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最终收益率表问题

txnelson 先生非常友好地与我分享了一个很棒的脚本，该脚本将通过/失败分配给列，然后最终根据整行分配最终的通过/失败。 我想将最后一列分成子列，虽然我知道如何在概念上完成它，但我无法弄清楚正确的语法..

 

这段代码找到所有列中的最小值，但是我想添加子集

Eval( Parse( "dt2 << New Column( \!"Row Pass/Fail\!", 公式( If( Min(" || foundCols || " ) == 1, 1, 0 ) )" ) );

 

列：
行通过/失败

Test_1 通过/失败

Test_2 通过/失败

Test_3a 通过/失败

Test_3b 通过/失败

Test_3c 通过/失败

Test_3d 通过/失败

Test_3e 通过/失败

Test_4_State_1 通过/失败

Test_4_State_3 通过/失败

Test_4_State_4 通过/失败

 

 

 colList = dt2 << Get Column Names( numeric, string );
 
 foundCols = "";
 
 For( i = 1, i <= N Items( colList ), i++,
 
  Spec = Column( dt2, colList[i] ) << Get Property( "Spec Limits" );
  
  dt2 << Clear Select;
  
  If( Is Empty( Spec ) == 0,
  
   dt2 << New Column( (Column( dt2, colList[i] ) << Get Name) || " Pass/Fail" );
   
   If( Is Missing( Try( Spec["LSL"], . ) ) == 0,
   
    dt2 << Select Where( As Column( dt2, colList[i] ) < Spec["LSL"] );
    
    Try( Column( dt2, N Cols( dt2 ) )[dt2 << get selected rows] = 0 );
   );
   
   If( Is Missing( Try( Spec["USL"], . ) ) == 0,
   
    dt2 << Select Where( As Column( dt2, colList[i] ) > Spec["USL"], Current Selection( "Extend" ) );
    
    Try( Column( dt2, N Cols( dt2 ) )[dt2 << Get Selected Rows] = 0 );
   );
   
   dt2 << invert row selection;
   
   Try( Column( dt2, N Cols( dt2 ) )[dt2 << Get Selected Rows] = 1 );
   
   Column( dt2, N Cols( dt2 ) ) << Set Property( "Value Labels", {1 = "Pass", 0 = "Fail"} );
   
   If( foundCols == "",
    foundCols = ":Name(\!"" || (Column( dt2, colList[i] ) << Get Name) || " Pass/Fail\!")",
    foundCols = foundCols || ", " || ":Name(\!"" || (Column( dt2, colList[i] ) << Get Name) || " Pass/Fail\!")"
   );
  );
 );

 

回复：最终收益率表问题

我不得不有点创意，但它做到了......它看起来有点笨拙，但它完成了工作.. 谢谢

 

  ycols_3 = dt3 << Get Column Names( Numeric, Continuous );
  
  yPassFail = {};
  
  For( i = 1 , i <= N Items( ycols_3 ), i++,
   If( Contains( ycols_3[i], "Pass/Fail" ),
    Insert Into( yPassFail, ycols_3[i]);
   )
  );
  
  Show( yPassFail );
  
  txList = {};
  
  For( i = 1 , i <= N Items( yPassFail ), i++,
   If( Contains( yPassFail[i], "TX S12 State 01" ),
    Insert Into( txList , yPassFail[i]);
   )
  );
  
  rxList = {};
  
  For( i = 1 , i <= N Items( yPassFail ), i++,
   If( Contains( yPassFail[i], "RX S21 State 03" ),
    Insert Into( rxList , yPassFail[i]);
   )
  );
  
  deltaAmpList = {};
  
  For( i = 1 , i <= N Items( yPassFail ), i++,
   If( Contains( yPassFail[i], "Delta Amp" ),
    Insert Into( deltaAmpList , yPassFail[i]);
   )
  );
  
  txList = Char( txList );
  rxList = Char( rxList );
  deltaAmpList = Char( deltaAmpList );
  
  Eval( Parse( "dt3 << New Column( \!"Tx_Spar\!", Formula( If( Min(" || txList || " ) == 1, 1, 0 ) ) )" ) );
  Eval( Parse( "dt3 << New Column( \!"Rx_Spar\!", Formula( If( Min(" || rxList || " ) == 1, 1, 0 ) ) )" ) );
  Eval( Parse( "dt3 << New Column( \!"Rx_SparDelta\!", Formula( If( Min(" || deltaAmpList  || " ) == 1, 1, 0 ) ) )" ) );

 

 

回复：最终收益率表问题

我假设有问题的变量是“FoundCols”。您将其设置为“”的值，这使得变量“FoundCols”成为一个字符串，而不是一个列表。稍后在您的代码中，您检查变量“FoundCols”是否等于 {}，事实并非如此。它等于“”。如果要将“FoundCols”初始化为空列表，请指定：

FoundCols = {};

回复：最终收益率表问题

不幸的是，错误消息“N Items() argument must be a list{579}”仍然是同样的问题

 

我相信您在我开始沿着这条道路前行时提供了这段代码。

 

哦，我们现在有 JMP 15，所以如果有另一种方法，我会洗耳恭听（眼睛）......

