blocks|key|732267|text|is.finite工作在vector上，而不是data.frame对象上。因此，我们可以使用data.frame循环使用lapply，并且只得到‘有限’值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|732268|lapply(df,+function(x)+x[is.finite(x)])|code-block|syntax|javascript|732269|如果每个列的Inf、-Inf值不同，那么上面的代码将有一个具有不相等length的元素的list。所以，最好还是把它作为一个list。如果我们想要一个data.frame，它应该有相同的长度。|732270|如果要删除包含NA或Inf/-Inf值的行|732271|df[Reduce(`&`,+lapply(df,+function(x)+!is.na(x)++&+is.finite(x))),]|732272|或者@nicola的一个紧凑的选项|732273|df[Reduce(`&`,+lapply(df,+is.finite)),]|732274|如果我们准备好使用一个包，一个紧凑的选项将是NaRV.omit。|732275|library(IDPmisc)
NaRV.omit(df)|732276|数据|732277|set.seed(24)
df+<-+as.data.frame(matrix(sample(c(1:5,+NA,+-Inf,+Inf),+
++++++++++++++++++++++20*5,+replace=TRUE),+ncol=5))|732278|entityMap^0|0|9|C|6|N|A|1A|A|1O|6|0|0|6|3|A|4|Y|6|18|4|1Q|4|23|A|0|0|0|0|0|M|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]|$9|1R|A|1S|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1T|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|1U|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|1V|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|1W|8|@]|D|@]|E|$I|J]]|$1|U|3|V|5|6|7|1X|8|@$9|1Y|A|1Z|B|C]]|D|@]|E|$]]|$1|W|3|X|5|H|7|20|8|@]|D|@]|E|$I|J]]|$1|Y|3|Z|5|6|7|21|8|@]|D|@]|E|$]]|$1|10|3|11|5|H|7|22|8|@]|D|@]|E|$I|J]]|$1|12|3|-4|5|6|7|23|8|@]|D|@]|E|$]]]|13|$]]

The <code>is.finite</code> works on <code>vector</code> and not on <code>data.frame</code> object. So, we can loop through the <code>data.frame</code> using <code>lapply</code> and get only the 'finite' values.

<pre><code>lapply(df, function(x) x[is.finite(x)])
</code></pre>

If the number of <code>Inf</code>, <code>-Inf</code> values are different for each column, the above code will have a <code>list</code> with elements having unequal <code>length</code>. So, it may be better to leave it as a <code>list</code>. If we want a <code>data.frame</code>, it should have equal lengths.

<hr>

If we want to remove rows contain any NA or Inf/-Inf values

<pre><code>df[Reduce(`&amp;`, lapply(df, function(x) !is.na(x) &amp; is.finite(x))),]
</code></pre>

Or a compact option by @nicola

<pre><code>df[Reduce(`&amp;`, lapply(df, is.finite)),]
</code></pre>

<hr>

If we are ready to use a package, a compact option would be <code>NaRV.omit</code>

<pre><code>library(IDPmisc)
NaRV.omit(df)
</code></pre>

<h3>data</h3>

<pre><code>set.seed(24)
df &lt;- as.data.frame(matrix(sample(c(1:5, NA, -Inf, Inf), 
 20*5, replace=TRUE), ncol=5))
</code></pre>

blocks|key|732297|text|要删除带有%2B/-Inf的行，我建议如下：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|732298|df+<-+df[!is.infinite(rowSums(df)),]|code-block|syntax|javascript|732299|或者，相当的，|732300|df+<-+df[is.finite(rowSums(df)),]|732301|第二个选项(带有is.finite()和不带否定词的选项)也删除包含NA值的行，以防这一操作尚未完成。|732302|entityMap^0|8|3|0|0|0|0|8|B|Y|2|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

To remove the rows with +/-<code>Inf</code> I'd suggest the following:

