blocks|key|354478|text|给定数据框架，df=data.frame(type=c('search','NULL','public','search','home','home','search'),x=c(0,0,0,1,0,1,0))|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|354479|如果您想知道第1列中每个值中有多少值在0的第2列中有一个值，则可以使用：table(df)[,1]|354480|只要你只和1和0一起工作，就能得到答案：|354481|++home+++NULL+public+search+
+++++1++++++1++++++1++++++2|code-block|syntax|javascript|354482|entityMap^0|7|2Q|0|10|D|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|W|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|X|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

Given the data frame,
<code>df=data.frame(type=c('search','NULL','public','search','home','home','search'),x=c(0,0,0,1,0,1,0))</code>

If you want to know how many of each value in column 1 have a value in column 2 of zero then you can use:
<code>table(df)[,1]</code>

as long as you are only working with 1's and 0's to get the answer:

<pre><code> home NULL public search 
 1 1 1 2
</code></pre>

blocks|key|423121|text|我假设您希望在满足特定条件(当变量具有某种值时)找到行数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|423122|如果是这样的话，那么我想您将"x“作为列中表示的变量。"x“可以接受多个值。假设当x为0时，您想要找到数据中有多少行。这可以通过以下方式做到：|423123|nrow(subset(data,+x=="0")|code-block|syntax|javascript|423124|“‘data”是R中数据集的对象名。|423125|编辑：|offset|length|style|BOLD|423126|我现在看到你编辑的数据了。你可以用这个来解决你的问题：|423127|table(data$type,+data$x)|423128|entityMap^0|0|0|0|0|0|3|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|10|8|@$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

I am assuming that you want to find the number of rows when a particular condition (when a variable is having some value) is met.

If this is the case, then I suppose you have "x" as a variable represented in a column. "x" can take multiple values. Suppose you want to find how many rows are there in your data when x is 0. This could be done by:

<pre><code>nrow(subset(data, x=="0")
</code></pre>

'data' is the object name for your dataset in R

EDIT:

I am seeing your edited dataframe now. You could use this to solve your problem:

<pre><code>table(data$type, data$x)
</code></pre>

blocks|key|354515|text|您还可以使用dplyr包来完成这一任务：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|354516|library(dplyr)

df2+<-+df+%25>%25+group_by(x,type)+%25>%25+tally()|code-block|syntax|javascript|354517|这意味着：|354518|++x+++type+n
1+0+++home+1
2+0+++NULL+1
3+0+public+1
4+0+search+2
5+1+++home+1
6+1+search+1|354519|entityMap^0|6|5|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

You could also do this with the <code>dplyr</code> package:

<pre><code>library(dplyr)

df2 &lt;- df %&gt;% group_by(x,type) %&gt;% tally()
</code></pre>

which gives:

<pre><code> x type n
1 0 home 1
2 0 NULL 1
3 0 public 1
4 0 search 2
5 1 home 1
6 1 search 1
</code></pre>

blocks|key|5537397|text|您还可以使用sqldf包：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5537398|library(sqldf)
df+<-+data.frame(type=c('search','NULL','public','search','home','home','search'),x=c(0,0,0,1,0,1,0))
sqldf("SELECT+type,+COUNT(*)+FROM+df+WHERE+x=0+GROUP+BY+type")|code-block|syntax|javascript|5537399|其结果如下：|5537400|++++type+COUNT(*)
1+++NULL++++++++1
2+++home++++++++1
3+public++++++++1
4+search++++++++2|5537401|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You could also use the sqldf package:

<pre><code>library(sqldf)
df &lt;- data.frame(type=c('search','NULL','public','search','home','home','search'),x=c(0,0,0,1,0,1,0))
sqldf("SELECT type, COUNT(*) FROM df WHERE x=0 GROUP BY type")
</code></pre>

which gives the following result:

<pre><code> type COUNT(*)
1 NULL 1
2 home 1
3 public 1
4 search 2
</code></pre>

blocks|key|350988|text|鉴于您的数据是作为数据框架结构的，下面的代码比上面给出的答案有更好的运行时间：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|350989|nrow(data[data$x=="0"])|code-block|syntax|javascript|350990|可以使用以下方法测试运行时：|350991|ptm+<-+proc.time()
nrow(subset(data,+x+==+"0"))
proc.time()+-+ptm

ptm+<-+proc.time()
nrow(data[data$x=="0"]))
proc.time()+-+ptm|350992|在我的例子中，运行时间大约快15倍，有100万行。|350993|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Given your data is structured as a data frame, the following code has a better running time than the answers given above:

<pre><code>nrow(data[data$x=="0"])
</code></pre>

You can test your run time using:

<pre><code>ptm &lt;- proc.time()
nrow(subset(data, x == "0"))
proc.time() - ptm

ptm &lt;- proc.time()
nrow(data[data$x=="0"]))
proc.time() - ptm
</code></pre>

In my case, the running time was about 15 times faster, with 1 million rows.

I would like to count how many rows there are per <code>type</code> if they meet the condition <code>x == 0</code>. Sort of like a group by in SQL 

Here is an example of the data 

<pre><code> type x 
search 0 
NULL 0 
public 0
search 1
home 0
home 1
search 0
</code></pre>

Conditional count and group by in R

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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如果每个type符合条件x == 0，我想计算它们有多少行。有点像SQL中的组下面是数据的一个示例  type    x    search    0 NULL      0 public    0search    1home      0home      1search    0

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