blocks|key|1053769|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1053770|m+=+rand(500,5)*100;+//your+dataset
m(m(:,ii)+<=+80)+=+1;
m(m(:,ii)+>+80)+=+2;|code-block|syntax|javascript|1053771|其中ii是您想要更改的列。例如，ii+=3|offset|length|style|CODE|1053772|m(m(:,3)+<=+80)+=+1;
m(m(:,3)+>+80)+=+2;|1053773|entityMap^0|0|0|2|2|G|2|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|P|$]]

Try this:

<pre><code>m = rand(500,5)*100; //your dataset
m(m(:,ii) &lt;= 80) = 1;
m(m(:,ii) &gt; 80) = 2;
</code></pre>

where <code>ii</code> is your the column you want to change. E.g. <code>ii</code> = 3

<pre><code>m(m(:,3) &lt;= 80) = 1;
m(m(:,3) &gt; 80) = 2;
</code></pre>

blocks|key|1057558|text|这可以通过多种方式完成。如您所说，您可以使用嵌套的for循环，也可以像Rodrigo建议的那样使用条件赋值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1057559|首先，这是您的数据：|1057560|%25//+your+matrix
M+=+rand(500,5)*100;|code-block|syntax|javascript|1057561|您可能想知道循环的行数和列数.|1057562|%25//+get+the+size+for+the+loops
[num_rows,num_columns]+=+size(M);|1057563|为了简单地遍历一列，以下内容应该可以工作：|1057564|%25//+loop+through+one+column+(column+#2),+save+one+result
col+=+2;
for+row+=+1:num_rows
++++if+M(row,col)+<=+80
++++++++age+=+1;
++++else
++++++++age+=+2;
++++end
end|1057565|但是，您说要去创建数据，所以您可能想要更改保存所有结果。上面的示例将留给您一个名为'age‘的变量，它只存储循环中的最后一个值。|1057566|以下内容应该允许您保存各个列的所有结果：|1057567|%25//+loop+through+one+column+(column+#2),+save+all+results
col+=+2;
%25//+initialize+age+array
age+=+zeros(500,1);
%25//+do+the+loop
for+row+=+1:num_rows
++++if+M(row,col)+<=+80
+++++++++age(row)+=+1;
++++else
++++++++age(row)+=+2;
++++end
end|1057568|要循环遍历所有列，需要另一个for循环：|1057569|%25//+loop+through+all+columns,+save+all+results
%25//+initialize+age+array
age+=+zeros(500,5);
%25//+loop+through+each+column
for+col+=+1:num_columns
++++%25//+loop+through+each+row
++++for+row+=+1:num_rows
++++++++if+M(row,col)+<=+80
++++++++++++age(row,col)+=+1;
++++++++else
++++++++++++age(row,col)+=+2;
++++++++end
++++end
end|1057570|最后，现在您已经看到了这一点，对于许多人来说，首选的方法是利用MATLAB的条件赋值技巧。下面将产生与最后一个代码片段相同的结果：|1057571|%25//+now+without+loops
age+=+zeros(500,5);
age(M+<=+80)+=+1;
age(M+>+80)+=+2;|1057572|(请注意，我在评论中使用了%25和//+.您可以忽略//，因为我只是添加了它，这样Stack+Exchange才能识别我的注释)|1057573|要回答后续的问题，没有必要添加关于young+=+1和old+=+2的内容。但是，它是首选的，因为它允许您删除幻数。|offset|length|style|CODE|1057574|编辑以回答后续：|1057575|要将结果保存在原始数组中，可以执行以下几项操作：|1057576|可以在末尾将新数组分配给旧数组的列。|ordered-list-item|1057577|可以在循环时分配新值。|1057578|您可以使用上面提到的条件替换Rodrigo。|1057579|第一个很容易..。使用上面最后两个过程中的一个，然后执行以下操作：|1057580|M(:,col_to_replace)+=+age(:,col_to_replace_with);|1057581|或者您可以在一起添加一个新列：|1057582|M(:,6)+=+age(:,col_of_interest);|1057583|或者，您只需更改循环，以便将原始值替换为它们的离散值：|1057584|%25//+loop+through+all+columns,+save+all+results+in+original+locations
%25//+loop+through+each+column
for+col+=+1:num_columns
++++%25//+loop+through+each+row
++++for+row+=+1:num_rows
++++++++if+M(row,col)+<=+80
++++++++++++M(row,col)+=+1;
++++++++else
++++++++++++M(row,col)+=+2;
++++++++end
++++end
