Implementing np.subtract.outer for index of array([[array .... structure in python -

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मेरे पास सूचकांक का निम्न सेट है; 

  ind = array ([[array ( [[1, 2, 3, 6]), [सरणी ([2, 3, 7] [0, 1, 4]), [सरणी ([0, 1, 2, 5]), ), [सरणी ([2, 5, 6, 7, 10]), [सरणी ([0, 4, 5, 8])], [सरणी ([1, 4, 5, 6, 9])], [सरणी ([2, 5, 6, 7, 10]) ], [सरणी ([5, 8, 9, 10, 13]), [सरणी ([3, 6, 7, 11])], [सरणी ([4, 8, 9, 12])], [ सरणी ([6, 9, 10, 11, 14])], [सरणी ([7, 10, 11, 15]), [सरणी ([8, 12, 13])], [सरणी ([9, 12, 13, 14]), [सरणी ([10, 13, 14, 15]), [सरणी ([11, 14, 15])]])   
  X , जो नीचे दिखाया गया है  
  एक्स = सरणी ([[0., 0.], [0.33333333, 0.], [0.66666667, 0.], [0, 0, 03333333], [0.33333333, 0.33333333], [0.66666667, 0.33333333], [1.33333333], [0., 0.66666667], [0.33333333, 0.66666667], [0.66666667, 0.66666667], [1., 0.66666667], [0., 1], [0.33333333, 1.], [0.66666667, 1.], [1., 1.]])   < P>  forloop  टी का उपयोग किए बिना एक कारगर तरीका है ओ निम्नलिखित मैट्रिक्स प्राप्त करें;  
  ए = [एक्स [0,0] -एक्स [0,0] एक्स [0,0] -एक्स [1,0] एक्स [0,0] -एक्स [4, 0] एक्स [1,0] -एक्स [0,0] एक्स [1,0] -एक्स [1,0] एक्स [1,0] -एक्स [2,0] एक्स [1,0] -एक्स [ 5,0] एक्स [2,0] -एक्स [1,0] एक्स [2,0] -एक्स [2,0] एक्स [2,0] -एक्स [3,0] एक्स [2,0] - एक्स [6,0] ... ... ... एक्स [15,0] -एक्स [11,0] एक्स [15,0] -एक्स [14,0] एक्स [15,0] -एक्स [ 15,0]]   
 मैंने पहले ही np.subtract.outer और np.subtract की कोशिश की है।  इंडस्ट्रीज़  संरचना के साथ काम करना बहुत मुश्किल है (बस का उपयोग करके  X [ind, 0]  कहते हैं)। मैं  इंडस्ट्रीज़  संरचना को बदलना पसंद नहीं करता मैट्रिक्स  ए  को प्राप्त करने के लिए  forloop  का उपयोग किए बिना क्लीवर मार्ग की तलाश कर रहा है।  
   
 
 
 यदि आप इंडेक्स की सरणी को बदलते हैं जैसे:  
 / P> 
  ind = ind.flatten ()   
 आप जो चाहते हैं वह प्राप्त कर सकते हैं:  
  a = X [: , 0] ans = [एनपी.ओन्स (लेन (आई)) * एक [0] -ए [i] I के लिए में]   
  इंडस्ट्रीज़ की अपनी मूल परिभाषा के साथ  आपको इसके बजाय  i [0]  का उपयोग करना चाहिए।  
 
 संपादित करें: इसमें सुधार किया जा सकता है (जैसा कि @जैम द्वारा सुझाया गया है):  < पूर्व>  ans = [एक [0] - एक [i] के लिए I में]

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