blocks|key|861181|text|当然，您可以对每一行进行采样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|861182|sapply+(1:4,+function+(row)+df1[row,]<<-sample(df1[row,]))|code-block|syntax|javascript|861183|它会自动调整行，这样每一行中的1数就不会改变。|offset|length|style|CODE|861184|entityMap^0|0|0|F|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|$]]

Of course you can sample each row:

<pre><code>sapply (1:4, function (row) df1[row,]&lt;&lt;-sample(df1[row,]))
</code></pre>

will shuffle the rows itself, so the number of <code>1</code>'s in each row doesn't change. Small changes and it also works great with columns, but this is a exercise for the reader :-P

blocks|key|861220|text|您还可以在R包picante中使用randomizeMatrix函数|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|861221|示例：|861222|test+<-+matrix(c(1,1,0,1,0,1,0,0,1,0,0,1,0,1,0,0),nrow=4,ncol=4)
>+test
+++++[,1]+[,2]+[,3]+[,4]
[1,]++++1++++0++++1++++0
[2,]++++1++++1++++0++++1
[3,]++++0++++0++++0++++0
[4,]++++1++++0++++1++++0

randomizeMatrix(test,null.model+=+"frequency",iterations+=+1000)

+++++[,1]+[,2]+[,3]+[,4]
[1,]++++0++++1++++0++++1
[2,]++++1++++0++++0++++0
[3,]++++1++++0++++1++++0
[4,]++++1++++0++++1++++0

randomizeMatrix(test,null.model+=+"richness",iterations+=+1000)

+++++[,1]+[,2]+[,3]+[,4]
[1,]++++1++++0++++0++++1
[2,]++++1++++1++++0++++1
[3,]++++0++++0++++0++++0
[4,]++++1++++0++++1++++0
>+|code-block|syntax|javascript|861223|选项null.model="frequency"维护列和，而richness维护行和。虽然主要用于在群落生态学中随机化物种存在缺失数据集，但它在这里工作得很好。|861224|此功能还具有其他空型号选项，有关+picante+documentation的更多详细信息(第36页)，请查看以下链接|861225|entityMap|0|LINK|mutability|MUTABLE|url|http://cran.r-project.org/web/packages/picante/picante.pdf^0|7|7|H|F|0|0|0|2|M|U|8|0|H|7|G|M|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|13|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|1A|8|@$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|Q|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|X]]]]

you can also use the <code>randomizeMatrix</code> function in the R package <code>picante</code>

example:

<pre><code>test &lt;- matrix(c(1,1,0,1,0,1,0,0,1,0,0,1,0,1,0,0),nrow=4,ncol=4)
&gt; test
 [,1] [,2] [,3] [,4]
[1,] 1 0 1 0
[2,] 1 1 0 1
[3,] 0 0 0 0
[4,] 1 0 1 0

randomizeMatrix(test,null.model = "frequency",iterations = 1000)

 [,1] [,2] [,3] [,4]
[1,] 0 1 0 1
[2,] 1 0 0 0
[3,] 1 0 1 0
[4,] 1 0 1 0

randomizeMatrix(test,null.model = "richness",iterations = 1000)

 [,1] [,2] [,3] [,4]
[1,] 1 0 0 1
[2,] 1 1 0 1
[3,] 0 0 0 0
[4,] 1 0 1 0
&gt; 
</code></pre>

The option <code>null.model="frequency"</code> maintains column sums and <code>richness</code> maintains row sums.
Though mainly used for randomizing species presence absence datasets in community ecology it works well here. 

