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将缺失值替换为 0

你好，

我正在尝试将数据表中的所有缺失值替换为 0。下面的脚本不起作用。我使用 JMP 13。还有其他方法可以做到这一点吗？

dt <SPAN class="token operator">=</SPAN> <SPAN class="token function">Current Data Table</SPAN><SPAN class="token punctuation">(</SPAN><SPAN class="token punctuation">)</SPAN><SPAN class="token punctuation">;</SPAN>
nc <SPAN class="token operator">=</SPAN> dt <SPAN class="token operator"><</SPAN><SPAN class="token operator"><</SPAN><SPAN class="token messages"> get column names</SPAN><SPAN class="token punctuation">(</SPAN> Numeric <SPAN class="token punctuation">)</SPAN><SPAN class="token punctuation">;</SPAN>
<SPAN class="token function">For</SPAN><SPAN class="token punctuation">(</SPAN> i <SPAN class="token operator">=</SPAN> <SPAN class="token number">1</SPAN><SPAN class="token punctuation">,</SPAN> i <SPAN class="token operator"><=</SPAN> <SPAN class="token function">N Items</SPAN><SPAN class="token punctuation">(</SPAN> nc <SPAN class="token punctuation">)</SPAN><SPAN class="token punctuation">,</SPAN> i<SPAN class="token operator">++</SPAN><SPAN class="token punctuation">,</SPAN>
  nc<SPAN class="token punctuation">[</SPAN>i<SPAN class="token punctuation">]</SPAN><SPAN class="token punctuation">[</SPAN>dt <SPAN class="token operator"><</SPAN><SPAN class="token operator"><</SPAN><SPAN class="token messages"> get rows where</SPAN><SPAN class="token punctuation">(</SPAN> <SPAN class="token function">Is Missing</SPAN><SPAN class="token punctuation">(</SPAN> nc<SPAN class="token punctuation">[</SPAN>i<SPAN class="token punctuation">]</SPAN><SPAN class="token punctuation">[</SPAN><SPAN class="token punctuation">]</SPAN> <SPAN class="token punctuation">)</SPAN> <SPAN class="token punctuation">)</SPAN><SPAN class="token punctuation">]</SPAN> <SPAN class="token operator">=</SPAN> <SPAN class="token number">0</SPAN>
<SPAN class="token punctuation">)</SPAN><SPAN class="token punctuation">;</SPAN>

谢谢！

回复：用 0 替换缺失值

>> 我可以用“N/A”字符串替换数字列中的缺失值吗？

使用值标签来做到这一点。右键单击列标题，转到列属性，选择值标签。然后在值中输入句点，在标签字段中输入 N/A，如下所示。然后点击添加，然后就ok了。

要编写脚本，请按照上述步骤的日志记录器进行操作...

数据表( "DT name" ):第 904 列 << 值标签( {"." = "N/A"} ) << 使用值标签( 1 );

回复：用 0 替换缺失值

在我可以用“N/A”字符串替换数字列中的缺失值吗？

回复：用 0 替换缺失值

亲爱的伊恩和克雷奇，

非常感谢您的提交。我运行了一些初步测试，Ian 的解决方案比 Recode 快 4 倍以上，Craige 的解决方案几乎快 10 倍。如果有人想比较所有 3 个解决方案，请尝试以下操作：

//Recode solution

nr = 10000;  // Number of rows
nc = 10000;  // Number of columns
nm = 0.9;  // Proportion of missing values

// Make a table and inject some missing values
mat = J(nr*nc, 1, RandomNormal());
mat[RandomIndex(nr*nc, nr*nc*nm)] = .;
dt = AsTable(Shape(mat, nr, nc));
dt << setName("Data with some missingness");

replaceMissingStart = HP Time();

// replace missings with zeros
nc = dt << get column names( Numeric );

For( i = 1, i <= N Items( nc ), i++,
 col = Column(dt,nc[i]);
 colname = col << getname();

    Eval(Parse(EvalInsert("\[
  dt << Recode Column(
   :Name("^colname^"),
   {Map Value( _rcOrig, {., 0}, Unmatched( _rcNow ) )},
   Update Properties( 1 ),
   Target Column( :Name("^colname^") )
  );
 ]\")));
 
 //Code from the Community
 //Column( nc[i] )[dmdt << get rows where( Is Missing( As Column( nc[i] ) ) )] = 0;

 // standardize column attributes, change all to nominal
 //col << Set Modeling Type( "Nominal" ) 
);

replaceMissingStop = HP Time();
totalReplaceTime = replaceMissingStop-replaceMissingStart;
show(totalReplaceTime/1000000);

Russ Wolfinger 提到他认为矩阵可以更快地解决问题，我想你们两个也是这么想的！

回复：用 0 替换缺失值

将解决方案与数据表下标相结合......

nr = 10000;  // Number of rows
nc = 10000;  // Number of columns
nm = 0.9;  // Proportion of missing values

// Make a table and inject some missing values
mat = J(nr*nc, 1, RandomNormal());
mat[RandomIndex(nr*nc, nr*nc*nm)] = .;
dt = AsTable(Shape(mat, nr, nc));
dt << setName("Data with some missingness");


replaceMissingStart = HP Time();

nc = dt << get column names( Numeric );
mat = dt[0,nc];
mat[Loc(mat, .)] = 0;
dt[0,nc] = mat;


replaceMissingStop = HP Time();
totalReplaceTime = replaceMissingStop-replaceMissingStart;
show(totalReplaceTime/1000000);

