blocks|key|3107802|text|有几种不同的方法可以做到这一点。我更喜欢使用apply，因为它很容易扩展：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3107803|##Generate+some+data
dd+=+data.frame(a+=+1:4,+b=+1:0,+c=0:3)

##Go+through+each+row+and+determine+if+a+value+is+zero
row_sub+=+apply(dd,+1,+function(row)+all(row+!=0+))
##Subset+as+usual
dd[row_sub,]|code-block|syntax|javascript|3107804|entityMap^0|M|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

There are a few different ways of doing this. I prefer using <code>apply</code>, since it's easily extendable:

<pre><code>##Generate some data
dd = data.frame(a = 1:4, b= 1:0, c=0:3)

##Go through each row and determine if a value is zero
row_sub = apply(dd, 1, function(row) all(row !=0 ))
##Subset as usual
dd[row_sub,]
</code></pre>

blocks|key|233912|text|嗯，你可以把你的0换成NA，然后使用这些解决方案中的一个，但是为了不同，你可能会注意到，如果一个数字大于0，那么它只有一个有限的对数，所以rowSums的log只有在一行没有零的情况下才是有限的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|233913|dfr[is.finite(rowSums(log(dfr[-1]))),]|code-block|syntax|javascript|233914|entityMap^0|8|1|B|2|1G|1|1X|7|25|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|L|$]]

Well, you could swap your <code>0</code>'s for <code>NA</code> and then use one of those solutions, but for sake of a difference, you could notice that a number will only have a finite logarithm if it is greater than <code>0</code>, so that <code>rowSums</code> of the <code>log</code> will only be finite if there are no zeros in a row.

<pre><code>dfr[is.finite(rowSums(log(dfr[-1]))),]
</code></pre>

blocks|key|233921|text|我可能会同意Joran的建议，用NAs替换0，然后使用你提到的内置函数。如果您不能/不想这样做，一种方法是使用any()查找包含0的行并子集出这些行：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|233922|set.seed(42)
#Fake+data
x+<-+data.frame(a+=+sample(0:2,+5,+TRUE),+b+=+sample(0:2,+5,+TRUE))
>+x
++a+b
1+2+1
2+2+2
3+0+0
4+2+1
5+1+2
#Subset+out+any+rows+with+a+0+in+them
#Note+the+negation+with+!+around+the+apply+function
x[!(apply(x,+1,+function(y)+any(y+==+0))),]
++a+b
1+2+1
2+2+2
4+2+1
5+1+2|code-block|syntax|javascript|233923|要实现Joran的方法，下面这样的代码应该可以让您开始：|233924|x[x==0]+<-+NA|233925|entityMap^0|1J|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|$]]

I would probably go with Joran's suggestion of replacing 0's with NAs and then using the built in functions you mentioned. If you can't/don't want to do that, one approach is to use <code>any()</code> to find rows that contain 0's and subset those out:

<pre><code>set.seed(42)
#Fake data
x &lt;- data.frame(a = sample(0:2, 5, TRUE), b = sample(0:2, 5, TRUE))
&gt; x
 a b
1 2 1
2 2 2
3 0 0
4 2 1
5 1 2
#Subset out any rows with a 0 in them
#Note the negation with ! around the apply function
x[!(apply(x, 1, function(y) any(y == 0))),]
 a b
1 2 1
2 2 2
4 2 1
5 1 2
</code></pre>

To implement Joran's method, something like this should get you started:

<pre><code>x[x==0] &lt;- NA
</code></pre>

blocks|key|230092|text|我更喜欢对csgillespie的方法进行简单的修改，而不需要函数定义：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|230093|d[apply(d!=0,+1,+all),]|code-block|syntax|javascript|230094|其中d是您的数据框。|offset|length|style|CODE|230095|entityMap^0|0|0|2|1|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I prefer a simple adaptation of csgillespie's method, foregoing the need of a function definition:

<pre><code>d[apply(d!=0, 1, all),]
</code></pre>

where <code>d</code> is your data frame.

blocks|key|232944|text|我会做以下事情。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|232945|将零设置为NA。|232946|+data[data==0]+<-+NA
+data|code-block|syntax|javascript|232947|删除与NA关联的行。|232948|+data2<-data[complete.cases(data),]|232949|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I would do the following.

