$1|I|3|J|5|K|7|15|8|@]|D|@]|E|$L|M]]|$1|N|3|O|5|6|7|16|8|@$9|17|A|18|B|H]]|D|@]|E|$]]|$1|P|3|Q|5|K|7|19|8|@]|D|@]|E|$L|M]]|$1|R|3|S|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|T|3|U|5|K|7|1D|8|@]|D|@]|E|$L|M]]|$1|V|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|W|$]]

1

For each <code>Y</code>, we check if any value in <code>X</code> (other than in the same row) matches.

<pre><code>sapply(1:NROW(df), function(i) df$Y[i] %in% df$X[-i])
#[1] FALSE TRUE FALSE
</code></pre>

If indices are necessary, wrap the whole thing in <code>which</code>

<pre><code>which(sapply(1:NROW(df), function(i) df$Y[i] %in% df$X[-i]))
#[1] 2
</code></pre>

2 (not tested well)

<pre><code>df &lt;- data.frame(X=c('a','b','c'), Y=c('a','c','d'), stringsAsFactors = FALSE)
temp = outer(df$X, df$Y, "==") #Check equality among values of X and Y
diag(temp) = FALSE #Set diagonal values as FALSE (for same row)
colSums(temp) &gt; 0
#[1] FALSE TRUE FALSE
</code></pre>

blocks|key|262735|text|#+Assume+you+want+to+use+position+4,+value+'c',+to+find+all+the+rows+that+Y+is+'c'
df+<-+data.frame(X+=+c('a',+'b',+'d',+'c'),
+++++++++++++++++Y+=+c('a',+'c',+'c',+'d'))


row+<-+4+#+assume+the+desire+row+is+position+4

val+<-+as.character(+df[(row),'X']+)+#+get+the+character+and+turn+it+into+character+type

df[df$Y+==+val,]
#+Result
#+++X+Y
#+2+b+c
#+3+d+c|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|262736|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># Assume you want to use position 4, value 'c', to find all the rows that Y is 'c'
df &lt;- data.frame(X = c('a', 'b', 'd', 'c'),
 Y = c('a', 'c', 'c', 'd'))


row &lt;- 4 # assume the desire row is position 4

val &lt;- as.character( df[(row),'X'] ) # get the character and turn it into character type

df[df$Y == val,]
# Result
# X Y
# 2 b c
# 3 d c
</code></pre>

blocks|key|1048909|text|我觉得这个应该管用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1048910|df+<-+data.frame(X=+c(1,2,3,4,5,3,2,1),+Y+=+c(1,2,3,4,5,6,7,8))
which(with(df,++(X+%25in%25+Y)+&+(X+!=+Y)))|code-block|syntax|javascript|1048911|在原始data.frame上工作，如果我们设置了stringsasfactors=FALSE|1048912|df+<-+data.frame(X=c('a','b','c'),+Y=c('a','c','d'),+stringsAsFactors+=+F)++++
which(with(df,++(X+%25in%25+Y)+&+(X+!=+Y)))|1048913|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I think this should work.

<pre><code>df &lt;- data.frame(X= c(1,2,3,4,5,3,2,1), Y = c(1,2,3,4,5,6,7,8))
which(with(df, (X %in% Y) &amp; (X != Y)))
</code></pre>

Works on the original data.frame, if we set stringsasfactors=FALSE

<pre><code>df &lt;- data.frame(X=c('a','b','c'), Y=c('a','c','d'), stringsAsFactors = F) 
which(with(df, (X %in% Y) &amp; (X != Y)))
</code></pre>

blocks|key|514145|text|很复杂，但我还是把它放在这里。即使X中有重复的值，这也应该有效。例如，使用以下数据格式的df2。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|514146|df2+=+data.frame(X=c('a','b','c','a','d'),+Y=c('a','c','d','e','b'))
++X+Y
1+a+a
2+b+c
3+c+d
4+a+e
5+d+b


##+Specifying+the+same+factor+levels+allows+us+to+get+a+square+matrix+
df2$X+=+factor(df2$X,levels=union(df2$X,df2$Y))
df2$Y+=+factor(df2$Y,levels=union(df2$X,df2$Y))
m+=+as.matrix(table(df2))
valY+=+rowSums(m)*colSums(m)-diag(m)
which(df2$Y+%25in%25+names(valY)[as.logical(valY)])
[1]+1+2+3+5|code-block|syntax|javascript|514147|entityMap^0|18|3|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|$]]

Quite convoluted but I'll put it here anyway. This should work even if there are repeated values in X.
For example with the following dataframe <code>df2</code>:

<pre><code>df2 = data.frame(X=c('a','b','c','a','d'), Y=c('a','c','d','e','b'))
 X Y
1 a a
2 b c
3 c d
4 a e
5 d b


## Specifying the same factor levels allows us to get a square matrix 
df2$X = factor(df2$X,levels=union(df2$X,df2$Y))
df2$Y = factor(df2$Y,levels=union(df2$X,df2$Y))
m = as.matrix(table(df2))
valY = rowSums(m)*colSums(m)-diag(m)
which(df2$Y %in% names(valY)[as.logical(valY)])
[1] 1 2 3 5
</code></pre>

blocks|key|514161|text|本质上，您想知道Y是否在X中，但是当X+==+Y时，您希望条件是FALSE|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|514162|df$Z+<-+with(df,+(Y+!=+X)+&+(Y+%25in%25+X))|code-block|syntax|javascript|514163|entityMap^0|8|1|C|1|I|6|W|5|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]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

Essentially you want to know whether <code>Y</code> is in <code>X</code> but you want the condition to be <code>FALSE</code> when <code>X == Y</code>:

<pre><code>df$Z &lt;- with(df, (Y != X) &amp; (Y %in% X))
</code></pre>

I have the following <code>df</code> and use-case, I'd like to find and set something in all rows for which exist another row satisfying a condition e.g.

<pre><code>df &lt;- data.frame(X=c('a','b','c'), Y=c('a','c','d'))
&gt; df
 X Y
1 a a
2 b c
3 c d
</code></pre>

I'd like to find those rows whos Y value is the same as X value in another row. In the example above would be row #2 is true because <code>Y = c</code> and row #3 has <code>X = c</code>. Note that row #1 does not satisfy the condition.

Something like:

<pre><code>df$Flag &lt;- find(df, Y == X_in_another_row(df))
</code></pre>

Data frame: How to compare current row to some other rows without looping?

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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我有下面的df和用例，我想在所有行中找到并设置一些东西，这些行存在另一个满足条件的行。df <- data.frame(X=c('a','b','c'), Y=c('a','c','d'))> df X Y1 a a2 b c3 c d我想在(另一个行)中找到Y值与X值相同的行。在上面...

问数据帧:如何将当前行与其他行进行比较而不进行循环？
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问数据帧:如何将当前行与其他行进行比较而不进行循环？EN