大约10秒。

回复：用 0 替换缺失值

FWIW，这种方法可能会更好。对于 10,000 x 10,000，在我的笔记本电脑上花费了大约 10 秒（包括制作起始表）：

NamesDefaultToHere(1);

// Size the problem . . .
nr = 10000;  // Number of rows
nc = 10000;  // Number of columns
nm = 0.1;  // Proportion of missing values

// Make a table and inject some missing values
mat = J(nr*nc, 1, RandomNormal());
mat[RandomIndex(nr*nc, nr*nc*nm)] = .;
dt = AsTable(Shape(mat, nr, nc));
dt << setName("Data with some missingness");

// Start here: Get values into a matrix (assumes contiguous numeric columns)
mat = dt << getAsMatrix;
// Reshape as a vector
mat = Shape(mat, nr*nc, 1);
// Replace missing
mat[Loc(mat, .)] = 0;
// Reshape back to a matrix
mat = Shape(mat, nc, nc);
// Make a new table
dt2 = AsTable(Shape(mat, nr, nc));
dt2 << setName("Data with missing set to zero values");

回复：用 0 替换缺失值

大家好，

我一直在研究这个问题，涉及数百列和数百行缺失。我对表现并不满意，尝试了很多方法。看来 Recode 解决这个问题的速度比当前解决方案快约 150 倍。与可能关心速度的其他人共享代码。如果您找到更快的东西，请告诉我。每一毫秒都很重要！

</CODE> <PRE><CODE class=" language-jsl">replaceMissingStart = HP Time(); // replace missings with zeros nc = dmdt << get column names( Numeric ); For( i = 1, i <= N Items( nc ), i++, col = Column(dt,nc[i]); colname = col << getname(); Eval(Parse(EvalInsert("\[ dmdt << Recode Column( :Name("^colname^"), {Map Value( _rcOrig, {., 0}, Unmatched( _rcNow ) )}, Update Properties( 1 ), Target Column( :Name("^colname^") ) ); ]\"))); //Code from the Community //Column( nc[i] )[dmdt << get rows where( Is Missing( As Column( nc[i] ) ) )] = 0; // standardize column attributes, change all to nominal col << Set Modeling Type( "Nominal" ) ); replaceMissingStop = HP Time(); totalReplaceTime = replaceMissingStop-replaceMissingStart; show(totalReplaceTime/1000000);

</CODE> 感谢 Rebecca Lyzinski 的想法！</LINGO-BODY><LINGO-SUB id="lingo-sub-228122" slang="en-US" mode="NONE">回复：用 0 替换缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-228122" slang="en-US" mode="NONE">我经常使用这个脚本。非常感谢您提供它。</LINGO-BODY><LINGO-SUB id="lingo-sub-65695" slang="en-US" mode="NONE">回复：用 0 替换缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-65695" slang="en-US" mode="NONE">尝试在脚本中更具体一些。 为其中包含特定列的数据表提供句柄。就像 dt=table();或类似的东西。 然后引用表中的列 dt::nameOFCol   您的脚本可能正在尝试修改此步骤运行时当前表中不存在的列。</LINGO-BODY><LINGO-SUB id="lingo-sub-65039" slang="en-US" mode="NONE">回复：用 0 替换缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-65039" slang="en-US" mode="NONE">像这样的事情很难从描述中诊断出来。您可能必须发布您的脚本（或至少足够的脚本来复制问题），以便我们帮助解决它。</LINGO-BODY><LINGO-SUB id="lingo-sub-64972" slang="en-US" mode="NONE">回复：用 0 替换缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-64972" slang="en-US" mode="NONE">嗨卡梅伦， 我重新启动 JMP，脚本在特定数据表上运行良好。缺失值被 0 替换。但是当我集成到脚本的其余部分时，它给出了以下错误：在“Column”的访问或评估中找不到列，Column/*###*/(nc[i]) 在编写这段脚本之前，我有一个原始数据表，我根据所选列对其进行子集化、堆叠数据、汇总 anf，然后将其制成最终数据表。我正在尝试用 0 替换最终数据表中的缺失值。    我对 JMP 脚本相当陌生，不确定为什么集成会出现问题。知道为什么会发生这种情况吗？  </LINGO-BODY><LINGO-SUB id="lingo-sub-64943" slang="en-US" mode="NONE">回复：用 0 替换缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-64943" slang="en-US" mode="NONE">你好<A href="https://community.jmp.com/t5/user/viewprofilepage/user-id/12425" target="_blank">@nshivan</A> ,我在版本 13 和 14 中运行 Jim 的代码，没有出现任何错误。也许重新启动 JMP 并再次运行它？</LINGO-BODY><LINGO-SUB id="lingo-sub-64931" slang="en-US" mode="NONE">回复：用 0 替换缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-64931" slang="en-US" mode="NONE">嗨吉姆，   我尝试了上述方法，但出现以下错误。     <PRE><CODE class=" language-jsl">could not find column in access or evaluation of 'Column' , Column/*###*/(nc[i])

知道发生了什么事吗？我使用 JMP 13，我不确定该语法是否不适用于该版本。

谢谢，

氮

回复：用 0 替换缺失值

这是您想要执行的操作的正确语法

dt = Current Data Table();
nc = dt << get column names( Numeric );
For( i = 1, i <= N Items( nc ), i++,
 Column( nc[i] )[dt << get rows where( Is Missing( As Column( nc[i] ) ) )] = 0
);

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