blocks|key|1478356|text|你可以试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1478357|v1+<-+c(1:ncol(dat))+
v2+<-+v1+%2B+c(1L,+1L,+-2L)+
m+<-+cbind(v1,v2)
fit+<-+lapply(1:length(dat),function(x)+lm(dat[,m[x,1]]~dat[,m[x,2]]))
rsq+<-+sapply(1:length(dat),+function(x)+summary(fit[[x]])$r.squared)|code-block|syntax|javascript|1478358|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|$]]

You can try this:

<pre><code>v1 &lt;- c(1:ncol(dat)) 
v2 &lt;- v1 + c(1L, 1L, -2L) 
m &lt;- cbind(v1,v2)
fit &lt;- lapply(1:length(dat),function(x) lm(dat[,m[x,1]]~dat[,m[x,2]]))
rsq &lt;- sapply(1:length(dat), function(x) summary(fit[[x]])$r.squared)
</code></pre>

blocks|key|688652|text|它应该能发挥作用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|688653|results+<-+apply(combn(colnames(dat),+2),+2,+function(x)summary(lm(dat[,+x[1]]+~+dat[,+x[2]]))$r.squared)|code-block|syntax|javascript|688654|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|$]]

It should work:

<pre><code>results &lt;- apply(combn(colnames(dat), 2), 2, function(x)summary(lm(dat[, x[1]] ~ dat[, x[2]]))$r.squared)
</code></pre>

blocks|key|1478363|text|或者是一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1478364|results+<-+sapply(1:ncol(dat),+function(x)+summary(+lm(+dat[+,+x+]+~+dat[+,ifelse(+x%25%253+!=+0,+x%2B1,+x-2)])+)$r.squared+)|code-block|syntax|javascript|1478365|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|$]]

Or in one line:

<pre><code>results &lt;- sapply(1:ncol(dat), function(x) summary( lm( dat[ , x ] ~ dat[ ,ifelse( x%%3 != 0, x+1, x-2)]) )$r.squared )
</code></pre>

blocks|key|688688|text|如果您只需要r平方值，您可以使用cor函数给出相关矩阵。r2只是该矩阵中值的平方。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|688689|entityMap^0|G|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

If you just need the r-squared value, you can use the <code>cor</code> function to give the correlation matrix. The r2 is just the square of the values in that matrix.

blocks|key|695963|text|使用dplyr包的替代过程：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|695964|set.seed(1234)
dat+<-+data.frame(replicate(18,rnorm(10)))

library(dplyr)


data.frame(colnames+=+names(dat))+%25>%25++++++++#+get+the+names+of+columns
++mutate(group+=+cumsum(ifelse(row_number()+%25in%25+seq(1,ncol(dat),3),1,0)))+%25>%25++#+create+group+id+based+on+3+consecutive+columns
++group_by(group)+%25>%25++++++++++++++++++++++++#+for+each+group+id
++do({cb+=+combn(.$colnames,2)+++++++++++++++#+create+combinations+of+column+names
++++++data.frame(col1+=+cb[1,],
+++++++++++++++++col2+=+cb[2,])})+%25>%25
++mutate(formula+=+paste(col1,"~",col2))+%25>%25+#+create+a+formula+for+each+combination
++rowwise()+%25>%25++++++++++++++++++++++++++++++#+for+each+row/formula
++do(data.frame(formula+=+.$formula,
++++++++++++++++r.sq+=+summary(lm(.$formula,+data=dat))$r.squared))+#+create+model+and+get+r+squared