<pre><code>df &lt;- df[!is.infinite(rowSums(df)),]
</code></pre>

or, equivalently,

<pre><code>df &lt;- df[is.finite(rowSums(df)),]
</code></pre>

The second option (the one with <code>is.finite()</code> and without the negation) removes also rows containing <code>NA</code> values in case that this has not already been done.

blocks|key|732311|text|为了保留没有Inf的行，我们可以这样做：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|732312|df[apply(df,+1,+function(x)+all(is.finite(x))),+]|code-block|syntax|javascript|732313|另外，NA的处理也是这样的，因为：|732314|值为NA的行索引将删除结果中的此行。|732315|此外，结果中没有带有NaN的行。|732316|set.seed(24)
df+<-+as.data.frame(matrix(sample(c(0:9,+NA,+-Inf,+Inf,+NaN),++20*5,+replace=TRUE),+ncol=5))
df2+<-+df[apply(df,+1,+function(x)+all(is.finite(x))),+]|732317|以下是不同的is.~-functions的结果：|732318|x+<-+c(42,+NA,+NaN,+Inf)
is.finite(x)
#+[1]++TRUE+FALSE+FALSE+FALSE
is.na(x)
#+[1]+FALSE++TRUE++TRUE+FALSE
is.nan(x)
#+[1]+FALSE+FALSE++TRUE+FALSE|732319|entityMap^0|6|3|0|0|3|2|0|2|2|0|A|3|0|0|6|4|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|U|3|V|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|X|$]]

To keep the rows without <code>Inf</code> we can do:

<pre><code>df[apply(df, 1, function(x) all(is.finite(x))), ]
</code></pre>

Also <code>NA</code>s are handled by this because of: 
a rowindex with value <code>NA</code> will remove this row in the result.

Also rows with <code>NaN</code> are not in the result.

<pre><code>set.seed(24)
df &lt;- as.data.frame(matrix(sample(c(0:9, NA, -Inf, Inf, NaN), 20*5, replace=TRUE), ncol=5))
df2 &lt;- df[apply(df, 1, function(x) all(is.finite(x))), ]
</code></pre>

Here are the results of the different <code>is.~</code>-functions:

<pre><code>x &lt;- c(42, NA, NaN, Inf)
is.finite(x)
# [1] TRUE FALSE FALSE FALSE
is.na(x)
# [1] FALSE TRUE TRUE FALSE
is.nan(x)
# [1] FALSE FALSE TRUE FALSE
</code></pre>

blocks|key|732329|text|我有这个问题，上面的解决方案对我都没有用。我使用以下方法删除了我的dataframe第15和第16列中的%2B/-Inf行。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|732330|d<-subset(c,+c[,15:16]!="-Inf")+
e<-subset(d,+d[,15:16]!="Inf")|code-block|syntax|javascript|732331|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 had this problem and none of the above solutions worked for me. I used the following to remove rows with +/-Inf in columns 15 and 16 of my dataframe. 

<pre><code>d&lt;-subset(c, c[,15:16]!="-Inf") 
e&lt;-subset(d, d[,15:16]!="Inf")
</code></pre>

blocks|key|1494721|text|根据数据的不同，有几个选项使用dplyr::filter()和is.finite()或is.infinite()的作用域变体，这些选项可能有用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1494722|library(dplyr)

#+sample+data
df+<-+data_frame(a+=+c(1,+2,+3,+NA),+b+=+c(5,+Inf,+8,+8),+c+=+c(9,+10,+Inf,+11),+d+=+c('a',+'b',+'c',+'d'))

#+across+all+columns:
df+%25>%25+
++filter_all(all_vars(!is.infinite(.)))

#+note+that+is.finite()+does+not+work+with+NA+or+strings:
df+%25>%25+
++filter_all(all_vars(is.finite(.)))

#+checking+only+numeric+columns:
df+%25>%25+
++filter_if(~is.numeric(.),+all_vars(!is.infinite(.)))

