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POcomparing strings and calculating occurance
 

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  1. POcomparing strings and calculating occurance
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    <p>I don't know how to attack this problem!!!</p> <p>I have 3 lists with a word, a tag and a number it appears on a document:</p> <pre><code>v1 = [['be', 'VSIS3S0', 1], ['scott', 'NP00000', 2], ['north', 'NCMS000', 1], ['revolution', 'NP00000', 1], ['name', 'VMP00SM', 1]] v2 = [['mechanic', 'NCMS000', 1], ['be', 'VSIS3S0', 1], ['tool', 'AQ0CS0', 1], ['sam', 'NP00000', 1], ['frida', 'NP00000', 1]] v3 = [['be', 'VSIP3S0', 1], ['scott', 'NP00000', 1], ['who', 'NP00000', 1]] </code></pre> <p>How can I build a function that receiving these lists compare each word so that, for example, the word <code>be</code> in <code>v1</code> appears in the three lists once, in that case append onto a result list <code>(1 * log(3/3))</code>, where 1 -> maximum value of occurrance(which is the 3rd element of the sublist), log numerator 3 -> constant, log denominator 3 -> because the word appears on <code>v1</code>, <code>v2</code> and <code>v3</code>.</p> <p>Next we have <code>scott</code> -> In this case append onto the result list <code>(2 * log(3/2))</code>, 2 -> maximum word´s value of occurrance, log numerator 3 -> constant, log denominator 2 -> because the word 'scott' appears on <code>v1</code> and <code>v2</code>.</p> <p>Next we have <code>north</code> -> In this case append onto the result list <code>(1 * log(3/1))</code>, 1 -> maximum word´s value of occurrance, log numerator 3 -> constant, log denominator 1 -> because the word 'north' appears only <code>v1</code>.</p> <p>Next we have <code>revolution</code> -> In this case append onto the result list <code>(1 * log(3/1))</code>, 1 -> maximum word´s value of occurrance, log numerator 3 -> constant, log denominator 1 -> because the word 'north' appears only <code>v1</code>.</p> <p>Next we have <code>name</code> -> In this case append onto the result list <code>(1 * log(3/1))</code>, 1 -> maximum word´s value of occurrance, log numerator 3 -> constant, log denominator 1 -> because the word 'name' appears only <code>v1</code>.</p> <p>Furthermore we have to do the same with <code>v2</code> by comparing <code>mechanic</code>, <code>be</code>, <code>tool</code>, etc. with the other words, calculate the max value of occurance and multuplying it <code>w/ log(3/?)</code> depending if the word appears or not in <code>v1</code> and <code>v3</code>.</p> <p>This is my attempt for <code>v1</code>:</p> <pre><code>def f1(v1, v2, v3): res =[] for e in v1: if e != 0: if e in v2 and e in v3: res.append(0) elif e in v2: res.append(e * math.log(3/2)) else: res.append(e * math.log(3)) return res </code></pre> <p>returning: <code>[0, 2.1972245773362196, 0, 0, 0, 0]</code></p> <p>and this is obviously not the result</p> <p>it sholud return something like:</p> <pre><code>[['be', 0.47], ['scott', 0.35 ], ['north', 0.47], ['revolution', 0.47], ['north', 0.47]] </code></pre> <p>Hope u can help me w/ this, I have a week trying to solve it!</p> <p>Thank you</p>
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