blocks|key|4617576|text|你想的太多了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4617577|sum(is.na(df$col))|code-block|syntax|javascript|4617578|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|$]]

You're over-thinking the problem:

<pre><code>sum(is.na(df$col))
</code></pre>

blocks|key|4617704|text|如果您要查找数据帧中每一列的NA计数，则：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4617705|na_count+<-sapply(x,+function(y)+sum(length(which(is.na(y)))))|code-block|syntax|javascript|4617706|应该会给你一个列表，列出每一列的计数。|4617707|na_count+<-+data.frame(na_count)|4617708|应该在数据帧中很好地输出数据，如下所示：|4617709|----------------------
%7C+row.names+%7C+na_count
------------------------
%7C+column_1++%7C+count|4617710|entityMap^0|E|2|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|T|$]]

If you are looking for <code>NA</code> counts for each column in a dataframe then:

<pre><code>na_count &lt;-sapply(x, function(y) sum(length(which(is.na(y)))))
</code></pre>

should give you a list with the counts for each column.

<pre><code>na_count &lt;- data.frame(na_count)
</code></pre>

Should output the data nicely in a dataframe like:

<pre><code>----------------------
| row.names | na_count
------------------------
| column_1 | count
</code></pre>

blocks|key|111634|text|尝试使用colSums函数|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|111635|df+<-+data.frame(x+=+c(1,2,NA),+y+=+rep(NA,+3))

colSums(is.na(df))

#x+y+
#1+3+|code-block|syntax|javascript|111636|entityMap^0|4|7|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|$]]

Try the <code>colSums</code> function

<pre><code>df &lt;- data.frame(x = c(1,2,NA), y = rep(NA, 3))

colSums(is.na(df))

#x y 
#1 3 
</code></pre>

blocks|key|111473|text|如果您希望计算整个数据帧中的NAs数量，您还可以使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|111474|sum(is.na(df))|code-block|syntax|javascript|111475|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|$]]

If you are looking to count the number of NAs in the entire dataframe you could also use

<pre><code>sum(is.na(df))
</code></pre>

blocks|key|4617886|text|获得所有列的NA计数的一个快速而简单的Tidyverse解决方案是使用summarise_all()，我认为这比使用purrr或sapply更容易阅读|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4617887|library(tidyverse)
#+Example+data
df+<-+tibble(col1+=+c(1,+2,+3,+NA),+
+++++++++++++col2+=+c(NA,+NA,+"a",+"b"))

df+%25>%25+summarise_all(~+sum(is.na(.)))
#>+#+A+tibble:+1+x+2
#>++++col1++col2
#>+++<int>+<int>
#>+1+++++1+++++2|code-block|syntax|javascript|4617888|或者使用更现代的across()函数：|4617889|df+%25>%25+summarise(across(everything(),+~+sum(is.na(.))))|4617890|entityMap^0|6|2|Z|F|1M|5|1S|6|0|0|8|8|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|P|$]]

A quick and easy Tidyverse solution to get a <code>NA</code> count for all columns is to use <code>summarise_all()</code> which I think makes a much easier to read solution than using <code>purrr</code> or <code>sapply</code>
<pre><code>library(tidyverse)
# Example data
df &lt;- tibble(col1 = c(1, 2, 3, NA), 
 col2 = c(NA, NA, &quot;a&quot;, &quot;b&quot;))

df %&gt;% summarise_all(~ sum(is.na(.)))
#&gt; # A tibble: 1 x 2
#&gt; col1 col2
#&gt; &lt;int&gt; &lt;int&gt;
#&gt; 1 1 2
</code></pre>
Or using the more modern <code>across()</code> function:
<pre class="lang-r prettyprint-override"><code>df %&gt;% summarise(across(everything(), ~ sum(is.na(.))))
</code></pre>

blocks|key|4617668|text|在summary()输出中，该函数还计算NAs，因此如果想要计算多个变量的NAs之和，可以使用此函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4617669|entityMap^0|1|9|K|2|11|2|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

In the <code>summary()</code> output, the function also counts the <code>NA</code>s so one can use this function if one wants the sum of <code>NA</code>s in several variables.

blocks|key|111626|text|一种计算数据帧每一列中空值数量的整形方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|111627|library(tidyverse)
library(purrr)

df+%25>%25
++++map_df(function(x)+sum(is.na(x)))+%25>%25
++++gather(feature,+num_nulls)+%25>%25
++++print(n+=+100)|code-block|syntax|javascript|111628|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|$]]

A tidyverse way to count the number of nulls in every column of a dataframe:

<pre><code>library(tidyverse)
library(purrr)

df %&gt;%
 map_df(function(x) sum(is.na(x))) %&gt;%
 gather(feature, num_nulls) %&gt;%
 print(n = 100)
</code></pre>

blocks|key|111607|text|此表单与Kevin+Ogoros的表单略有不同：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|111608|na_count+<-function+(x)+sapply(x,+function(y)+sum(is.na(y)))|code-block|syntax|javascript|111609|返回NA计数为命名整型数组|111610|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

This form, slightly changed from Kevin Ogoros's one:

<pre><code>na_count &lt;-function (x) sapply(x, function(y) sum(is.na(y)))
</code></pre>

returns NA counts as named int array

blocks|key|4617902|text|sapply(name+of+the+data,+function(x)+sum(is.na(x)))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4617903|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>sapply(name of the data, function(x) sum(is.na(x)))
</code></pre>

blocks|key|111586|text|用户rrs的答案是正确的，但它只告诉您要传递的数据帧的特定列中NA值的数量，以获得整个数据帧的NA值的数量，尝试如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|111587|apply(<name+of+dataFrame>,+2<for+getting+column+stats>,+function(x)+{sum(is.na(x))})|code-block|syntax|javascript|111588|这样就行了。|111589|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

User rrs answer is right but that only tells you the number of NA values in the particular column of the data frame that you are passing to get the number of NA values for the whole data frame try this:

<pre><code>apply(&lt;name of dataFrame&gt;, 2&lt;for getting column stats&gt;, function(x) {sum(is.na(x))})
</code></pre>

This does the trick

blocks|key|111619|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|111620|length(df$col[is.na(df$col)])|code-block|syntax|javascript|111621|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|$]]

Try this:

<pre><code>length(df$col[is.na(df$col)])
</code></pre>

blocks|key|4617851|text|类似于Hute37的答案，但使用的是purrr包。我认为这种整齐的方法比AbiK提出的答案更简单。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4617852|library(purrr)
map_dbl(df,+~sum(is.na(.)))|code-block|syntax|javascript|4617853|注意:波浪号(~)创建一个匿名函数。和“.”指匿名函数的输入，在本例中为data.frame+df。|4617854|entityMap^0|I|5|0|0|7|1|1B|2|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|N|$]]

Similar to hute37's answer but using the <code>purrr</code> package. I think this tidyverse approach is simpler than the answer proposed by AbiK.

<pre><code>library(purrr)
map_dbl(df, ~sum(is.na(.)))
</code></pre>

Note: the tilde (<code>~</code>) creates an anonymous function. And the '.' refers to the input for the anonymous function, in this case the data.frame <code>df</code>.

blocks|key|111770|text|如果你想在每一列中一个接一个地打印空值，那么你可以使用下面的方法。简单的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|111771|lapply(df,+function(x)+{+length(which(is.na(x)))})|code-block|syntax|javascript|111772|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|$]]

If you're looking for null values in each column to be printed one after the other then you can use this. Simple solution.
<pre><code>lapply(df, function(x) { length(which(is.na(x)))})
</code></pre>

blocks|key|111657|text|您可以使用它来计算每列中的NA或空格的数量|type|unstyled|depth|inlineStyleRanges|entityRanges|data|111658|colSums(is.na(data_set_name)%7Cdata_set_name+==+'')|code-block|syntax|javascript|111659|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|$]]

You can use this to count number of NA or blanks in every column

<pre><code>colSums(is.na(data_set_name)|data_set_name == '')
</code></pre>

blocks|key|4617925|text|为了完整性，您还可以在table中使用useNA参数。例如，table(df$col,+useNA="always")将计算所有非NA案例和NA案例。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4617926|entityMap^0|J|5|U|T|1T|2|1Y|2|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

In the interests of completeness you can also use the <code>useNA</code> argument in table. For example <code>table(df$col, useNA=&quot;always&quot;)</code> will count all of non <code>NA</code> cases and the <code>NA</code> ones.

I want to count the number of <code>NA</code> values in a data frame column. Say my data frame is called <code>df</code>, and the name of the column I am considering is <code>col</code>. The way I have come up with is following: 

<pre><code>sapply(df$col, function(x) sum(length(which(is.na(x))))) 
</code></pre>

Is this a good/most efficient way to do this?

Determine the number of NA values in a column

我想要计算数据框列中的NA值的数量。假设我的数据框名为df，我考虑的列名为col。我想出的方法如下：sapply(df$col, function(x) sum(length(which(is.na(x)))))  这是一个好的/最有效的方法吗？

问确定一列中NA值的数量
EN

回答 15

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问确定一列中NA值的数量EN

回答 15

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问确定一列中NA值的数量
EN