#+checking+only+select+columns,+in+this+case+a+through+c:
df+%25>%25+
++filter_at(vars(a:c),+all_vars(!is.infinite(.)))|code-block|syntax|javascript|1494723|entityMap^0|F|F|V|B|17|D|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|L|$]]

Depending on the data, there are a couple options using scoped variants of <code>dplyr::filter()</code> and <code>is.finite()</code> or <code>is.infinite()</code> that might be useful:

<pre><code>library(dplyr)

# sample data
df &lt;- data_frame(a = c(1, 2, 3, NA), b = c(5, Inf, 8, 8), c = c(9, 10, Inf, 11), d = c('a', 'b', 'c', 'd'))

# across all columns:
df %&gt;% 
 filter_all(all_vars(!is.infinite(.)))

# note that is.finite() does not work with NA or strings:
df %&gt;% 
 filter_all(all_vars(is.finite(.)))

# checking only numeric columns:
df %&gt;% 
 filter_if(~is.numeric(.), all_vars(!is.infinite(.)))

# checking only select columns, in this case a through c:
df %&gt;% 
 filter_at(vars(a:c), all_vars(!is.infinite(.)))
</code></pre>

blocks|key|732384|text|为了dplyr+1.0.0，我花了一段时间才完成这个工作，所以我想我应该使用c_across提供新版本的@sbha解决方案，因为filter_all、filter_if已经不再受欢迎了。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|732385|library(dplyr)
df+<-+tibble(a+=+c(1,+2,+3,+NA),+b+=+c(5,+Inf,+8,+8),+c+=+c(9,+10,+Inf,+11),+d+=+c('a',+'b',+'c',+'d'))
#+++++++a+++++b+++++c+d++++
#+++<dbl>+<dbl>+<dbl>+<chr>
#+1+++++1+++++5+++++9+a++++
#+2+++++2+++Inf++++10+b++++
#+3+++++3+++++8+++Inf+c++++
#+4++++NA+++++8++++11+d+

df+%25>%25+
++rowwise+%25>%25+
++filter(!all(is.infinite(c_across(where(is.numeric)))))
#+#+A+tibble:+4+x+4
#+#+Rowwise:+
#+++++++a+++++b+++++c+d++++
#+++<dbl>+<dbl>+<dbl>+<chr>
#+1+++++1+++++5+++++9+a++++
#+2+++++2+++Inf++++10+b++++
#+3+++++3+++++8+++Inf+c++++
#+4++++NA+++++8++++11+d+

df+%25>%25+
++rowwise+%25>%25+
++filter(!any(is.infinite(c_across(where(is.numeric)))))
#+#+A+tibble:+2+x+4
#+#+Rowwise:+
#+++++++a+++++b+++++c+d++++
#+++<dbl>+<dbl>+<dbl>+<chr>
#+1+++++1+++++5+++++9+a++++
#+2++++NA+++++8++++11+d+

df+%25>%25+
++rowwise+%25>%25+
++filter(!any(is.infinite(c_across(a:c))))

#+#+A+tibble:+2+x+4
#+#+Rowwise:+
#+++++++a+++++b+++++c+d++++
#+++<dbl>+<dbl>+<dbl>+<chr>
#+1+++++1+++++5+++++9+a++++
#+2++++NA+++++8++++11+d+|code-block|syntax|javascript|732386|老实说，我认为@sbha的答案更简单！|732387|entityMap^0|2|B|12|8|1S|A|23|9|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|D]|$9|U|A|V|B|D]|$9|W|A|X|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|Y|8|@]|E|@]|F|$J|K]]|$1|L|3|M|5|6|7|Z|8|@]|E|@]|F|$]]|$1|N|3|-4|5|6|7|10|8|@]|E|@]|F|$]]]|O|$]]