Set the zero to NA.

<pre><code> data[data==0] &lt;- NA
 data
</code></pre>

Delete the rows associated with NA.

<pre><code> data2&lt;-data[complete.cases(data),]
</code></pre>

blocks|key|233945|text|您可以使用dplyr包中的过滤器。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|233946|让我们将您的数据帧命名为df|233947|library(dplyr)+df1+<-+filter(df,+Mac1+>+0,+Mac2+>+0,+Mac3+>+0,+Mac4+>+0)|offset|length|style|CODE|233948|df1将只包含条目大于零的行。希望这能有所帮助。|233949|entityMap^0|0|0|0|20|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@$F|Q|G|R|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|S|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|M|$]]

You can use filter from dplyr package.

Let's call your data frame df 

<code>library(dplyr)
df1 &lt;- filter(df, Mac1 &gt; 0, Mac2 &gt; 0, Mac3 &gt; 0, Mac4 &gt; 0)</code>

df1 will have only rows with entries above zero. Hope this helps.

blocks|key|2894223|text|在基数R中，我们可以使用grep选择要测试的列，将数据与0进行比较，使用rowSums选择所有非零值的行。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2894224|cols+<-+grep("%5EMac",+names(df))
df[rowSums(df[cols]+!=+0)+==+length(cols),+]

#++++++++++DateTime+Mac1+Mac2+Mac3+Mac4
#1+2011-04-02+06:05+++21+++21+++21+++21
#2+2011-04-02+06:10+++22+++22+++22+++22
#3+2011-04-02+06:20+++24+++24+++24+++24|code-block|syntax|javascript|2894225|使用反转逻辑执行此操作，但输出相同|2894226|df[rowSums(df[cols]+==+0)+==+0,+]|2894227|2894228|在dplyr中，我们可以使用filter_at测试特定的列，并使用all_vars选择所有值都不等于0的行。|2894229|library(dplyr)
df+%25>%25++filter_at(vars(starts_with("Mac")),+all_vars(.+!=+0))|2894230|BOLD|2894231|df+<-+structure(list(DateTime+=+structure(1:6,+.Label+=+c("2011-04-02+06:00",+
"2011-04-02+06:05",+"2011-04-02+06:10",+"2011-04-02+06:15",+"2011-04-02+06:20",+
"2011-04-02+06:25"),+class+=+"factor"),+Mac1+=+c(20L,+21L,+22L,+
23L,+24L,+0L),+Mac2+=+c(0L,+21L,+22L,+23L,+24L,+25L),+Mac3+=+c(20L,+
21L,+22L,+0L,+24L,+25L),+Mac4+=+c(20L,+21L,+22L,+23L,+24L,+0L
)),+class+=+"data.frame",+row.names+=+c(NA,+-6L))|2894232|entityMap^0|C|4|10|7|0|0|0|0|0|1|5|E|9|X|8|0|0|0|4|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|15|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|17|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|R|3|S|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|T|3|E|5|6|7|1G|8|@$9|1H|A|1I|B|U]]|D|@]|E|$]]|$1|V|3|W|5|H|7|1J|8|@]|D|@]|E|$I|J]]|$1|X|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|Y|$]]

In base R, we can select the columns which we want to test using <code>grep</code>, compare the data with 0, use <code>rowSums</code> to select rows which has all non-zero values.

<pre><code>cols &lt;- grep("^Mac", names(df))
df[rowSums(df[cols] != 0) == length(cols), ]

# DateTime Mac1 Mac2 Mac3 Mac4
#1 2011-04-02 06:05 21 21 21 21
#2 2011-04-02 06:10 22 22 22 22
#3 2011-04-02 06:20 24 24 24 24
</code></pre>

Doing this with inverted logic but giving the same output

<pre><code>df[rowSums(df[cols] == 0) == 0, ]
</code></pre>

<hr>

In <code>dplyr</code>, we can use <code>filter_at</code> to test for specific columns and use <code>all_vars</code> to select rows where all the values are not equal to 0. 

