Formatting a table to look nice












5












$begingroup$


my current code is:



binsize = 21;

data = {535, 481, 554, 567, 565, 513, 526, 506, 565, 475, 552, 533, 

   474, 556, 520, 508, 597, 479, 537, 499, 546, 473, 579, 526, 594, 

   477, 518, 538, 497, 565};

firstbin = 472;

a = BinCounts[data, {firstbin, Max[data] + binsize, binsize}];

b = Range[firstbin, Max[data] + binsize, binsize];

Transpose[{Take[b, Length[a]], a}] // TableForm


Which does give me a nice table, however I would like the bins to be labeled something like "472-492 6" instead of just "472 6"



any advice would be great, thank you.






table formatting







share|improve this question











$endgroup$

















    5












    $begingroup$


    my current code is:



    binsize = 21;
    
data = {535, 481, 554, 567, 565, 513, 526, 506, 565, 475, 552, 533, 
    
   474, 556, 520, 508, 597, 479, 537, 499, 546, 473, 579, 526, 594, 
    
   477, 518, 538, 497, 565};
    
firstbin = 472;
    
a = BinCounts[data, {firstbin, Max[data] + binsize, binsize}];
    
b = Range[firstbin, Max[data] + binsize, binsize];
    
Transpose[{Take[b, Length[a]], a}] // TableForm


    Which does give me a nice table, however I would like the bins to be labeled something like "472-492 6" instead of just "472 6"



    any advice would be great, thank you.






    table formatting







    share|improve this question











    $endgroup$















      5












      5








      5


      1



      $begingroup$


      my current code is:



      binsize = 21;
      
data = {535, 481, 554, 567, 565, 513, 526, 506, 565, 475, 552, 533, 
      
   474, 556, 520, 508, 597, 479, 537, 499, 546, 473, 579, 526, 594, 
      
   477, 518, 538, 497, 565};
      
firstbin = 472;
      
a = BinCounts[data, {firstbin, Max[data] + binsize, binsize}];
      
b = Range[firstbin, Max[data] + binsize, binsize];
      
Transpose[{Take[b, Length[a]], a}] // TableForm


      Which does give me a nice table, however I would like the bins to be labeled something like "472-492 6" instead of just "472 6"



      any advice would be great, thank you.






      table formatting







      share|improve this question











      $endgroup$




      my current code is:



      binsize = 21;
      
data = {535, 481, 554, 567, 565, 513, 526, 506, 565, 475, 552, 533, 
      
   474, 556, 520, 508, 597, 479, 537, 499, 546, 473, 579, 526, 594, 
      
   477, 518, 538, 497, 565};
      
firstbin = 472;
      
a = BinCounts[data, {firstbin, Max[data] + binsize, binsize}];
      
b = Range[firstbin, Max[data] + binsize, binsize];
      
Transpose[{Take[b, Length[a]], a}] // TableForm


      Which does give me a nice table, however I would like the bins to be labeled something like "472-492 6" instead of just "472 6"



      any advice would be great, thank you.






      table formatting




      table formatting






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 9 hours ago







      Wombles

















      asked 9 hours ago









      WomblesWombles

      434




      434






















          2 Answers
          2






          active

          oldest

          votes


















          7












          $begingroup$

          You can use HistogramList to get bin limits and bin counts in one step and process the output to get the desired structure:



          {binlims, bincounts} = HistogramList[data, {firstbin, Max[data] + binsize, binsize}];
          
bins = Row[{#, #2 - 1}, "-"] & @@@ Partition[binlims, 2, 1];
          
TableForm[Transpose[{bins, bincounts}]]


          enter image description here



          Alternatively, you can use MovingMap, Developer`PartitionMap or
          the (undocumented) 6-argument form of Partition to get the first column:



          bins2 = MovingMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 1]
          
bins3 = Developer`PartitionMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 2, 1];
          
bins4 = Partition[binlims, 2, 1, {1, -1}, {}, Row[{#, #2 - 1}, "-"] &];
          
bins == bins2 == bins3 == bins4



          True




          Finally, you can also use a combination of StringRiffle and ToString in place of Rowas follows:



          bins5 = Partition[binlims, 2, 1, {1, -1}, {}, StringRiffle[ToString/@{#, #2 - 1}, " - "]&]
          
