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POPython dictionary comprehension very slow
 

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  1. POPython dictionary comprehension very slow
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    <p>I have a dictionary <code>d1</code> and a list <code>l1</code>.</p> <p>The dictionary keys are strings, and the values are Objects I have defined myself. If it helps, I can describe the Object in more detail but for now, the objects have a list attribute <code>names</code>, and some of the elements of <code>name</code> may or may not appear in <code>l1</code>. </p> <p>What I wanted to do was to throw away any element of the dictionary <code>d1</code>, in which the <code>name</code> attribute of the object in said element does not contain any of the elements that appear in <code>l1</code>.</p> <p>As a trivial example:</p> <pre><code>l1 = ['cat', 'dog', 'mouse', 'horse', 'elephant', 'zebra', 'lion', 'snake', 'fly'] d1 = {'1':['dog', 'mouse', 'horse','orange', 'lemon'], '2':['apple', 'pear','cat', 'mouse', 'horse'], '3':['kiwi', 'lime','cat', 'dog', 'mouse'], '4':['carrot','potato','cat', 'dog', 'horse'], '5':['chair', 'table', 'knife']} </code></pre> <p>so the resulting dictionary will be more or less the same but the elements of each list will be the key-value pairs from <code>1</code> to <code>4</code> excluding the fruit and vegetables, and will not contain a 5th key-value par as none of the furniture values appear in <code>l1</code>.</p> <p>To do this I used a nested list/dictionary comprehension which looked like this:</p> <pre><code>d2 = {k: [a for a in l1 if a in d1[k]] for k in d1.keys()} print(d2) &gt;&gt;&gt;&gt;{'1': ['dog', 'mouse', 'horse'], '3': ['cat', 'dog', 'mouse'], '2': ['cat', 'mouse', 'horse'], '5': [], '4': ['cat', 'dog', 'horse']} d2 = {k: v for k,v in d2.iteritems() if len(v)&gt;0} print(d2) &gt;&gt;&gt;&gt;{'1': ['dog', 'mouse', 'horse'], '3': ['cat', 'dog', 'mouse'], '2': ['cat', 'mouse', 'horse'], '4': ['cat', 'dog', 'horse'],} </code></pre> <p>This seems to work, but for big dictionaries, 7000+ items, it takes around 20 seconds to work through. In and of itself, not horrible, but I need to do this inside a loop that will iterate 10,000 times, so currently it's not feasible. Any suggestions on how to do this quickly?</p>
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    1. USDavy Kavanagh
    1. USDavy Kavanagh
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      1. VTUpMod
    2. VO
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    1. CONote to everyone: He is using python 2.7 not 3 due to use of `itertitems`, don't let the `print()` fool you
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      1. USjamylak
    2. COpython 2.7 has dict comprehensions?
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      1. USClaudiu
    3. CO@Claudiu Yes they were backported
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      1. USjamylak
 

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