blocks|key|1225133|text|我不知道我是否正确地理解了你的问题，但也许这里的这个问题是有帮助的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1225134|value+=+round(runif(20,+min+=+0.1,+max+=+0.5),+1)

csumVec+=+numeric(length(value))
counterVec+=+numeric(length(value))
startIndex+=+1
csum+=+0
counter+=+1

for(i+in+1:length(value))+{
++csum+=+csum+%2B+value[i]
++if(csum+>+1)+{
++++counterVec[startIndex:i]+=+counter
++++csumVec[startIndex:i]+=+csum-value[i]
++++startIndex+=+i
++++counter+=+counter%2B1
++++csum+=+value[i]
++}
++if(i+==+length(value))+{
++++counterVec[startIndex:i]+=+counter
++++csumVec[startIndex:i]+=+csum
++}
}

cbind(value,+counterVec,+csumVec)|code-block|syntax|javascript|1225135|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|$]]

I don't know if I understood your problem correctly, but maybe this one here helps:
<pre><code>value = round(runif(20, min = 0.1, max = 0.5), 1)

csumVec = numeric(length(value))
counterVec = numeric(length(value))
startIndex = 1
csum = 0
counter = 1

for(i in 1:length(value)) {
 csum = csum + value[i]
 if(csum &gt; 1) {
 counterVec[startIndex:i] = counter
 csumVec[startIndex:i] = csum-value[i]
 startIndex = i
 counter = counter+1
 csum = value[i]
 }
 if(i == length(value)) {
 counterVec[startIndex:i] = counter
 csumVec[startIndex:i] = csum
 }
}

cbind(value, counterVec, csumVec)
</code></pre>

blocks|key|913520|text|似乎您可以计算累积和，除以1，并取floor()+(四舍五入)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|913521|floor(cumsum(value)+/+1)
##+[1]+0+0+0+1+1+3|code-block|syntax|javascript|913522|这是正确的，只是它是基于0的，并且counter不会增加1。|913523|counter0+=+floor(cumsum(value)+/+1)
counter+=+match(counter0,+unique(counter0))
counter
##+[1]+1+1+1+2+2+3|913524|得到“棘手”部分后，我将使用dplyr+(library(dplyr))作为其余部分|913525|##+library(dplyr)
tibble(value,+counter)+%7C>
++++mutate(cum_sum+=+cumsum(value))+%7C>
++++group_by(counter)+%7C>
++++mutate(cumsum+=+max(cumsum(value)))
##+#+A+tibble:+6+×+3
##+#+Groups:+++counter+[3]
##+++value+counter+cumsum
##+++<dbl>+++<int>++<dbl>
##+1+++0.3+++++++1++++0.9
##+2+++0.3+++++++1++++0.9
##+3+++0.3+++++++1++++0.9
##+4+++0.3+++++++2++++0.4
##+5+++0.1+++++++2++++0.4
##+6+++2+++++++++3++++2|913526|或者在(更一般的)函数中捕获棘手的部分。|913527|cumgroup+<-+function(x,+upper+=+1)+{
++++counter0+=+floor(cumsum(x)+/+upper)
++++match(counter0,+unique(counter0))
}|913528|并将其并入dplyr解决方案|913529|tibble(value)+%7C>
++++mutate(counter+=+cumgroup(value))+%7C>
++++group_by(counter)+%7C>
++++mutate(cumsum+=+max(cumsum(value)))|913530|或者取决于你到底想要什么|913531|tibble(value)+%7C>
++++mutate(
++++++++cumsum+=+cumsum(value)+%25%25+1,
++++++++counter+=+cumgroup(value)
++++)+%7C>
++++group_by(counter)+%7C>+
++++mutate(cumsum+=+max(cumsum))+%7C>
++++select(value,+counter,+cumsum)|913532|entityMap^0|H|7|0|0|H|7|0|0|L|E|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1D|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1E|8|@$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1I|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|1K|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|H|7|1L|8|@]|D|@]|E|$I|J]]|$1|10|3|11|5|6|7|1M|8|@]|D|@]|E|$]]|$1|12|3|13|5|H|7|1N|8|@]|D|@]|E|$I|J]]|$1|14|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]]|15|$]]

It seems like you can calculate the cumulative sum, divide by 1, and take the <code>floor()</code> (round down)
<pre><code>floor(cumsum(value) / 1)
## [1] 0 0 0 1 1 3
</code></pre>
This is correct, except that it is 0-based and the <code>counter</code> does not increment by 1. Fix these by matching the result above with their unique values
<pre><code>counter0 = floor(cumsum(value) / 1)
counter = match(counter0, unique(counter0))
counter
## [1] 1 1 1 2 2 3
</code></pre>
Having got the 'tricky' part, I'd use dplyr (<code>library(dplyr)</code>) for the rest
<pre><code>## library(dplyr)
tibble(value, counter) |&gt;
 mutate(cum_sum = cumsum(value)) |&gt;
 group_by(counter) |&gt;
 mutate(cumsum = max(cumsum(value)))
## # A tibble: 6 × 3
## # Groups: counter [3]
## value counter cumsum
## &lt;dbl&gt; &lt;int&gt; &lt;dbl&gt;
## 1 0.3 1 0.9
## 2 0.3 1 0.9
## 3 0.3 1 0.9
## 4 0.3 2 0.4
## 5 0.1 2 0.4
## 6 2 3 2
</code></pre>
or perhaps capturing the tricky part in a (more general) function
<pre><code>cumgroup &lt;- function(x, upper = 1) {
 counter0 = floor(cumsum(x) / upper)
 match(counter0, unique(counter0))
}
</code></pre>
and incorporating into the dplyr solution
<pre><code>tibble(value) |&gt;
 mutate(counter = cumgroup(value)) |&gt;
 group_by(counter) |&gt;
 mutate(cumsum = max(cumsum(value)))
</code></pre>
or depending on what precisely you want
<pre><code>tibble(value) |&gt;
 mutate(
 cumsum = cumsum(value) %% 1,
 counter = cumgroup(value)
 ) |&gt;
 group_by(counter) |&gt; 
 mutate(cumsum = max(cumsum)) |&gt;
 select(value, counter, cumsum)
</code></pre>

I have a dataframe with numerical values in one row. Now I want to calculate the cumsum of those rows, until &gt;= 1.
If this point is reached -&gt; print for all those rows a counter, write in every row the cumsum for its counter, then look for the cumsum of the next rows.
Should look somewhow like this:
<pre><code>value counter cumsum
0.3 1 0.9
0.3 1 0.9
0.3 1 0.9
0.3 2 0.4
0.1 2 0.4
2 3 2
</code></pre>
My problem is how to tell R to stop the cumsum, if &gt;= than 1. Any ideas?
Thank you in advance.

R loop until condition matched, then go on

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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我在一行中有一个带有数值的数据格式。现在我要计算这些行的累积和，直到>= 1。如果达到这一点，所有这些行都打印为计数器，在每一行中写入其计数器的累计和，然后查找下一行的累积和。应该是这样的：value    counter   cumsum0.3      1         0.90.3      1        ...

问循环，直到条件匹配，然后继续。
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问循环，直到条件匹配，然后继续。EN