TableForm[Transpose[{bins5, bincounts}]]


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much!
            $endgroup$
            – Wombles
            8 hours ago










          • $begingroup$
            @Wombles, you are welcome.
            $endgroup$
            – kglr
            8 hours ago



















          3












          $begingroup$

          Here is one way:



          bb = b - 1;
          
c = Complement[bb, {Min[bb]}];
          
Transpose[{Take[b, Length[a]], ConstantArray["---", Length[a]], c, a}] // TableForm


          Here is the output:



          enter image description here



          With MarcoB's hint, and some experimenting:



          c = Complement[b - 1, {Min[b - 1]}];
          
y = Map[ToString, Take[b, Length[a]]];
          
z = Map[ToString, c]; 
          
Transpose@{Map[StringJoin, Transpose[{y, ConstantArray["---", Length[a]], z}]], a} // TableForm


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
            $endgroup$
            – mjw
            8 hours ago






          • 1




            $begingroup$
            You might want ToString and StringJoin
            $endgroup$
            – MarcoB
            8 hours ago










          • $begingroup$
            @MarcoB, Yes, Thank you! That's what I was looking for! I actually found ToString and figured there was a way of joining the columns ... Probably there is a more efficient way then what I've put together. At least it is a start.
            $endgroup$
            – mjw
            4 hours ago












          • $begingroup$
            Row as suggested in kglr's answer, can also work as a substitute for StringJoin.
            $endgroup$
            – mjw
            3 hours ago













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          2 Answers
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          2 Answers
          2






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          7












          $begingroup$

          You can use HistogramList to get bin limits and bin counts in one step and process the output to get the desired structure:



          {binlims, bincounts} = HistogramList[data, {firstbin, Max[data] + binsize, binsize}];
          
bins = Row[{#, #2 - 1}, "-"] & @@@ Partition[binlims, 2, 1];
          
TableForm[Transpose[{bins, bincounts}]]


          enter image description here



          Alternatively, you can use MovingMap, Developer`PartitionMap or
          the (undocumented) 6-argument form of Partition to get the first column:



          bins2 = MovingMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 1]
          
bins3 = Developer`PartitionMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 2, 1];
          
bins4 = Partition[binlims, 2, 1, {1, -1}, {}, Row[{#, #2 - 1}, "-"] &];
          
bins == bins2 == bins3 == bins4



          True




          Finally, you can also use a combination of StringRiffle and ToString in place of Rowas follows:



          bins5 = Partition[binlims, 2, 1, {1, -1}, {}, StringRiffle[ToString/@{#, #2 - 1}, " - "]&]
          
TableForm[Transpose[{bins5, bincounts}]]


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much!
            $endgroup$
            – Wombles
            8 hours ago










          • $begingroup$
            @Wombles, you are welcome.
            $endgroup$
            – kglr
            8 hours ago
















          7












          $begingroup$

          You can use HistogramList to get bin limits and bin counts in one step and process the output to get the desired structure:



          {binlims, bincounts} = HistogramList[data, {firstbin, Max[data] + binsize, binsize}];
          
bins = Row[{#, #2 - 1}, "-"] & @@@ Partition[binlims, 2, 1];
          
TableForm[Transpose[{bins, bincounts}]]


          enter image description here



          Alternatively, you can use MovingMap, Developer`PartitionMap or
          the (undocumented) 6-argument form of Partition to get the first column:



          bins2 = MovingMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 1]
          
bins3 = Developer`PartitionMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 2, 1];
          
bins4 = Partition[binlims, 2, 1, {1, -1}, {}, Row[{#, #2 - 1}, "-"] &];
          
bins == bins2 == bins3 == bins4



          True




          Finally, you can also use a combination of StringRiffle and ToString in place of Rowas follows:



          bins5 = Partition[binlims, 2, 1, {1, -1}, {}, StringRiffle[ToString/@{#, #2 - 1}, " - "]&]
          
