entityMap|blocks|key|bqfrk|text|NA序列的开始可以这样计算：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3jsk3|library("data.table")
DT+<-+data.table(a+=+c(0,+NA,+NA,+0,+NA,+1,+1,+NA,+0,+NA,+1,+NA,+NA,+NA,+0,+1,+1,+0,+NA,+0))

r+<-+DT[,+rle(is.na(a))]
R+<-+data.table(r$values,+r$lengths,+start=c(1,+1%2Bhead(cumsum(r$lengths),+-1)))

i+<-+R[(V1),+start]
j+<-+R[(V1),+start%2BV2-1]
i[(DT[i-1,+a]+%2B+DT[j%2B1,+a])==1]
#+result:+[1]++5++8+10+12|code-block|syntax|javascript|ne9^0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|J|9|@]|A|@]|B|$F|G]]|$2|H|4|-4|6|7|8|K|9|@]|A|@]|B|$]]]]

The begin of the NA-sequences can calculated so:
<pre><code>library(&quot;data.table&quot;)
DT &lt;- data.table(a = c(0, NA, NA, 0, NA, 1, 1, NA, 0, NA, 1, NA, NA, NA, 0, 1, 1, 0, NA, 0))

r &lt;- DT[, rle(is.na(a))]
R &lt;- data.table(r$values, r$lengths, start=c(1, 1+head(cumsum(r$lengths), -1)))

i &lt;- R[(V1), start]
j &lt;- R[(V1), start+V2-1]
i[(DT[i-1, a] + DT[j+1, a])==1]
# result: [1] 5 8 10 12
</code></pre>

entityMap|blocks|key|13v2j|text|您可以尝试使用approx|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2eb7n|DT[,b+:=+approx((1:.N)[!is.na(a)],na.omit(a),1:.N)$y]|code-block|syntax|javascript|5je2q|然后再申请|e4p6h|DT[,+which(is.na(a)+&+b>0+&+b<1)]|a6ur6|或|87eb3|DT[,+which(is.na(a)+&+between(b,+0,+1,+FALSE))]|4n780|这给了我们|8ltki|[1]++5++8+10+12+13+14|7k4sh^0|7|6|0|0|0|0|0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|Y|9|@$A|Z|B|10|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|11|9|@]|E|@]|F|$J|K]]|$2|L|4|M|6|7|8|12|9|@]|E|@]|F|$]]|$2|N|4|O|6|I|8|13|9|@]|E|@]|F|$J|K]]|$2|P|4|Q|6|7|8|14|9|@]|E|@]|F|$]]|$2|R|4|S|6|I|8|15|9|@]|E|@]|F|$J|K]]|$2|T|4|U|6|7|8|16|9|@]|E|@]|F|$]]|$2|V|4|W|6|I|8|17|9|@]|E|@]|F|$J|K]]|$2|X|4|-4|6|7|8|18|9|@]|E|@]|F|$]]]]

You can try using <code>approx</code>
<pre><code>DT[,b := approx((1:.N)[!is.na(a)],na.omit(a),1:.N)$y]
</code></pre>
and then apply
<pre><code>DT[, which(is.na(a) &amp; b&gt;0 &amp; b&lt;1)]
</code></pre>
or
<pre><code>DT[, which(is.na(a) &amp; between(b, 0, 1, FALSE))]
</code></pre>
which gives
<pre><code>[1] 5 8 10 12 13 14
</code></pre>

entityMap|0|type|LINK|mutability|MUTABLE|data|url|https://stackoverflow.com/questions/7735647/replacing-nas-with-latest-non-na-value|blocks|key|1n0p7|text|正如Dirk+Eddelbuettel在这里：Replacing+NAs+with+latest+non-NA+value中描述的那样，zoo包及其na.locf()函数可以为您提供帮助。|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|enmqa|library(data.table)
library(zoo)

DT+<-+data.table(a+=+c(0,+NA,+NA,+0,+NA,+1,+1,+NA,+0,+NA,+1,+NA,+NA,+NA,+0,+1,+1,+0,+NA,+0))

