blocks|key|545812|text|让我假设您的data.frame是不同数据类型的混合，并不是所有列都需要修改。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|545813|要只修改第12到18列(总共21列)，只需执行以下操作|545814|df[,+12:18][df[,+12:18]+==+0]+<-+NA|code-block|syntax|javascript|545815|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|$]]

Let me assume that your data.frame is a mix of different datatypes and not all columns need to be modified.

to modify only columns 12 to 18 (of the total 21), just do this

<pre><code>df[, 12:18][df[, 12:18] == 0] &lt;- NA
</code></pre>

blocks|key|329082|text|dplyr::na_if()是一个选项：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|329083|library(dplyr)++

df+<-+data_frame(col1+=+c(1,+2,+3,+0),
+++++++++++++++++col2+=+c(0,+2,+3,+4),
+++++++++++++++++col3+=+c(1,+0,+3,+0),
+++++++++++++++++col4+=+c('a',+'b',+'c',+'d'))

na_if(df,+0)
#+A+tibble:+4+x+4
+++col1++col2++col3+col4+
++<dbl>+<dbl>+<dbl>+<chr>
1+++++1++++NA+++++1+a++++
2+++++2+++++2++++NA+b++++
3+++++3+++++3+++++3+c++++
4++++NA+++++4++++NA+d|code-block|syntax|javascript|329084|entityMap^0|0|E|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|$]]

<code>dplyr::na_if()</code> is an option:

<pre><code>library(dplyr) 

df &lt;- data_frame(col1 = c(1, 2, 3, 0),
 col2 = c(0, 2, 3, 4),
 col3 = c(1, 0, 3, 0),
 col4 = c('a', 'b', 'c', 'd'))

na_if(df, 0)
# A tibble: 4 x 4
 col1 col2 col3 col4 
 &lt;dbl&gt; &lt;dbl&gt; &lt;dbl&gt; &lt;chr&gt;
1 1 NA 1 a 
2 2 2 NA b 
3 3 3 3 c 
4 NA 4 NA d
</code></pre>

blocks|key|545691|text|另一种不使用[<-函数的方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|545692|一个样本数据帧dat+(无耻地复制自@Chase的答案)：|545693|dat

++x+y
1+0+2
2+1+2
3+1+1
4+2+1
5+0+0|code-block|syntax|javascript|545694|可以通过is.na<-函数将零替换为NA：|545695|is.na(dat)+<-+!dat


dat

+++x++y
1+NA++2
2++1++2
3++1++1
4++2++1
5+NA+NA|545696|entityMap^0|6|3|0|7|3|0|0|4|7|I|2|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

An alternative way without the <code>[&lt;-</code> function:

A sample data frame <code>dat</code> (shamelessly copied from @Chase's answer):

<pre><code>dat

 x y
1 0 2
2 1 2
3 1 1
4 2 1
5 0 0
</code></pre>

Zeroes can be replaced with <code>NA</code> by the <code>is.na&lt;-</code> function:

<pre><code>is.na(dat) &lt;- !dat


dat

 x y
1 NA 2
2 1 2
3 1 1
4 2 1
5 NA NA
</code></pre>

blocks|key|545645|text|#Sample+data
set.seed(1)
dat+<-+data.frame(x+=+sample(0:2,+5,+TRUE),+y+=+sample(0:2,+5,+TRUE))
#-----
++x+y
1+0+2
2+1+2
3+1+1
4+2+1
5+0+0

#replace+zeros+with+NA
dat[dat==0]+<-+NA
#-----
+++x++y
1+NA++2
2++1++2
3++1++1
4++2++1
5+NA+NA|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|545646|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>#Sample data
set.seed(1)
dat &lt;- data.frame(x = sample(0:2, 5, TRUE), y = sample(0:2, 5, TRUE))
#-----
 x y
1 0 2
2 1 2
3 1 1
4 2 1
5 0 0