<pre><code>library(dplyr)
df %&gt;% filter_at(vars(starts_with("Mac")), all_vars(. != 0))
</code></pre>

data

<pre><code>df &lt;- structure(list(DateTime = structure(1:6, .Label = c("2011-04-02 06:00", 
"2011-04-02 06:05", "2011-04-02 06:10", "2011-04-02 06:15", "2011-04-02 06:20", 
"2011-04-02 06:25"), class = "factor"), Mac1 = c(20L, 21L, 22L, 
23L, 24L, 0L), Mac2 = c(0L, 21L, 22L, 23L, 24L, 25L), Mac3 = c(20L, 
21L, 22L, 0L, 24L, 25L), Mac4 = c(20L, 21L, 22L, 23L, 24L, 0L
)), class = "data.frame", row.names = c(NA, -6L))
</code></pre>

blocks|key|233985|text|使用tidyverse/dplyr，您还可以删除变量子集中任何零值的行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|233986|#+variables+starting+with+Mac+must+be+non-zero
filter_at(df,+vars(starts_with("Mac")),+all_vars((.)+!=+0))

#+variables+x,+y,+and+z+must+be+non-zero
filter_at(df,+vars(x,+y,+z),+all_vars((.)+!=+0))

#+all+numeric+variables+must+be+non-zero
filter_if(df,+is.numeric,+all_vars((.)+!=+0))|code-block|syntax|javascript|233987|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|$]]

Using tidyverse/dplyr, you can also remove rows with any zero value in a subset of variables:

<pre><code># variables starting with Mac must be non-zero
filter_at(df, vars(starts_with("Mac")), all_vars((.) != 0))

# variables x, y, and z must be non-zero
filter_at(df, vars(x, y, z), all_vars((.) != 0))

# all numeric variables must be non-zero
filter_if(df, is.numeric, all_vars((.) != 0))
</code></pre>

blocks|key|2894315|text|由于dplyr+1.0.0弃用了@+variants很好地展示的作用域变体，下面是新语法的更新。这可能很有用，因为在这种情况下，across()无法工作，我花了一些时间才弄清楚解决方案，如下所示。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2894316|目标是提取在一列中至少包含一个0的所有行。|2894317|df+%25>%25+
++rowwise()+%25>%25+
++filter(any(c_across(everything(.))+==+0))|code-block|syntax|javascript|2894318|有了数据|2894319|df+<-+data.frame(a+=+1:4,+b=+1:0,+c=0:3)
df+<-+rbind(df,+c(0,0,0))
df+<-+rbind(df,+c(9,9,9))

#+A+tibble:+4+x+3
#+Rowwise:+
++++++a+++++b+++++c
++<dbl>+<dbl>+<dbl>
1+++++1+++++1+++++0
2+++++2+++++0+++++1
3+++++4+++++0+++++3
4+++++0+++++0+++++0|2894320|因此，它不会正确地返回包含所有9的最后一行。|2894321|entityMap^0|1S|8|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|11|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|T|$]]

As dplyr 1.0.0 deprecated the scoped variants which @Feng Mai nicely showed, here is an update with the new syntax. This might be useful because in this case, <code>across()</code> doesn't work, and it took me some time to figure out the solution as follows.
The goal was to extract all rows that contain at least one 0 in a column.
<pre><code>df %&gt;% 
 rowwise() %&gt;% 
 filter(any(c_across(everything(.)) == 0))
</code></pre>
with the data
<pre><code>df &lt;- data.frame(a = 1:4, b= 1:0, c=0:3)
df &lt;- rbind(df, c(0,0,0))
df &lt;- rbind(df, c(9,9,9))

# A tibble: 4 x 3
# Rowwise: 
 a b c
 &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt;
1 1 1 0
2 2 0 1
3 4 0 3
4 0 0 0
</code></pre>
So it correctly doesn't return the last row containing all 9s.