 

谢谢

 

 colsToPassFail = dt3 << get column names( numeric, String );
 
 Show( colsToPassFail );
  
  foundCols = "";
  
  For( i = 1, i <= N Items( colsToPassFail  ), i++,
   Spec = Column( dt3, colsToPassFail[i] ) << Get Property( "Spec Limits" );
   dt3 << Clear Select;
   If( Is Empty( spec ) == 0,
    dt3 << New Column( ( Column( dt3, colsToPassFail[i] ) << Get Name) || " Pass/Fail" );
    If( Is Missing( Try( spec["LSL"], . ) ) == 0,
     dt3 << Select Where( As Column( dt3, colsToPassFail[i] ) < spec["LSL"] );
     Try( Column( dt3, N Cols( dt3 ) )[dt3 << Get Selected Rows] = 0 );
    );
    If( Is Missing( Try( spec["USL"], . ) ) == 0,
     dt3 << Select Where( As Column( dt3, colsToPassFail[i] ) > spec["USL"], Current Selection( "extend" ) );
     Try( Column( dt3, N Cols( dt3 ) )[dt3 << Get Selected Rows] = 0 );
    );
    dt3 << Invert Row Selection;
    Try( Column( dt3, N Cols( dt3 ) )[dt3 << Get Selected Rows] = 1 );
    Column( dt3, N Cols( dt3 ) ) << set property( "Value Labels", {1 = "Pass", 0 = "Fail"} );
    If( foundCols == {},
     foundCols = ":Name(\!"" || (Column( dt3, colsToPassFail[i] ) << Get Name) || " Pass/Fail\!")",
     foundCols = foundCols || ", " || ":Name(\!"" || (Column( dt3, colsToPassFail[i] ) << Get Name) || " Pass/Fail\!")"
    );
   );
  );
  
  dt3 << Clear Select;
  
  txList = {};
  
  For( i = 1 ,i <= N Items( foundCols ), i++,
   If( Contains( foundCols[i], "TX" ),
    Insert Into( txList, foundCols[i]);
   )
  );

回复：最终收益率表问题

我意识到我在上一个条目中犯了一个错误。填充 NewList 的代码需要更改为：

newList = {};
For( i=1,i<=N Items(foundCols),i++,
 If(contains(foundCols[i],"Tx"),
  insert into( newList, foundCols[i]);
 )
);

试试这个，看看是否有帮助

回复：最终收益率表问题

我很抱歉，自从我最初发布以来，我终于有时间重新审视它。 我收到“N Items() 参数必须是一个列表”的错误。

foundCols = "";是原始代码，不在列表中。

这是我认为应该发生的

 

 

listCols = { foundCols };
listCols = Eval List ( listCols );

 

 

此时我将所有 foundCols 放入列表中但是当我运行下面的脚本时 newList 不会填充..

 

newList = {};

For( i = 1, i <= N Items( listCols ), i++,
&nbsp;&nbsp;&nbsp; If( Contains( listCols, "TX" ),
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Insert Into( newList, listCols[i]);
&nbsp;&nbsp;&nbsp; )
);

 

提前致谢

 

回复：最终收益率表问题

我认为您可以做的是简单地创建一个仅包含具有 Rx 3 名称的列的新列表，然后使用该列表创建一个新列以在新列中设置值。就像是：

newList = {};
For( i=1,i<=N Items(foundCols),i++,
 If(contains(foundCols,"Rx 3"),
  insert into( newList, foundCols[i]);
 )
);

Eval( Parse( "dt2 << New Column( \!"Rx 3 Row Pass/Fail\!", Formula( If( Min(" || newList || " ) == 1, 1, 0 ) ) )" ) );

回复：最终收益率表问题

然而，这就是我目前正在做的事情，因为分析在测试名称中具有相同 Rx 的多个产品线，其中一个在 Rx 中有 300 种不同的条件，我希望动态地执行此操作，以便它适用于任何现有和即将推出的产品。 .

