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將缺失值替換為 0

你好，

我正在嘗試將數據表中的所有缺失值替換為 0。下面的腳本不起作用。我使用 JMP 13。還有其他方法可以做到這一點嗎？

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>

謝謝！

回复：用 0 替換缺失值

>> 我可以用“N/A”字符串替換數字列中的缺失值嗎？

使用值標籤來做到這一點。右鍵單擊列標題，轉到列屬性，選擇值標籤。然後在值中輸入句點，在標籤字段中輸入 N/A，如下所示。然後點擊添加，然後就ok了。

要編寫腳本，請按照上述步驟的日誌記錄器進行操作...

數據表( "DT name" ):第 904 列 << 值標籤( {"." = "N/A"} ) << 使用值標籤( 1 );

回复：用 0 替換缺失值

在我可以用“N/A”字符串替換數字列中的缺失值嗎？

回复：用 0 替換缺失值

親愛的伊恩和克雷奇，

非常感謝您的提交。我進行了一些初步測試，Ian 的解決方案比 Recode 快 4 倍以上，Craige 的解決方案幾乎快 10 倍。如果有人想比較所有 3 個解決方案，請嘗試以下操作：

//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 提到他認為矩陣可以更快地解決問題，我想你們兩個也是這麼想的！

回复：用 0 替換缺失值

將解決方案與數據表下標相結合......

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);

大約10秒。

回复：用 0 替換缺失值

FWIW，這種方法可能會更好。對於 10,000 x 10,000，在我的筆記本電腦上花費了大約 10 秒（包括製作起始表）：

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 替換缺失值

大家好，

我一直在研究這個問題，涉及數百列和數百行缺失。我對錶現並不滿意，嘗試了很多方法。看來 Recode 解決這個問題的速度比當前解決方案快約 150 倍。與可能關心速度的其他人共享代碼。如果您找到更快的東西，請告訴我。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> 感謝 Rebecca Lyzinski 的想法！</LINGO-BODY><LINGO-SUB id="lingo-sub-228122" slang="en-US" mode="NONE">回复：用 0 替換缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-228122" slang="en-US" mode="NONE">我經常使用這個腳本。非常感謝您提供它。</LINGO-BODY><LINGO-SUB id="lingo-sub-65695" slang="en-US" mode="NONE">回复：用 0 替換缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-65695" slang="en-US" mode="NONE">嘗試在腳本中更具體一些。 為其中包含特定列的數據表提供句柄。就像 dt=table();或類似的東西。 然後引用表中的列 dt::nameOFCol   您的腳本可能正在嘗試修改此步驟運行時當前表中不存在的列。</LINGO-BODY><LINGO-SUB id="lingo-sub-65039" slang="en-US" mode="NONE">回复：用 0 替換缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-65039" slang="en-US" mode="NONE">像這樣的事情很難從描述中診斷出來。您可能必鬚髮布您的腳本（或至少足夠的腳本來複製問題），以便我們幫助解決它。</LINGO-BODY><LINGO-SUB id="lingo-sub-64972" slang="en-US" mode="NONE">回复：用 0 替換缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-64972" slang="en-US" mode="NONE">嗨卡梅倫， 我重新啟動 JMP，腳本在特定數據表上運行良好。缺失值被 0 替換。但是當我集成到腳本的其餘部分時，它給出了以下錯誤：在“Column”的訪問或評估中找不到列，Column/*###*/(nc[i]) 在編寫這段腳本之前，我有一個原始數據表，我根據所選列對其進行子集化、堆疊數據、匯總 anf，然後將其製成最終數據表。我正在嘗試用 0 替換最終數據表中的缺失值。    我對 JMP 腳本相當陌生，不確定為什麼集成會出現問題。知道為什麼會發生這種情況嗎？  </LINGO-BODY><LINGO-SUB id="lingo-sub-64943" slang="en-US" mode="NONE">回复：用 0 替換缺失值</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 中運行 Jim 的代碼，沒有出現任何錯誤。也許重新啟動 JMP 並再次運行它？</LINGO-BODY><LINGO-SUB id="lingo-sub-64931" slang="en-US" mode="NONE">回复：用 0 替換缺失值</LINGO-SUB><LINGO-BODY id="lingo-body-64931" slang="en-US" mode="NONE">嗨吉姆，   我嘗試了上述方法，但出現以下錯誤。     <PRE><CODE class=" language-jsl">could not find column in access or evaluation of 'Column' , Column/*###*/(nc[i])

知道發生了什麼事嗎？我使用 JMP 13，我不確定該語法是否不適用於該版本。

謝謝，

氮

回复：用 0 替換缺失值

這是您想要執行的操作的正確語法

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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