non_nas+<-+DT[!is.na(a),+a]
successor+<-+c(non_nas[-1],+0)
diff+<-+abs(non_nas+-+successor)
DT[!is.na(a),+diff:=diff]|code-block|syntax|javascript|6r5sb|这将为您提供如下数据表：|ckc9u|+++++a+diff
+1:++0++++0
+2:+NA+++NA
+3:+NA+++NA
+4:++0++++1
+5:+NA+++NA
+6:++1++++0
+7:++1++++1
+8:+NA+++NA
+9:++0++++1
10:+NA+++NA
11:++1++++1
12:+NA+++NA
13:+NA+++NA
14:+NA+++NA
15:++0++++1
16:++1++++0
17:++1++++1
18:++0++++0
19:+NA+++NA
20:++0++++0|18lfj|这里的想法是，diff列中的每个'1‘告诉您，在执行下面的NAs之后，'a’中的值将发生变化。|748u7|现在，您希望删除'diff‘列中的NAs。为了清楚起见，我们将结果放入新的列'b‘中。这就是zoo包发挥作用的地方：|diae9|DT[,+b:=na.locf(diff)]|afknc|这会导致|3qr3i|+++++a+diff+b
+1:++0++++0+0
+2:+NA+++NA+0
+3:+NA+++NA+0
+4:++0++++1+1
+5:+NA+++NA+1
+6:++1++++0+0
+7:++1++++1+1
+8:+NA+++NA+1
+9:++0++++1+1
10:+NA+++NA+1
11:++1++++1+1
12:+NA+++NA+1
13:+NA+++NA+1
14:+NA+++NA+1
15:++0++++1+1
16:++1++++0+0
17:++1++++1+1
18:++0++++0+0
19:+NA+++NA+0
20:++0++++0+0|1or9h|最终|2n55s|DT[is.na(a)+&+b+==+1,+which+=+TRUE]|bgfpn|将为您提供：|cercm|[1]++5++8+10+12+13+14|9hii^0|1W|3|22|9|N|12|0|0|0|0|0|0|1A|3|0|0|0|0|0|0|0|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|1E|G|@$H|1F|I|1G|J|K]|$H|1H|I|1I|J|K]]|L|@$H|1J|I|1K|A|1L]]|6|$]]|$A|M|C|N|2|O|F|1M|G|@]|L|@]|6|$P|Q]]|$A|R|C|S|2|E|F|1N|G|@]|L|@]|6|$]]|$A|T|C|U|2|O|F|1O|G|@]|L|@]|6|$P|Q]]|$A|V|C|W|2|E|F|1P|G|@]|L|@]|6|$]]|$A|X|C|Y|2|E|F|1Q|G|@$H|1R|I|1S|J|K]]|L|@]|6|$]]|$A|Z|C|10|2|O|F|1T|G|@]|L|@]|6|$P|Q]]|$A|11|C|12|2|E|F|1U|G|@]|L|@]|6|$]]|$A|13|C|14|2|O|F|1V|G|@]|L|@]|6|$P|Q]]|$A|15|C|16|2|E|F|1W|G|@]|L|@]|6|$]]|$A|17|C|18|2|O|F|1X|G|@]|L|@]|6|$P|Q]]|$A|19|C|1A|2|E|F|1Y|G|@]|L|@]|6|$]]|$A|1B|C|1C|2|O|F|1Z|G|@]|L|@]|6|$P|Q]]|$A|1D|C|-4|2|E|F|20|G|@]|L|@]|6|$]]]]

The <code>zoo</code> package and its <code>na.locf()</code> function could help you as described by Dirk Eddelbuettel in here: <a href="https://stackoverflow.com/questions/7735647/replacing-nas-with-latest-non-na-value">Replacing NAs with latest non-NA value</a>.
<pre><code>library(data.table)
library(zoo)

DT &lt;- data.table(a = c(0, NA, NA, 0, NA, 1, 1, NA, 0, NA, 1, NA, NA, NA, 0, 1, 1, 0, NA, 0))

non_nas &lt;- DT[!is.na(a), a]
successor &lt;- c(non_nas[-1], 0)
diff &lt;- abs(non_nas - successor)
DT[!is.na(a), diff:=diff]
</code></pre>
This will give you a data table as follows:
<pre><code> a diff
 1: 0 0
 2: NA NA
 3: NA NA
 4: 0 1
 5: NA NA
 6: 1 0
 7: 1 1
 8: NA NA
 9: 0 1
10: NA NA
11: 1 1
12: NA NA
13: NA NA
14: NA NA
15: 0 1
16: 1 0
17: 1 1
18: 0 0
19: NA NA
20: 0 0
</code></pre>
The idea here is that every '1' in the diff column tells you that the value in 'a' is going to change after the NAs below.
Now you want to get rid of the NAs in the 'diff' column. For clarity, we put the result into the new column 'b'. This is where the <code>zoo</code> package comes into play:
<pre><code>DT[, b:=na.locf(diff)]
</code></pre>
This results in
<pre><code> a diff b
 1: 0 0 0
 2: NA NA 0
 3: NA NA 0
 4: 0 1 1
 5: NA NA 1
 6: 1 0 0
 7: 1 1 1
 8: NA NA 1
 9: 0 1 1
10: NA NA 1
11: 1 1 1
12: NA NA 1
13: NA NA 1
14: NA NA 1
15: 0 1 1
16: 1 0 0
17: 1 1 1
18: 0 0 0
19: NA NA 0
20: 0 0 0
</code></pre>
Eventually
<pre><code>DT[is.na(a) &amp; b == 1, which = TRUE]
</code></pre>
will give you:
<pre><code>[1] 5 8 10 12 13 14
</code></pre>

I want to identify rows that contains NA, and is between a zero and a one. Consider this data.table:
<pre><code>DT &lt;- data.table(a = c(0, NA, NA, 0, NA, 1, 1, NA, 0, NA, 1, NA, NA, NA, 0, 1, 1, 0, NA, 0))

# DT
# a
# 1: 0
# 2: NA
# 3: NA
# 4: 0
# 5: NA
# 6: 1
# 7: 1
# 8: NA
# 9: 0
# 10: NA
# 11: 1
# 12: NA
# 13: NA
# 14: NA
# 15: 0
# 16: 1
# 17: 1
# 18: 0
# 19: NA
# 20: 0
</code></pre>
How to identify row no. 5, 8, 10 and 12:14?

Find rows with NA between 0s and 1s

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我希望标识包含NA且介于0和1之间的行。考虑一下这个data.table： DT <- data.table(a = c(0, NA, NA, 0, NA, 1, 1, NA, 0, NA, 1, NA, NA, NA, 0, 1, 1, 0, NA, 0))# DT# a# 1: 0# 2: NA# 3: NA#...

问查找NA介于0和1之间的行
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问查找NA介于0和1之间的行EN