#++++++formula+++++++++r.sq
#++++++++(chr)++++++++(dbl)
#+1++++X1+~+X2+3.072421e-02
#+2++++X1+~+X3+3.056746e-01
#+3++++X2+~+X3+7.708176e-02
#+4++++X4+~+X5+7.293980e-01
#+5++++X4+~+X6+3.244157e-01
#+6++++X5+~+X6+2.231886e-01
#+7++++X7+~+X8+6.637355e-03
#+8++++X7+~+X9+1.497414e-06
#+9++++X8+~+X9+9.758725e-02
#+10+X10+~+X11+2.728225e-01
#+11+X10+~+X12+5.973809e-02
#+12+X11+~+X12+1.196112e-01
#+13+X13+~+X14+5.541950e-02
#+14+X13+~+X15+3.488573e-02
#+15+X14+~+X15+2.519877e-02
#+16+X16+~+X17+7.004510e-04
#+17+X16+~+X18+8.827935e-02
#+18+X17+~+X18+1.112862e-01|code-block|syntax|javascript|695965|如果您愿意，可以将mutate(group+=+cumsum(ifelse(row_number()+%25in%25+seq(1,ncol(dat),3),1,0)))+(基于3列的窗口创建对)替换为mutate(group+=+ntile(row_number(),6))+(创建6组连续的3列)。|695966|entityMap^0|2|5|0|0|9|20|2P|11|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

An alternative process using <code>dplyr</code> package:

<pre><code>set.seed(1234)
dat &lt;- data.frame(replicate(18,rnorm(10)))

library(dplyr)


data.frame(colnames = names(dat)) %&gt;% # get the names of columns
 mutate(group = cumsum(ifelse(row_number() %in% seq(1,ncol(dat),3),1,0))) %&gt;% # create group id based on 3 consecutive columns
 group_by(group) %&gt;% # for each group id
 do({cb = combn(.$colnames,2) # create combinations of column names
 data.frame(col1 = cb[1,],
 col2 = cb[2,])}) %&gt;%
 mutate(formula = paste(col1,"~",col2)) %&gt;% # create a formula for each combination
 rowwise() %&gt;% # for each row/formula
 do(data.frame(formula = .$formula,
 r.sq = summary(lm(.$formula, data=dat))$r.squared)) # create model and get r squared


# formula r.sq
# (chr) (dbl)
# 1 X1 ~ X2 3.072421e-02
# 2 X1 ~ X3 3.056746e-01
# 3 X2 ~ X3 7.708176e-02
# 4 X4 ~ X5 7.293980e-01
# 5 X4 ~ X6 3.244157e-01
# 6 X5 ~ X6 2.231886e-01
# 7 X7 ~ X8 6.637355e-03
# 8 X7 ~ X9 1.497414e-06
# 9 X8 ~ X9 9.758725e-02
# 10 X10 ~ X11 2.728225e-01
# 11 X10 ~ X12 5.973809e-02
# 12 X11 ~ X12 1.196112e-01
# 13 X13 ~ X14 5.541950e-02
# 14 X13 ~ X15 3.488573e-02
# 15 X14 ~ X15 2.519877e-02
# 16 X16 ~ X17 7.004510e-04
# 17 X16 ~ X18 8.827935e-02
# 18 X17 ~ X18 1.112862e-01
</code></pre>

If you prefer you can replace <code>mutate(group = cumsum(ifelse(row_number() %in% seq(1,ncol(dat),3),1,0)))</code> (create pairs based on a window of 3 consecutive columns) with
 <code>mutate(group = ntile(row_number(),6))</code> (create 6 groups of 3 consecutive columns).

I am computing the r-squared for multiple pairs of columns in a data frame. I can do this by individually writing out the code for each pair but I wanted to automate this using apply or some other vectorized approach based on the pattern of columns I am choosing from the data frame.

Sample data:

<pre><code>set.seed(1234)
dat &lt;- data.frame(replicate(18,rnorm(10)))
</code></pre>

To get the r-squared for column 1 v. 2:

<pre><code>fit &lt;- lm(dat[,1] ~ dat[,2])
summary(fit)$r.squared
</code></pre>

But I would like to do all of the following combinations: {1, 2}, {2, 3}, {3, 1}, {4, 5}, {5, 6}, {6, 4}... etc. through the 18th column.

In other words, all combinations of three with a window moving over to the next set of three each time. This way I can just call the function once on the whole data frame and get all the r-squared values at once instead of repeating the code 18 times.

Subsetting dataframe in R based on specifc pattern of columns

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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我正在计算数据帧中多对列的r平方。我可以通过为每对单独写出代码来做到这一点，但是我想使用apply或其他一些矢量化方法来实现这一点，这是基于我从数据框架中选择的列的模式。样本数据：set.seed(1234)dat <- data.frame(replicate(18,rnorm(10)))为了得到第1节第2节的r平方...

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