TableForm[Transpose[{bins5, bincounts}]]


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            Thank you very much!
            $endgroup$
            – Wombles
            8 hours ago










          • $begingroup$
            @Wombles, you are welcome.
            $endgroup$
            – kglr
            8 hours ago














          7












          7








          7





          $begingroup$

          You can use HistogramList to get bin limits and bin counts in one step and process the output to get the desired structure:



          {binlims, bincounts} = HistogramList[data, {firstbin, Max[data] + binsize, binsize}];
          
bins = Row[{#, #2 - 1}, "-"] & @@@ Partition[binlims, 2, 1];
          
TableForm[Transpose[{bins, bincounts}]]


          enter image description here



          Alternatively, you can use MovingMap, Developer`PartitionMap or
          the (undocumented) 6-argument form of Partition to get the first column:



          bins2 = MovingMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 1]
          
bins3 = Developer`PartitionMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 2, 1];
          
bins4 = Partition[binlims, 2, 1, {1, -1}, {}, Row[{#, #2 - 1}, "-"] &];
          
bins == bins2 == bins3 == bins4



          True




          Finally, you can also use a combination of StringRiffle and ToString in place of Rowas follows:



          bins5 = Partition[binlims, 2, 1, {1, -1}, {}, StringRiffle[ToString/@{#, #2 - 1}, " - "]&]
          
TableForm[Transpose[{bins5, bincounts}]]


          enter image description here






          share|improve this answer











          $endgroup$



          You can use HistogramList to get bin limits and bin counts in one step and process the output to get the desired structure:



          {binlims, bincounts} = HistogramList[data, {firstbin, Max[data] + binsize, binsize}];
          
bins = Row[{#, #2 - 1}, "-"] & @@@ Partition[binlims, 2, 1];
          
TableForm[Transpose[{bins, bincounts}]]


          enter image description here



          Alternatively, you can use MovingMap, Developer`PartitionMap or
          the (undocumented) 6-argument form of Partition to get the first column:



          bins2 = MovingMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 1]
          
bins3 = Developer`PartitionMap[Row[{First@#, Last@# - 1}, "-"] &, binlims, 2, 1];
          
bins4 = Partition[binlims, 2, 1, {1, -1}, {}, Row[{#, #2 - 1}, "-"] &];
          
bins == bins2 == bins3 == bins4



          True




          Finally, you can also use a combination of StringRiffle and ToString in place of Rowas follows:



          bins5 = Partition[binlims, 2, 1, {1, -1}, {}, StringRiffle[ToString/@{#, #2 - 1}, " - "]&]
          
TableForm[Transpose[{bins5, bincounts}]]


          enter image description here







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 3 hours ago

























          answered 8 hours ago









          kglrkglr

          187k10203421




          187k10203421












          • $begingroup$
            Thank you very much!
            $endgroup$
            – Wombles
            8 hours ago










          • $begingroup$
            @Wombles, you are welcome.
            $endgroup$
            – kglr
            8 hours ago


















          • $begingroup$
            Thank you very much!
            $endgroup$
            – Wombles
            8 hours ago










          • $begingroup$
            @Wombles, you are welcome.
            $endgroup$
            – kglr
            8 hours ago
















          $begingroup$
          Thank you very much!
          $endgroup$
          – Wombles
          8 hours ago




          $begingroup$
          Thank you very much!
          $endgroup$
          – Wombles
          8 hours ago












          $begingroup$
          @Wombles, you are welcome.
          $endgroup$
          – kglr
          8 hours ago




          $begingroup$
          @Wombles, you are welcome.
          $endgroup$
          – kglr
          8 hours ago











          3












          $begingroup$

          Here is one way:



          bb = b - 1;
          
c = Complement[bb, {Min[bb]}];
          
