python - Count length of string without vowels -

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मेरे पास एक स्ट्रिंग है: - 

  a = 'abcderui'

मुझे स्ट्रिंग a सेवों की लंबाई की गणना करना चाहते हैं I

वर्तमान में मैं कर रहा हूं: - 

  a = सूची (a) के लिए i में: अगर मैं स्वर में हूँ: यदि कोई स्ट्रिंग बड़ा नहीं है तो सबसे अच्छा विकल्प है:  
   
 
 
 / P> 
  & gt; & gt; & gt; A = 'abcderui' & gt; & gt; & gt; लेन (ए। ट्रांसस्लेट (कोई नहीं, 'एईओयू')) 4  
 
  समय:  
 
  & gt; & gt; & gt; ए = 'एबीसीडीआरयूई' * 100 & gt; & gt; & gt; % Timeit len ​​(a.translate (कोई भी नहीं, 'एईआईओयू')) 1000000 छोरों, सर्वोत्तम 3: 1.86 μs प्रति लूप & gt; & gt; & gt; % Timeit राशि (यदि सी 'aeiou' में नहीं है तो सी के लिए 1) 10000 छोरों, सर्वोत्तम 3: 53.2 μs प्रति लूप & gt; & gt; & gt; % Timeit len ​​(nonvowels.findall (a)) 10000 छोरों, सबसे अच्छा 3: 65.3 μs प्रति लूप & gt; & gt; & gt; % Timeit len ​​(vowels.sub ('', a)) 10000 छोरों, सबसे अच्छा 3: 72 प्रति लूप प्रति μs

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