This function has other null model options as well, check out following link for more details (page 36) of the <a href="http://cran.r-project.org/web/packages/picante/picante.pdf" rel="nofollow"> <code>picante</code> documentation</a>

blocks|key|861278|text|数据帧中的随机样本和排列如果是矩阵形式的索引，请使用基本包中的sample函数data.frame+=+sample(1:nrow(df1)，size=1*nrow(df1))随机样本和排列|type|unstyled|depth|inlineStyleRanges|entityRanges|data|861279|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Random Samples and Permutations ina dataframe
If it is in matrix form convert into data.frame
use the sample function from the base package
indexes = sample(1:nrow(df1), size=1*nrow(df1))
Random Samples and Permutations

blocks|key|861329|text|这是使用包dplyr混洗data.frame的另一种方法|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|861330|按行排列：|861331|df2+<-+slice(df1,+sample(1:n()))|code-block|syntax|javascript|861332|或|861333|df2+<-+sample_frac(df1,+1L)|861334|按列排列：|861335|df2+<-+select(df1,+one_of(sample(names(df1))))+|861336|entityMap^0|5|5|C|A|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|13|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|15|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|16|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|V|$]]

This is another way to shuffle the <code>data.frame</code> using package <code>dplyr</code>:

row-wise:

<pre><code>df2 &lt;- slice(df1, sample(1:n()))
</code></pre>

or

<pre><code>df2 &lt;- sample_frac(df1, 1L)
</code></pre>

column-wise:

<pre><code>df2 &lt;- select(df1, one_of(sample(names(df1)))) 
</code></pre>

blocks|key|644107|text|您还可以在数据框中对相同数量的项目进行“采样”，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|644108|nr<-dim(M)[1]
random_M+=+M[sample.int(nr),]|code-block|syntax|javascript|644109|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 also "sample" the same number of items in your data frame with something like this:

<pre><code>nr&lt;-dim(M)[1]
random_M = M[sample.int(nr),]
</code></pre>

blocks|key|861438|text|如果目标是随机地打乱每一列，上面的一些答案是不起作用的，因为这些列是联合打乱的(这保留了列间的相关性)。其他的则需要安装一个包。然而，存在一行代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|861439|df2+=+lapply(df1,+function(x)+{+sample(x)+})|code-block|syntax|javascript|861440|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 the goal is to randomly shuffle each column, some of the above answers don't work since the columns are shuffled jointly (this preserves inter-column correlations). Others require installing a package. Yet a one-liner exist:

<pre><code>df2 = lapply(df1, function(x) { sample(x) })
</code></pre>

I have a dataframe (df1) like this.

<pre><code> f1 f2 f3 f4 f5
d1 1 0 1 1 1 
d2 1 0 0 1 0
d3 0 0 0 1 1
d4 0 1 0 0 1
</code></pre>

The d1...d4 column is the rowname, the f1...f5 row is the columnname.

To do sample(df1), I get a new dataframe with count of 1 same as df1. So, the count of 1 is conserved for the whole dataframe but not for each row or each column.

Is it possible to do the randomization row-wise or column-wise?

I want to randomize the df1 column-wise for each column, i.e. the number of 1 in each column remains the same. and each column need to be changed by at least once. For example, I may have a randomized df2 like this: (Noted that the count of 1 in each column remains the same but the count of 1 in each row is different.

<pre><code> f1 f2 f3 f4 f5
d1 1 0 0 0 1 
d2 0 1 0 1 1
d3 1 0 0 1 1
d4 0 0 1 1 0
</code></pre>

Likewise, I also want to randomize the df1 row-wise for each row, i.e. the no. of 1 in each row remains the same, and each row need to be changed (but the no of changed entries could be different). For example, a randomized df3 could be something like this:

<pre><code> f1 f2 f3 f4 f5
d1 0 1 1 1 1 &lt;- two entries are different
d2 0 0 1 0 1 &lt;- four entries are different
d3 1 0 0 0 1 &lt;- two entries are different
d4 0 0 1 0 1 &lt;- two entries are different
</code></pre>

PS. Many thanks for the help from Gavin Simpson, Joris Meys and Chase for the previous answers to my previous question on randomizing two columns.

How to randomize (or permute) a dataframe rowwise and columnwise?

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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我有一个这样的数据帧(df1)。     f1   f2   f3   f4   f5d1   1    0    1    1    1  d2   1    0    0    1    0d3   0    0    0    1    1d4   0    1    0    0    1d1...d4列是行名...

问如何按行和按列随机化(或置换)数据帧？
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