end|1057585|最后，您可以只使用条件替换方法。下面的示例将用离散化值替换M的所有行和列：|1057586|M(M+<=+80)+=+1;
M(M+>+80)+=+2;|1057587|要回答您的具体示例，这将遍历第1列并将结果保存在第3列中：|1057588|%25//+loop+through+one+column+(column+#1),+save+all+results+in+another+column+(#3)
col+=+1;
save_col+=+3;
%25//+do+the+loop
for+row+=+1:num_rows
++++if+M(row,col)+<=+80
++++++++M(row,save_col)+=+1;
++++else
++++++++M(row,save_col)+=+2;
++++end
end|1057589|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Magic_number_%2528programming%2529#Unnamed_numerical_constants^0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|H|9|R|7|1J|2|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|2F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|2G|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|2H|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|2I|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|2J|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|2K|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|2L|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|2M|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|2N|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|2O|8|@]|9|@]|A|$G|H]]|$1|W|3|X|5|6|7|2P|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|F|7|2Q|8|@]|9|@]|A|$G|H]]|$1|10|3|11|5|6|7|2R|8|@]|9|@]|A|$]]|$1|12|3|13|5|F|7|2S|8|@]|9|@]|A|$G|H]]|$1|14|3|15|5|6|7|2T|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|2U|8|@$18|2V|19|2W|1A|1B]|$18|2X|19|2Y|1A|1B]]|9|@$18|2Z|19|30|1|31]]|A|$]]|$1|1C|3|1D|5|6|7|32|8|@]|9|@]|A|$]]|$1|1E|3|1F|5|6|7|33|8|@]|9|@]|A|$]]|$1|1G|3|1H|5|1I|7|34|8|@]|9|@]|A|$]]|$1|1J|3|1K|5|1I|7|35|8|@]|9|@]|A|$]]|$1|1L|3|1M|5|1I|7|36|8|@]|9|@]|A|$]]|$1|1N|3|1O|5|6|7|37|8|@]|9|@]|A|$]]|$1|1P|3|1Q|5|F|7|38|8|@]|9|@]|A|$G|H]]|$1|1R|3|1S|5|6|7|39|8|@]|9|@]|A|$]]|$1|1T|3|1U|5|F|7|3A|8|@]|9|@]|A|$G|H]]|$1|1V|3|1W|5|6|7|3B|8|@]|9|@]|A|$]]|$1|1X|3|1Y|5|F|7|3C|8|@]|9|@]|A|$G|H]]|$1|1Z|3|20|5|6|7|3D|8|@]|9|@]|A|$]]|$1|21|3|22|5|F|7|3E|8|@]|9|@]|A|$G|H]]|$1|23|3|24|5|6|7|3F|8|@]|9|@]|A|$]]|$1|25|3|26|5|F|7|3G|8|@]|9|@]|A|$G|H]]|$1|27|3|-4|5|6|7|3H|8|@]|9|@]|A|$]]]|28|$29|$5|2A|2B|2C|A|$2D|2E]]]]

This can be done many ways. You can use nested for loops, like you said, or conditional assignments as Rodrigo suggests.

First, here's your data:

<pre><code>%// your matrix
M = rand(500,5)*100;
</code></pre>

You'll want to know the number of rows and columns for the loop...

<pre><code>%// get the size for the loops
[num_rows,num_columns] = size(M);
</code></pre>

To simply loop through one column, the following should work:

<pre><code>%// loop through one column (column #2), save one result
col = 2;
for row = 1:num_rows
 if M(row,col) &lt;= 80
 age = 1;
 else
 age = 2;
 end
end
</code></pre>

But, you said you wanted to descretize the data, so you'll likely want to change save all the results. The above example will leave you with a variable called 'age' which will only store that last value from the loop.