It took me awhile to work this out for dplyr 1.0.0 so i thought i would put up the new version of @sbha solutions using <code>c_across</code> since <code>filter_all</code>, <code>filter_if</code> are getting deprecated.
<pre><code>library(dplyr)
df &lt;- tibble(a = c(1, 2, 3, NA), b = c(5, Inf, 8, 8), c = c(9, 10, Inf, 11), d = c('a', 'b', 'c', 'd'))
# a b c d 
# &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt;
# 1 1 5 9 a 
# 2 2 Inf 10 b 
# 3 3 8 Inf c 
# 4 NA 8 11 d 

df %&gt;% 
 rowwise %&gt;% 
 filter(!all(is.infinite(c_across(where(is.numeric)))))
# # A tibble: 4 x 4
# # Rowwise: 
# a b c d 
# &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt;
# 1 1 5 9 a 
# 2 2 Inf 10 b 
# 3 3 8 Inf c 
# 4 NA 8 11 d 

df %&gt;% 
 rowwise %&gt;% 
 filter(!any(is.infinite(c_across(where(is.numeric)))))
# # A tibble: 2 x 4
# # Rowwise: 
# a b c d 
# &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt;
# 1 1 5 9 a 
# 2 NA 8 11 d 

df %&gt;% 
 rowwise %&gt;% 
 filter(!any(is.infinite(c_across(a:c))))

# # A tibble: 2 x 4
# # Rowwise: 
# a b c d 
# &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt;
# 1 1 5 9 a 
# 2 NA 8 11 d 
</code></pre>
To be honest I think @sbha answer is simpler!

blocks|key|1494801|text|我认为自己对编码很陌生，无法让上面的建议与我的代码一起工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1494802|我找到了一种不太复杂的方法来减少2行数据，首先用Na替换Inf，然后选择包含完整数据的行：|1494803|Df[sapply(Df,+is.infinite)]+<-+NA
Df<-Df[complete.cases(Df),+]|code-block|syntax|javascript|1494804|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|$]]

I consider myself new to coding and I couldn't get the recommendations above to work with my code.
I found a less complicated way to reduce a dataframe with 2 lines, first by replacing Inf with Na, then by selecting rows with complete data:
<pre><code>Df[sapply(Df, is.infinite)] &lt;- NA
Df&lt;-Df[complete.cases(Df), ]
</code></pre>

blocks|key|732451|text|df[!is.infinite(df$x),]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|732452|其中x是df的列，包含无穷大的值。发布的第一个答案取决于行和，但是对于我自己的问题，df有不能添加的列。|unstyled|732453|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>df[!is.infinite(df$x),]
</code></pre>
wherein x is the column of df that contains the infinite values. The first answer posted was contingent on rowsums but for my own problem, the df had columns which could not be added.

I have a very large dataframe(df) with approximately 35-45 columns(variables) and rows greater than 300. Some of the rows contains NA,NaN,Inf,-Inf values in single or multiple variables and I have used 
<code>na.omit(df)</code> to remove rows with NA and NaN but I cant remove rows with Inf and -Inf values using na.omit function.

While searching I came across this thread <a href="https://stackoverflow.com/questions/22804473/remove-rows-with-inf-and-nan-in-r">Remove rows with Inf and NaN in R</a> and used the modified code <code>df[is.finite(df)]</code> but its not removing the rows with Inf and -Inf and also gives this error

<blockquote>
 Error in is.finite(df) : default method not implemented for type
 'list'
</blockquote>

EDITED

Remove the entire row even the corresponding one or multiple columns have inf and -inf

How to remove rows with inf from a dataframe 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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我有一个非常大的dataframe(df)，它有大约35-45列(变量)和大于300的行。有些行在单个变量或多个变量中包含NA、NaN、Inf、-Inf值，我使用na.omit(df)删除带有NA和NaN的行，但不能使用na.omit函数删除带有Inf和-Inf值的行。在搜索时，我遇到了这个线程，并使用了修改过的代码d...

问如何从R中的数据中删除带有inf的行
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问如何从R中的数据中删除带有inf的行EN