Transpose[{Take[b, Length[a]], ConstantArray["---", Length[a]], c, a}] // TableForm


          Here is the output:



          enter image description here



          With MarcoB's hint, and some experimenting:



          c = Complement[b - 1, {Min[b - 1]}];
          
y = Map[ToString, Take[b, Length[a]]];
          
z = Map[ToString, c]; 
          
Transpose@{Map[StringJoin, Transpose[{y, ConstantArray["---", Length[a]], z}]], a} // TableForm


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
            $endgroup$
            – mjw
            8 hours ago






          • 1




            $begingroup$
            You might want ToString and StringJoin
            $endgroup$
            – MarcoB
            8 hours ago










          • $begingroup$
            @MarcoB, Yes, Thank you! That's what I was looking for! I actually found ToString and figured there was a way of joining the columns ... Probably there is a more efficient way then what I've put together. At least it is a start.
            $endgroup$
            – mjw
            4 hours ago












          • $begingroup$
            Row as suggested in kglr's answer, can also work as a substitute for StringJoin.
            $endgroup$
            – mjw
            3 hours ago


















          3












          $begingroup$

          Here is one way:



          bb = b - 1;
          
c = Complement[bb, {Min[bb]}];
          
Transpose[{Take[b, Length[a]], ConstantArray["---", Length[a]], c, a}] // TableForm


          Here is the output:



          enter image description here



          With MarcoB's hint, and some experimenting:



          c = Complement[b - 1, {Min[b - 1]}];
          
y = Map[ToString, Take[b, Length[a]]];
          
z = Map[ToString, c]; 
          
Transpose@{Map[StringJoin, Transpose[{y, ConstantArray["---", Length[a]], z}]], a} // TableForm


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
            $endgroup$
            – mjw
            8 hours ago






          • 1




            $begingroup$
            You might want ToString and StringJoin
            $endgroup$
            – MarcoB
            8 hours ago










          • $begingroup$
            @MarcoB, Yes, Thank you! That's what I was looking for! I actually found ToString and figured there was a way of joining the columns ... Probably there is a more efficient way then what I've put together. At least it is a start.
            $endgroup$
            – mjw
            4 hours ago












          • $begingroup$
            Row as suggested in kglr's answer, can also work as a substitute for StringJoin.
            $endgroup$
            – mjw
            3 hours ago
















          3












          3








          3





          $begingroup$

          Here is one way:



          bb = b - 1;
          
c = Complement[bb, {Min[bb]}];
          
Transpose[{Take[b, Length[a]], ConstantArray["---", Length[a]], c, a}] // TableForm


          Here is the output:



          enter image description here



          With MarcoB's hint, and some experimenting:



          c = Complement[b - 1, {Min[b - 1]}];
          
y = Map[ToString, Take[b, Length[a]]];
          
z = Map[ToString, c]; 
          
Transpose@{Map[StringJoin, Transpose[{y, ConstantArray["---", Length[a]], z}]], a} // TableForm


          enter image description here






          share|improve this answer











          $endgroup$



          Here is one way:



          bb = b - 1;
          
c = Complement[bb, {Min[bb]}];
          
Transpose[{Take[b, Length[a]], ConstantArray["---", Length[a]], c, a}] // TableForm


          Here is the output:



          enter image description here



          With MarcoB's hint, and some experimenting:



          c = Complement[b - 1, {Min[b - 1]}];
          
y = Map[ToString, Take[b, Length[a]]];
          
z = Map[ToString, c]; 
          
Transpose@{Map[StringJoin, Transpose[{y, ConstantArray["---", Length[a]], z}]], a} // TableForm


          enter image description here







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 3 hours ago

























          answered 8 hours ago









          mjwmjw

          4428




          4428












          • $begingroup$
            I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
            $endgroup$
            – mjw
            8 hours ago






          • 1




            $begingroup$
            You might want ToString and StringJoin
            $endgroup$
            – MarcoB
            8 hours ago