blocks|key|2894334|text|以下是一种更新的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2894335|library(dplyr)

df+%25>%25
++filter(!if_any(starts_with("Mac"),+~+.+==+0))

++++++++++DateTime+Mac1+Mac2+Mac3+Mac4
1+2011-04-02+06:05+++21+++21+++21+++21
2+2011-04-02+06:10+++22+++22+++22+++22
3+2011-04-02+06:20+++24+++24+++24+++24|code-block|syntax|javascript|2894336|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|$]]

Here is a more up-to-date approach:
<pre><code>library(dplyr)

df %&gt;%
 filter(!if_any(starts_with(&quot;Mac&quot;), ~ . == 0))

 DateTime Mac1 Mac2 Mac3 Mac4
1 2011-04-02 06:05 21 21 21 21
2 2011-04-02 06:10 22 22 22 22
3 2011-04-02 06:20 24 24 24 24
</code></pre>

blocks|key|2894415|text|在被类似的问题弄得头破血流后，我想到了这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2894416|我想在第1列和第9列中保留零，因为它们在那里有意义，但在其余列中保留了零。此代码将在数据帧中的第2列到第8列产生大于零的值，使第1列和第9列保持不变。|2894417|dataset+%25>%25+filter(data.frame(dataset[,2:8]>0))|code-block|syntax|javascript|2894418|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

After breaking my head on a similar problem, I came up with this:
I wanted to leave the zeros in column 1 and 9 because they made sense there but leave the zeros out from the rest of the columns. This code would produce bigger than zero values in column 2 till 8 in the data frame, leaving column 1 and 9 intact.
<pre><code>dataset %&gt;% filter(data.frame(dataset[,2:8]&gt;0))
</code></pre>

blocks|key|2894430|text|最短的答案将从数字到逻辑的隐式强制中受益，使用if_any/if_all将零转换为假，将其他值转换为真|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2894431|library(dplyr)

df+%25>%25+filter(if_all(starts_with('Mac')))|code-block|syntax|javascript|2894432|entityMap^0|N|D|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

The shortest answer possible would benefit from implicit coercion of numeric to logical, with which zeroes are converted to FALSE and other values to TRUE, and using <code>if_any/if_all</code>
<pre><code>library(dplyr)

df %&gt;% filter(if_all(starts_with('Mac')))
</code></pre>

I have a problem to solve how to remove rows with a Zero value in R. In others hand, I can use <code>na.omit()</code> to delete all the NA values or use <code>complete.cases()</code> to delete rows that contains NA values.

Is there anyone know how to remove rows with a Zero Values in R?

For example :

Before

<pre><code>| DateTime | Mac1 | Mac2 | Mac3 | Mac4 |
----------------------------------------------------
| 2011-04-02 06:00 | 20 | 0 | 20 | 20 | 
| 2011-04-02 06:05 | 21 | 21 | 21 | 21 | 
| 2011-04-02 06:10 | 22 | 22 | 22 | 22 | 
| 2011-04-02 06:15 | 23 | 23 | 0 | 23 | 
| 2011-04-02 06:20 | 24 | 24 | 24 | 24 | 
| 2011-04-02 06:25 | 0 | 25 | 25 | 0 | 
</code></pre>

After

<pre><code>| DateTime | Mac1 | Mac2 | Mac3 | Mac4 |
----------------------------------------------------
| 2011-04-02 06:05 | 21 | 21 | 21 | 21 | 
| 2011-04-02 06:10 | 22 | 22 | 22 | 22 | 
| 2011-04-02 06:20 | 24 | 24 | 24 | 24 | 
</code></pre>

How to remove rows with any zero value

我有一个问题要解决，如何删除R中具有零值的行。换句话说，我可以使用na.omit()删除所有NA值，或者使用complete.cases()删除包含NA值的行。有没有人知道如何删除R中零值的行？例如：之前的|    DateTime      | Mac1  | Mac2  | Mac3  | Mac4  |-----...

问如何删除具有任何零值的行
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