blocks|key|1311939|text|您可以为长格式创建一个具有适当id和time列的新数据框，然后将其与原始数据框合并。这将为不匹配的值保留NA，然后可以用0替换这些值|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1311940|merge(foo,+
++++++with(foo,+
+++++++++++data.frame(id=rep(id,time),+time=sequence(time))
++++++),+
++++++all.y=TRUE
)
##++++id+time++y
##+1+++1++++1+NA
##+2+++1++++2++0
##+3+++2++++1+NA
##+4+++2++++2+NA
##+5+++2++++3++1
##+6+++3++++1+NA
##+7+++3++++2+NA
##+8+++3++++3+NA
##+9+++3++++4++0
##+10++4++++1+NA
##+11++4++++2++1
##+12++5++++1+NA
##+13++5++++2+NA
##+14++5++++3++0|code-block|syntax|javascript|1311941|类似的合并适用于第一次扩展。将不带time列的foo与上面创建的数据框合并：|1311942|merge(foo[c('id','y')],+
++++++with(foo,+
+++++++++++data.frame(id=rep(id,time),+time=sequence(time))
++++++)
)+
##++++id+y+time
##+1+++1+0++++1
##+2+++1+0++++2
##+3+++2+1++++1
##+4+++2+1++++2
##+5+++2+1++++3
##+6+++3+0++++1
##+7+++3+0++++2
##+8+++3+0++++3
##+9+++3+0++++4
##+10++4+1++++1
##+11++4+1++++2
##+12++5+0++++1
##+13++5+0++++2
##+14++5+0++++3|1311943|没有必要在后一个表达式中指定all+(或all.y)，因为每个匹配的time值都有多个id值，这些值都被展开了。在前一种情况下，两个数据帧的time值都是匹配的，如果不指定all+(或all.y)，就会得到原始数据。|1311944|entityMap^0|F|2|I|4|1G|2|1O|1|0|0|H|4|N|3|0|0|E|3|K|5|Y|4|17|2|1Y|4|2E|3|2K|5|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]|$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|1N|8|@]|D|@]|E|$]]]|R|$]]

You can create a new data frame with the proper <code>id</code> and <code>time</code> columns for the long format, then merge that with the original. This leaves <code>NA</code> for the unmatched values, which can then be substituted with <code>0</code>:

<pre><code>merge(foo, 
 with(foo, 
 data.frame(id=rep(id,time), time=sequence(time))
 ), 
 all.y=TRUE
)
## id time y
## 1 1 1 NA
## 2 1 2 0
## 3 2 1 NA
## 4 2 2 NA
## 5 2 3 1
## 6 3 1 NA
## 7 3 2 NA
## 8 3 3 NA
## 9 3 4 0
## 10 4 1 NA
## 11 4 2 1
## 12 5 1 NA
## 13 5 2 NA
## 14 5 3 0
</code></pre>

A similar merge works for the first expansion. Merge <code>foo</code> without the <code>time</code> column with the same created data frame as above:

<pre><code>merge(foo[c('id','y')], 
 with(foo, 
 data.frame(id=rep(id,time), time=sequence(time))
 )
) 
## id y time
## 1 1 0 1
## 2 1 0 2
## 3 2 1 1
## 4 2 1 2
## 5 2 1 3
## 6 3 0 1
## 7 3 0 2
## 8 3 0 3
## 9 3 0 4
## 10 4 1 1
## 11 4 1 2
## 12 5 0 1
## 13 5 0 2
## 14 5 0 3
</code></pre>

It's not necessary to specify <code>all</code> (or <code>all.y</code>) in the latter expression because there are multiple <code>time</code> values for each matching <code>id</code> value, and these are expanded. In the prior case, the <code>time</code> values were matched from both data frames, and without specifying <code>all</code> (or <code>all.y</code>) you would get your original data back.

blocks|key|1311952|text|初始扩展可以通过以下方式实现：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1311953|newdat+<-+transform(+
++foo[rep(rownames(foo),foo$time),],+
++time+=+sequence(foo$time)
)

#++++id+y+time
#1++++1+0++++1
#1.1++1+0++++2
#2++++2+1++++1
#2.1++2+1++++2
#2.2++2+1++++3
#+etc|code-block|syntax|javascript|1311954|要获得完整的解决方案，包括奖励部分，请执行以下操作：|1311955|newdat$y[-cumsum(foo$time)]+<-+0

#++++id+y+time
#1++++1+0++++1
#1.1++1+0++++2
#2++++2+0++++1
#2.1++2+0++++2
#2.2++2+1++++3
#etc|1311956|如果您真的很兴奋，您可以使用within一步完成所有这些操作|offset|length|style|CODE|1311957|within(
++foo[rep(rownames(foo),foo$time),],
++{
++++time+<-+sequence(foo$time)
++++y[-cumsum(foo$time)]+<-+0
++}
)|1311958|entityMap^0|0|0|0|0|E|6|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@$M|Z|N|10|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

The initial expansion can be achieved with:

<pre><code>newdat &lt;- transform( 
 foo[rep(rownames(foo),foo$time),], 
 time = sequence(foo$time)
)

# id y time
#1 1 0 1
#1.1 1 0 2
#2 2 1 1
#2.1 2 1 2
#2.2 2 1 3
# etc
</code></pre>

To get the complete solution, including the bonus part, then do:

<pre><code>newdat$y[-cumsum(foo$time)] &lt;- 0

# id y time
#1 1 0 1
#1.1 1 0 2
#2 2 0 1
#2.1 2 0 2
#2.2 2 1 3
#etc
</code></pre>

If you were really excitable, you could do it all in one step using <code>within</code>:

<pre><code>within(
 foo[rep(rownames(foo),foo$time),],
 {
 time &lt;- sequence(foo$time)
 y[-cumsum(foo$time)] &lt;- 0
 }
)
</code></pre>

blocks|key|5096264|text|使用dplyr+(和magritte以提高易读性)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5096265|library(magrittr)
library(dplyr)

foo[rep(1:nrow(foo),+foo$time),+]+%25>%25
++++group_by(id)+%25>%25
++++mutate(y+=+!duplicated(y,+fromLast+=+TRUE),
++++++++++++++++++time+=+1:n())|code-block|syntax|javascript|5096266|希望能有所帮助|5096267|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

With dplyr (and magritte for nice legibility):

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

foo[rep(1:nrow(foo), foo$time), ] %&gt;%
 group_by(id) %&gt;%
 mutate(y = !duplicated(y, fromLast = TRUE),
 time = 1:n())
</code></pre>

Hope it helps

blocks|key|1252085|text|如果你愿意使用"data.table"，你可以试试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1252086|library(data.table)
fooDT+<-+as.data.table(foo)
fooDT[,+list(time+=+sequence(time)),+by+=+list(id,+y)]
#+++++id+y+time
#++1:++1+0++++1
#++2:++1+0++++2
#++3:++2+1++++1
#++4:++2+1++++2
#++5:++2+1++++3
#++6:++3+0++++1
#++7:++3+0++++2
#++8:++3+0++++3
#++9:++3+0++++4
#+10:++4+1++++1
#+11:++4+1++++2
#+12:++5+0++++1
#+13:++5+0++++2
#+14:++5+0++++3|code-block|syntax|javascript|1252087|还有，关于奖金的问题：|1252088|fooDT[,+list(time+=+sequence(time)),+
++++++by+=+list(id,+y)][,+y+:=+{y[1:(.N-1)]+<-+0;+y},+
++++++++++++++++++++++++by+=+id][]
#+++++id+y+time
#++1:++1+0++++1
#++2:++1+0++++2
#++3:++2+0++++1
#++4:++2+0++++2
#++5:++2+1++++3
#++6:++3+0++++1
#++7:++3+0++++2
#++8:++3+0++++3
#++9:++3+0++++4
#+10:++4+0++++1
#+11:++4+1++++2
#+12:++5+0++++1
#+13:++5+0++++2
#+14:++5+0++++3|1252089|对于奖金问题，另一个选择是：|1252090|fooDT[,+list(time=seq_len(time)),+by=list(id,y)][y+==+1,+
++++++++++++++++y+:=+c(rep.int(0,+.N-1L),+1),+by=id][]|1252091|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

If you're willing to go with "data.table", you can try:

<pre><code>library(data.table)
fooDT &lt;- as.data.table(foo)
fooDT[, list(time = sequence(time)), by = list(id, y)]
# id y time
# 1: 1 0 1
# 2: 1 0 2
# 3: 2 1 1
# 4: 2 1 2
# 5: 2 1 3
# 6: 3 0 1
# 7: 3 0 2
# 8: 3 0 3
# 9: 3 0 4
# 10: 4 1 1
# 11: 4 1 2
# 12: 5 0 1
# 13: 5 0 2
# 14: 5 0 3
</code></pre>

And, for the bonus question:

<pre><code>fooDT[, list(time = sequence(time)), 
 by = list(id, y)][, y := {y[1:(.N-1)] &lt;- 0; y}, 
 by = id][]
# id y time
# 1: 1 0 1
# 2: 1 0 2
# 3: 2 0 1
# 4: 2 0 2
# 5: 2 1 3
# 6: 3 0 1
# 7: 3 0 2
# 8: 3 0 3
# 9: 3 0 4
# 10: 4 0 1
# 11: 4 1 2
# 12: 5 0 1
# 13: 5 0 2
# 14: 5 0 3
</code></pre>

For the bonus question, alternatively:

<pre><code>fooDT[, list(time=seq_len(time)), by=list(id,y)][y == 1, 
 y := c(rep.int(0, .N-1L), 1), by=id][]
</code></pre>

I would like to convert my data from a short format to a long format and I imagine there is a simple way to do it (possibly with reshape2, plyr, dplyr, etc?).

For example, I have:

<pre><code>foo &lt;- data.frame(id = 1:5, 
 y = c(0, 1, 0, 1, 0),
 time = c(2, 3, 4, 2, 3))

id y time
1 0 2
2 1 3
3 0 4
4 1 2
5 0 3
</code></pre>

I would like to expand/copy each row n times, where n is that row's value in the "time" column. However, I would also like the variable "time" to be incremented from 1 to n. That is, I would like to produce:

<pre><code>id y time
1 0 1
1 0 2
2 1 1
2 1 2
2 1 3
3 0 1
3 0 2
3 0 3
3 0 4
4 1 1
4 1 2
5 0 1
5 0 2
5 0 3
</code></pre>

As a bonus, I would also like to do a sort of incrementing of the variable "y" where, for those ids with y = 1, y is set to 0 until the largest value of "time". That is, I would like to produce:

<pre><code>id y time
1 0 1
1 0 2
2 0 1
2 0 2
2 1 3
3 0 1
3 0 2
3 0 3
3 0 4
4 0 1
4 1 2
5 0 1
5 0 2
5 0 3
</code></pre>

This seems like something that dplyr might already do, but I just don't know where to look. Regardless, any solution that avoids loops is helpful.

expand data.frame to long format and increment value

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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我想要将我的数据从短格式转换成长格式，我想有一种简单的方法可以做到这一点(可能使用reshape2、plyr、dplyr等？)。例如，我有：foo <- data.frame(id = 1:5, y = c(0, 1, 0, 1, 0), time = c(2, 3...

问将data.frame扩展为长格式和增量值
EN

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问将data.frame扩展为长格式和增量值EN