#replace zeros with NA
dat[dat==0] &lt;- NA
#-----
 x y
1 NA 2
2 1 2
3 1 1
4 2 1
5 NA NA
</code></pre>

blocks|key|545882|text|因为有人要这个的data.table版本，并且给定的data.frame解决方案不能与Data.Table一起使用，所以我提供了下面的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|545883|基本上，使用:=操作符-->+DT[x+==+0,+x+:=+NA]|offset|length|style|CODE|545884|library("data.table")

status+=+as.data.table(occupationalStatus)

head(status,+10)
++++origin+destination++N
+1:++++++1+++++++++++1+50
+2:++++++2+++++++++++1+16
+3:++++++3+++++++++++1+12
+4:++++++4+++++++++++1+11
+5:++++++5+++++++++++1++2
+6:++++++6+++++++++++1+12
+7:++++++7+++++++++++1++0
+8:++++++8+++++++++++1++0
+9:++++++1+++++++++++2+19
10:++++++2+++++++++++2+40


status[N+==+0,+N+:=+NA]

head(status,+10)
++++origin+destination++N
+1:++++++1+++++++++++1+50
+2:++++++2+++++++++++1+16
+3:++++++3+++++++++++1+12
+4:++++++4+++++++++++1+11
+5:++++++5+++++++++++1++2
+6:++++++6+++++++++++1+12
+7:++++++7+++++++++++1+NA
+8:++++++8+++++++++++1+NA
+9:++++++1+++++++++++2+19
10:++++++2+++++++++++2+40|code-block|syntax|javascript|545885|entityMap^0|0|6|2|F|J|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]|$D|S|E|T|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|U|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

Because someone asked for the Data.Table version of this, and because the given data.frame solution does not work with data.table, I am providing the solution below.

Basically, use the <code>:=</code> operator --> <code>DT[x == 0, x := NA]</code>

<pre><code>library("data.table")

status = as.data.table(occupationalStatus)

head(status, 10)
 origin destination N
 1: 1 1 50
 2: 2 1 16
 3: 3 1 12
 4: 4 1 11
 5: 5 1 2
 6: 6 1 12
 7: 7 1 0
 8: 8 1 0
 9: 1 2 19
10: 2 2 40


status[N == 0, N := NA]

head(status, 10)
 origin destination N
 1: 1 1 50
 2: 2 1 16
 3: 3 1 12
 4: 4 1 11
 5: 5 1 2
 6: 6 1 12
 7: 7 1 NA
 8: 8 1 NA
 9: 1 2 19
10: 2 2 40
</code></pre>

blocks|key|328917|text|您只能在数值字段中将0替换为NA+(即排除诸如因子之类的内容)，但它在逐列的基础上工作：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|328918|col[col+==+0+&+is.numeric(col)]+<-+NA|code-block|syntax|javascript|328919|通过一个函数，您可以将其应用于整个数据框：|328920|changetoNA+<-+function(colnum,df)+{
++++col+<-+df[,colnum]
++++if+(is.numeric(col))+{++#edit:+verifying+column+is+numeric
++++++++col[col+==+-1+&+is.numeric(col)]+<-+NA
++++}
++++return(col)
}
df+<-+data.frame(sapply(1:5,+changetoNA,+df))|328921|尽管您可以用数据框中的列数或1:ncol(df)替换1:5。|328922|entityMap^0|A|1|E|2|0|0|0|0|E|A|Q|3|0^^$0|@$1|2|3|4|5|6|7|S|8|@$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]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

You can replace <code>0</code> with <code>NA</code> only in numeric fields (i.e. excluding things like factors), but it works on a column-by-column basis:

<pre><code>col[col == 0 &amp; is.numeric(col)] &lt;- NA
</code></pre>

With a function, you can apply this to your whole data frame:

<pre><code>changetoNA &lt;- function(colnum,df) {
 col &lt;- df[,colnum]
 if (is.numeric(col)) { #edit: verifying column is numeric
 col[col == -1 &amp; is.numeric(col)] &lt;- NA
 }
 return(col)
}
df &lt;- data.frame(sapply(1:5, changetoNA, df))
</code></pre>

Although you could replace the <code>1:5</code> with the number of columns in your data frame, or with <code>1:ncol(df)</code>.

blocks|key|545991|text|如果您和我一样，想知道如何用NA替换dataframe中的所有值，那么它只是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|545992|​|545993|df[,]+<-+NA|code-block|syntax|javascript|545994|545995|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|N|8|@]|9|@]|A|$G|H]]|$1|I|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|K|$]]

If you are like me and landed here while wondering how to replace ALL values in a dataframe with NA, it's just:
<div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-html lang-html prettyprint-override"><code>df[,] &lt;- NA</code></pre>
</div>
</div>

I have a dataframe with some numeric columns. Some row has a 0 value which should be considered as null in statistical analysis. What is the fastest way to replace all the 0 value to NULL in R?

Replace all 0 values to NA

我有一个包含一些数值列的dataframe。某些行具有0值，在统计分析中应将其视为null。将R中的所有0值替换为NULL的最快方法是什么？

问将所有0值替换为NA
EN

回答 7

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问将所有0值替换为NAEN

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问将所有0值替换为NA
EN