          • $begingroup$
            @MarcoB, Yes, Thank you! That's what I was looking for! I actually found ToString and figured there was a way of joining the columns ... Probably there is a more efficient way then what I've put together. At least it is a start.
            $endgroup$
            – mjw
            4 hours ago












          • $begingroup$
            Row as suggested in kglr's answer, can also work as a substitute for StringJoin.
            $endgroup$
            – mjw
            3 hours ago




















          • $begingroup$
            I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
            $endgroup$
            – mjw
            8 hours ago






          • 1




            $begingroup$
            You might want ToString and StringJoin
            $endgroup$
            – MarcoB
            8 hours ago










          • $begingroup$
            @MarcoB, Yes, Thank you! That's what I was looking for! I actually found ToString and figured there was a way of joining the columns ... Probably there is a more efficient way then what I've put together. At least it is a start.
            $endgroup$
            – mjw
            4 hours ago












          • $begingroup$
            Row as suggested in kglr's answer, can also work as a substitute for StringJoin.
            $endgroup$
            – mjw
            3 hours ago


















          $begingroup$
          I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
          $endgroup$
          – mjw
          8 hours ago




          $begingroup$
          I wish I knew more about string manipulation to get the first three columns to combine to one column of text!
          $endgroup$
          – mjw
          8 hours ago




          1




          1




          $begingroup$
          You might want ToString and StringJoin
          $endgroup$
          – MarcoB
          8 hours ago




          $begingroup$
          You might want ToString and StringJoin
          $endgroup$
          – MarcoB
          8 hours ago












          $begingroup$
          @MarcoB, Yes, Thank you! That's what I was looking for! I actually found ToString and figured there was a way of joining the columns ... Probably there is a more efficient way then what I've put together. At least it is a start.
          $endgroup$
          – mjw
          4 hours ago






          $begingroup$
          @MarcoB, Yes, Thank you! That's what I was looking for! I actually found ToString and figured there was a way of joining the columns ... Probably there is a more efficient way then what I've put together. At least it is a start.
          $endgroup$
          – mjw
          4 hours ago














          $begingroup$
          Row as suggested in kglr's answer, can also work as a substitute for StringJoin.
          $endgroup$
          – mjw
          3 hours ago






          $begingroup$
          Row as suggested in kglr's answer, can also work as a substitute for StringJoin.
          $endgroup$
          – mjw
          3 hours ago




















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          GameSpot

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          body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} GameSpot 戰地風雲：惡名昭彰2在Gamespot的評論 网站类型 新聞 持有者 CBS 创始人 Pete Deemer Vince Broady Jon Epstein 网站 http://www.gamespot.com/ 注册 Optional (free and paid) 推出时间 1996年5月1日 [1] GameSpot （中国大陆：游戏基地），於1996年5月由Pete Deemer和Vince Broady創立，是一個提供新聞、評論、預告片、下載及其他的相關資訊的電子遊戲網站。GameSpot被一間後來被CNET收購的企業ZDNet所收購。根據Alexa，GameSpot.com是200個網路擁擠最嚴重的網站之一。 除了由GameSpot員工創作的內容，網站還允許用戶寫評論、網誌、之後在網路論壇分享。一些在CNET旗下的GameFAQs分享。 2004年， GameSpot被Spike TV的觀眾選上「電子遊戲賞節目」贏得「最傑出遊戲網站。 [2] 其他的遊戲網站還有IGN、1UP.com、GameSpy是它最大的競爭對手。2008年，根據Compete.com的統計，「gamespot.com」吸引了最少6000萬人的點擊率。 [3] GameSpot的主頁鏈結了到最近新聞、評論、預告、和一些有關遊戲機的入口：Wii、任天堂DS、電腦遊戲、Xbox 360、PSP、PlayStation 2、PlayStation 3。它還有一列「最受歡迎遊戲名單」，還有給用戶快速獲得遊戲資訊的搜尋器。GameSpot 還包括一些小範圍的遊戲機：任天堂64、GameCube、Game Boy Color、Game Boy Advance、Xbox、PlayStation、SEGA Saturn、Dreamcast、Neo Geo Pocket Color、N-Gage、手機遊戲。 目录 1 歷史 1.1 國際歷史 1.2 著名的員工 2 評論和分...
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          日野市