回复：最终收益率表问题

我会以交互方式创建一个新列，然后进入公式编辑器并仅选择要评估的 Rx 状态列

回复：最终收益率表问题

谢谢。 如果你不介意，你能帮我下一部分吗？

 

此代码采用其中包含“通过/失败”的所有列 {foundCols} 并找到最小值并 utlimitately 分配通过或失败。

 

<SPAN class="token function">Eval</SPAN><SPAN class="token punctuation">(</SPAN> <SPAN class="token function">Parse</SPAN><SPAN class="token punctuation">(</SPAN> <SPAN class="token string">"dt2 << New Column( \!"Row Pass/Fail\!", Formula( If( Min("</SPAN> <SPAN class="token operator">||</SPAN> foundCols <SPAN class="token operator">||</SPAN> <SPAN class="token string">" ) == 1, 1, 0 ) ) )"</SPAN> <SPAN class="token punctuation">)</SPAN> <SPAN class="token punctuation">)</SPAN><SPAN class="token punctuation">;</SPAN>

如何获得 foundCols 的不同组合，比如我只想要 Rx State 3 的列？

 

 

 

 foundCols = {:Name( "7200_VDD_Dig_Rx Pass/Fail" ), 
:Name( "7200_Analog_Rx Pass/Fail" ),
:Name( "CH9 Rx State 3 Pass/Fail" ),
:Name( "CH8 Rx State 3 Pass/Fail" ),
:Name( "CH7 Rx State 3 Pass/Fail" ),
:Name( "CH6 Rx State 3 Pass/Fail" ),
:Name( "CH5 Rx State 3 Pass/Fail" ),
:Name( "CH4 Rx State 3 Pass/Fail" ),
:Name( "CH3 Rx State 3 Pass/Fail" ),
:Name( "CH2 Rx State 3 Pass/Fail" ),
:Name( "CH15 Rx State 3 Pass/Fail" ),
:Name( "CH14 Rx State 3 Pass/Fail" ),
:Name( "CH13 Rx State 3 Pass/Fail" ),
:Name( "CH12 Rx State 3 Pass/Fail" ),
:Name( "CH11 Rx State 3 Pass/Fail" ),
:Name( "CH10 Rx State 3 Pass/Fail" ),
:Name( "CH1 Rx State 3 Pass/Fail" ),
:Name( "CH0 Rx State 3 Pass/Fail" ),
:Name( "CH0 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH1 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH10 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH11 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH12 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH13 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH14 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH15 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH2 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH3 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH4 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH5 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH6 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH7 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH8 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "CH9 Rx State 4 Delta Amp Pass/Fail" ),
:Name( "7100_VDD_Dig_Tx Pass/Fail" ),
:Name( "7100_Analog_Tx Pass/Fail" ),
:Name( "CH0 Tx State 1 Pass/Fail" ),
:Name( "CH1 Tx State 1 Pass/Fail" ),
:Name( "CH10 Tx State 1 Pass/Fail" ),
:Name( "CH11 Tx State 1 Pass/Fail" ),
:Name( "CH12 Tx State 1 Pass/Fail" ),
:Name( "CH13 Tx State 1 Pass/Fail" ),
:Name( "CH14 Tx State 1 Pass/Fail" ),
:Name( "CH15 Tx State 1 Pass/Fail" ),
:Name( "CH2 Tx State 1 Pass/Fail" ),
:Name( "CH3 Tx State 1 Pass/Fail" ),
:Name( "CH4 Tx State 1 Pass/Fail" ),
:Name( "CH5 Tx State 1 Pass/Fail" ),
:Name( "CH6 Tx State 1 Pass/Fail" ),
:Name( "CH7 Tx State 1 Pass/Fail" ),
:Name( "CH8 Tx State 1 Pass/Fail" ),
:Name( "CH9 Tx State 1 Pass/Fail" ),
:Name( "801 Pass/Fail" ),
:Name( "802 Pass/Fail" ),
:Name( "803 Pass/Fail" ),
:Name( "804 Pass/Fail" ),
:Name( "805 Pass/Fail" ),
:Name( "806 Pass/Fail" ),
:Name( "807 Pass/Fail" ),
:Name( "821 Pass/Fail" )};

 

回复：最终收益率表问题

您的 Eval(Parse()) 语句中存在一个简单的语法错误。一个“）”被遗漏了。这是一个工作版本

Eval( Parse( "dt2 << New Column( \!"Row Pass/Fail\!", Formula( If( Min(" || foundCols || " ) == 1, 1, 0 ) ) )" ) );