The following should allow you to save all the results from the individual column:

<pre><code>%// loop through one column (column #2), save all results
col = 2;
%// initialize age array
age = zeros(500,1);
%// do the loop
for row = 1:num_rows
 if M(row,col) &lt;= 80
 age(row) = 1;
 else
 age(row) = 2;
 end
end
</code></pre>

To loop through all the columns requires another for loop:

<pre><code>%// loop through all columns, save all results
%// initialize age array
age = zeros(500,5);
%// loop through each column
for col = 1:num_columns
 %// loop through each row
 for row = 1:num_rows
 if M(row,col) &lt;= 80
 age(row,col) = 1;
 else
 age(row,col) = 2;
 end
 end
end
</code></pre>

Finally, now that you've seen that, the preferred way for many is to take advantage of MATLAB's conditional assignment tricks. The following will produce the same result as the last code snippet:

<pre><code>%// now without loops
age = zeros(500,5);
age(M &lt;= 80) = 1;
age(M &gt; 80) = 2;
</code></pre>

(note that I've used % and // in my comments... you can ignore the // since I only added it so Stack Exchange would recognize my comments)

To answer your follow up, its not necessary to add the bit about <code>young = 1</code> and <code>old = 2</code>. It is preferred, however, since it allows you to remove the <a href="http://en.wikipedia.org/wiki/Magic_number_%28programming%29#Unnamed_numerical_constants" rel="nofollow">Magic Numbers</a>.

EDIT to answer follow-up:

To save the results in the original array, you can do a few things:

<ol>
<li>You can assign the new array to a column of the old array at the end</li>
<li>You can assign the new values as you loop</li>
<li>You can use the conditional replacement Rodrigo mentioned.</li>
</ol>

The first one is easy... use one of the last two procedures above, then do this:

<pre><code>M(:,col_to_replace) = age(:,col_to_replace_with);
</code></pre>

Or you could add a new column all together:

<pre><code>M(:,6) = age(:,col_of_interest);
</code></pre>

Alternatively, you can just change the loop so the original values are replaced with their discretized value:

<pre><code>%// loop through all columns, save all results in original locations
%// loop through each column
for col = 1:num_columns
 %// loop through each row
 for row = 1:num_rows
 if M(row,col) &lt;= 80
 M(row,col) = 1;
 else
 M(row,col) = 2;
 end
 end
end
</code></pre>

Finally, you can just use the conditional replacement method. The sample below will replace all the rows and columns of M with the discretized value:

<pre><code>M(M &lt;= 80) = 1;
M(M &gt; 80) = 2;
</code></pre>

To answer your specific example, this will loop through column 1 and save the result in column 3:

<pre><code>%// loop through one column (column #1), save all results in another column (#3)
col = 1;
save_col = 3;
%// do the loop
for row = 1:num_rows
 if M(row,col) &lt;= 80
 M(row,save_col) = 1;
 else
 M(row,save_col) = 2;
 end
end
</code></pre>

I have an dataset array <code>M</code> of size <code>500x5</code>, is there any way I can use a nested <code>for</code> loop to run through a particular column of the array? If so how would I go about doing that? I want the <code>if</code> statement in the loop to be something like:

<pre><code>if age &lt;= 80
 age = 1
else 
 age = 2
end 
</code></pre>

what would I put as the <code>for</code> loop? Would it be better to initialise variables as <code>young =1; old = 2;</code> and then have <code>age = young</code> in the <code>if</code> statement rather than <code>age = 1</code>? I am trying to discretise the data into either <code>1</code> or <code>2</code> with <code>1</code> being young and <code>2</code> being old.

Matlab - Nested loop for certain column

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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我有一个大小为M的dataset数组500x5，有任何方法可以使用嵌套的for循环来遍历数组的特定列吗？如果是这样的话，我该怎么做呢？我希望循环中的if语句类似于：if age <= 80 age = 1else age = 2end 我将把什么作为for循环？是否最好将变量初始化为young =1; old ...

问Matlab -某些列的嵌套循环
EN

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问Matlab -某些列的嵌套循环EN