blocks|key|3849231|text|这里有一个选项，假设您的data.frame名为"SODF"：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3849232|within(SODF[c(1,+seq(5,+nrow(SODF),+5)),+],+{+
++Price+<-+diff(c(0,+Price))+
})[-1,+]
#++++Price+Day
#+5+++++12+++5
#+10++++15++10
#+15++++15++15
#+20++++15++20|code-block|syntax|javascript|3849233|3849234|第一步是基本子集。根据您的描述和预期答案，您需要第一行，然后从第5行开始每隔5行：|3849235|>+SODF[c(1,+seq(5,+nrow(SODF),+5)),+]
+++Price+Day
1++++++2+++1
5+++++14+++5
10++++29++10
15++++44++15
20++++59++20|3849236|在那里，您可以在“价格”列上使用diff，但是由于diff将导致一个长度比您的输入短一个的向量，您需要“填充”输入向量，就像我使用diff(c(0,+Price))所做的那样。|3849237|#+Correct+values,+but+the+number+of+rows+needs+to+be+5
>+diff(SODF[c(1,+seq(5,+nrow(SODF),+5)),+"Price"])
[1]+12+15+15+15|3849238|然后，末尾的[-1,+]只删除无关的行。|3849239|更新|3849240|在下面的评论中，@geektrader在评论中指出(谢谢！)，这是使用：|3849241|SODF[c(1,+seq(5,+nrow(SODF),+5)),+]|3849242|作为您的输入data.frame，您可以考虑使用以下内容：|3849243|rbind(SODF[1,],+SODF[$Day+%25%25+5+==+0,]+)|3849244|这两种方法的不同之处在于，第一种方法只是根据行号进行子集，而第二种方法根据"Day“列中的值进行子集，提取其中"Day”是5的倍数的行。第二种方法可能很有用，例如，当数据集中有缺失的行时。|3849245|entityMap^0|C|A|0|0|0|0|0|G|4|P|4|1T|H|0|0|6|6|0|0|0|0|6|A|0|0|0^^$0|@$1|2|3|4|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|1E|8|@]|D|@]|E|$]]|$1|N|3|O|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|P|3|Q|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]]|D|@]|E|$]]|$1|R|3|S|5|H|7|1N|8|@]|D|@]|E|$I|J]]|$1|T|3|U|5|6|7|1O|8|@$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|V|3|W|5|6|7|1R|8|@]|D|@]|E|$]]|$1|X|3|Y|5|6|7|1S|8|@]|D|@]|E|$]]|$1|Z|3|10|5|H|7|1T|8|@]|D|@]|E|$I|J]]|$1|11|3|12|5|6|7|1U|8|@$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|13|3|14|5|H|7|1X|8|@]|D|@]|E|$I|J]]|$1|15|3|16|5|6|7|1Y|8|@]|D|@]|E|$]]|$1|17|3|-4|5|6|7|1Z|8|@]|D|@]|E|$]]]|18|$]]

Here's one option, assuming your <code>data.frame</code> is called "SODF":

<pre><code>within(SODF[c(1, seq(5, nrow(SODF), 5)), ], { 
 Price &lt;- diff(c(0, Price)) 
})[-1, ]
# Price Day
# 5 12 5
# 10 15 10
# 15 15 15
# 20 15 20
</code></pre>

<hr>

The first step is basic subsetting. According to your description and expected answer, you want the first row, and then every fifth row starting from row 5:

<pre><code>&gt; SODF[c(1, seq(5, nrow(SODF), 5)), ]
 Price Day
1 2 1
5 14 5
10 29 10
15 44 15
20 59 20
</code></pre>

From there, you can use <code>diff</code> on the "Price" column, but since <code>diff</code> will result in a vector that is one in length shorter than your input, you need to "pad" the input vector, which I did with <code>diff(c(0, Price))</code>.

<pre><code># Correct values, but the number of rows needs to be 5
&gt; diff(SODF[c(1, seq(5, nrow(SODF), 5)), "Price"])
[1] 12 15 15 15
</code></pre>

Then, the <code>[-1, ]</code> at the end just deletes the extraneous row.

<h2>Update</h2>

In the comments below, @geektrader points out in the comments (thanks!), an alternative to using:

<pre><code>SODF[c(1, seq(5, nrow(SODF), 5)), ]
</code></pre>

as your input <code>data.frame</code>, you may consider using the following instead:

<pre><code>rbind(SODF[1,], SODF[$Day %% 5 == 0,] )
</code></pre>

The difference in the two approaches is that the first approach simply subsets by row number, while the second approach subsets according to the value in the "Day" column, extracting rows where "Day" is a multiple of 5. This second approach might be useful, for instance, when there are missing rows in the dataset.

blocks|key|1918033|text|Ananda's是一种很好的方法(总是忘记自己的内心)。这是另一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1918034|dat2+<-+dat[seq(0,+nrow(dat),+by=5),+]
data.frame(Difference=diff(c(dat[1,1],+dat2[,+1])),+Day=dat2[,+2])|code-block|syntax|javascript|1918035|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|$]]

Ananda's is a nice approach (always forget about within myself). Here's another approach:

<pre><code>dat2 &lt;- dat[seq(0, nrow(dat), by=5), ]
data.frame(Difference=diff(c(dat[1,1], dat2[, 1])), Day=dat2[, 2])
</code></pre>

blocks|key|3849304|text|如果你有一个矩阵作为输入，这里是一个解决方案。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|3849305|在给定矩阵m、列col_id和数字间隔interv的情况下，后续函数将每个interv行的m矩阵的col_id列中的当前值与先前的值(之前的5行，明显相同的列)相减。|CODE|3849306|结果存储在一个名为diff的新列中，并附加到m矩阵的末尾。|3849307|简而言之，这种方法与@Ananda+Mahto使用的方法非常相似。|3849308|因此，这是一个函数：|3849309|subtract_column+<-+function(m,+col_id,+interv)+{
++select+<-+c(1,+seq(interv,+nrow(m),+interv))
++cbind(m[select[-1],+],+diff+=+diff(m[select,+col_id]))
}|code-block|syntax|javascript|3849310|示例：|3849311|#+this+emulates+your+data+as+a+matrix
price_vect+<-+c(2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,59)
day_vect+<-+1:20
matr+<-+do.call(cbind,+list(price+=+price_vect,+day+=+day_vect))
#+and+this+calls+the+function+above+and+does+the+job:
#+subtracts+every+5+rows+the+current+and+the+previous+(5+rows+back)+value+in+the+column+`price`+of+matrix+`matr`
subtract_column(matr,+'price',+5)|3849312|输出：|3849313|+++++price+day+diff
[1,]++++14+++5+++12
[2,]++++29++10+++15
[3,]++++44++15+++15
[4,]++++59++20+++15|3849314|entityMap^0|I|5|0|0|2B|5|1|8|6|J|6|11|6|19|1|1D|6|0|0|T|9|4|M|1|0|0|X|0|0|A|0|0|0|3|0|0|0|3|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|H]|$9|1B|A|1C|B|H]|$9|1D|A|1E|B|H]|$9|1F|A|1G|B|H]|$9|1H|A|1I|B|H]|$9|1J|A|1K|B|H]]|D|@]|E|$]]|$1|I|3|J|5|6|7|1L|8|@$9|1M|A|1N|B|C]|$9|1O|A|1P|B|H]|$9|1Q|A|1R|B|H]]|D|@]|E|$]]|$1|K|3|L|5|6|7|1S|8|@$9|1T|A|1U|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1V|8|@$9|1W|A|1X|B|C]]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1Y|8|@]|D|@]|E|$R|S]]|$1|T|3|U|5|6|7|1Z|8|@$9|20|A|21|B|C]]|D|@]|E|$]]|$1|V|3|W|5|Q|7|22|8|@]|D|@]|E|$R|S]]|$1|X|3|Y|5|6|7|23|8|@$9|24|A|25|B|C]]|D|@]|E|$]]|$1|Z|3|10|5|Q|7|26|8|@]|D|@]|E|$R|S]]|$1|11|3|-4|5|6|7|27|8|@]|D|@]|E|$]]]|12|$]]

Here a solution if you have a matrix as input.

The subsequent function, given a matrix <code>m</code>, a column <code>col_id</code> and a numeric interval <code>interv</code>, subtracts every <code>interv</code> rows the current value in the <code>col_id</code> column of the <code>m</code> matrix with the previous value (5 rows before, same column, obiviously).

The results are stored in a new column called <code>diff</code> and appended to the end of the <code>m</code> matrix.

In short, the approach is very similar to that used by @Ananda Mahto.

So, this is the function:

<pre><code>subtract_column &lt;- function(m, col_id, interv) {
 select &lt;- c(1, seq(interv, nrow(m), interv))
 cbind(m[select[-1], ], diff = diff(m[select, col_id]))
}
</code></pre>

Example:

<pre><code># this emulates your data as a matrix
price_vect &lt;- c(2,5,8,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,59)
day_vect &lt;- 1:20
matr &lt;- do.call(cbind, list(price = price_vect, day = day_vect))
# and this calls the function above and does the job:
# subtracts every 5 rows the current and the previous (5 rows back) value in the column `price` of matrix `matr`
subtract_column(matr, 'price', 5)
</code></pre>

Output:

<pre><code> price day diff
[1,] 14 5 12
[2,] 29 10 15
[3,] 44 15 15
[4,] 59 20 15
</code></pre>

So As you can see I have a price and Day columns below

<pre><code> Price Day
 2 1
 5 2
 8 3
 11 4
 14 5
 17 6
 20 7
 23 8
 26 9
 29 10
 32 11
 35 12
 38 13
 41 14
 44 15
 47 16
 50 17
 53 18
 56 19
 59 20
</code></pre>

I then want the output below

<pre><code> Difference Day
 12 5
 15 10
 15 15
 15 20
</code></pre>

So now I have the difference in prices every 5 days...it just basically subtracts the 5th day with the first day.....and then the 10th day with the 5th day etc....
I already made a code that will seperate my data into 5 day intervals...but I want the code that will let me minus the 5th with the 1st day....the 10th day with the 5th day...etc
So the code should look something like this

<pre><code>difference&lt;-tapply(Price[,1],Day, ____________)
</code></pre>

So basically Price[,1] will be my Price data.....while "Day" is the variable that I created that will let me seperate my Day data into 5 day intervals.....I'm thinking that in the blank section I could put in the function or another variable that will let me subtract the 5th day with the 1st day prices and then the 10th day and 5th day prices...etc.....you dont have to help me to seperate my Days into intervals...just how to do "difference" section....thanks guys

How to extract certain rows

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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如您所见，我在下面有一个价格和日期列 Price  Day    2   1    5   2    8   3    11  4    14  5    17  6    20  7    23  8    26  9    29  10    32  11    35  12    38  13    41  14...

问如何提取某些行
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