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          body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 日野市 日野市 日文轉寫  • 日文 日野市  • 平假名 ひのし  • 罗马字 Hino-shi 金剛寺（關東三十六不動尊靈場第9號）的不動堂 市旗 徽章 日野市在東京都的位置 日野市 日野市在日本的位置 坐标： 35°40′17″N 139°23′42″E  /  35.6714°N 139.395°E  / 35.6714; 139.395 国家   日本 地方 關東地方 都道府縣 東京都 接鄰行政區 府中市、國立市、立川市、昭島市、八王子市、多摩市 政府  • 市長 大坪冬彦 面积  •  总计 27.53 平方公里（10.63 平方英里） 人口 （2014年12月1日）  • 總計 183,323  • 密度 6,660/平方公里（17,200/平方英里） 象徵  • 市樹 樫  • 市花 菊花  • 市鳥 普通翠鳥 时区 日本標準時間 （UTC+9） 地方公共團體編號 13212-8 邮政编码 〒 191-8686 市役所地址 日野市神明一丁目12番1號 電話號碼 +81-42-585-1111 法人編號 1000020132128 網站 http://www.city.hino.lg.jp/ 人口：日野市官方網頁 日野市 （日语： 日野市 ／ ひのし   Hino shi   * / ? ）為一位于東京都（不含島嶼部分）中央地帶的城市。從東京站乘坐中央線特別快速列車45分鐘即可到該市。面積27.53km²。1963年（昭和38年）11月3日，該市開始實行市制，為全日本第559個市。汽車企業日野自動車的總部位於這裡。 往東京都特別區部的通勤率為20.9%，往八王子市的通勤率為12.0%（平成22年國勢調査）。 目录 1 概要 ...
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          Tu-95轟炸機

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          body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} 图-95 Ту–95 北约代号：熊（Bear） 一架在安格斯空军基地的图-95MS 概觀 類型 战略轰炸机 代號 北约代号： Bear （熊） 乘員 7名 駕駛員2名，機尾炮手1名，其他人員4名 首飛 1952年11月12日 服役 1956年 設計 圖波列夫設計局 產量 超过500架 現況 现役 主要用戶   蘇聯   俄羅斯 衍生機型 图-114、图-119、Tu-95 技术数据 長度 49.50米（162呎5吋） 翼展 51.10米（167呎8吋） 高度 12.12米（39呎9吋） 翼面積 310平方米（3,330平方呎） 空重 90,000公斤（198,000磅） 最大起飛重量 188,000公斤（414,500磅） 發動機 4具库兹涅佐夫NK-12MV型渦輪螺旋槳發動機 功率 4×11,000千瓦（14,800馬力） 性能數據 最大速度 925公里／時（500節，575哩／時） 爬升率 10米／秒（2,000呎／分） 最大升限 12,000米（39,000呎） 最大航程 15,000公里（8,100海浬，9,400哩） 翼負荷 606公斤／平方米（124磅／平方呎） 推重比 235 W／公斤（0.143馬力／磅） 武器装备 機炮 1具/2具AM-23型23公釐雷達控制機砲（機尾） 飛彈 空对地导弹： Kh-20、Kh-22 Kh-26、Kh-55 炸彈 60枚FAB-250炸彈，或 30枚FAB-500炸彈。 其他 载弹量（包括飞弹）：最多15,000公斤（33,000磅） Tu-95熊式MR型機 图-95 （俄语： Ту–95 ， 英语： Tu-95 ），北約代號： Bear （ 熊 ），是蘇聯圖波列夫設計局所研製，是全世界唯一服役的大型四渦輪螺旋槳發動機...
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