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POPython loop through list and shorten by one
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  1. POPython loop through list and shorten by one
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    copied!<p>I have a list:</p> <pre><code> mylist = ['apple', 'orange', 'dragon', 'panda'] </code></pre> <p>I want to be able to is loop over the list, do something on each element and then remove the element. I tried this:</p> <pre><code>for l in mylist: print l list.remove(l) </code></pre> <p>but the my output is:</p> <pre><code>apple dragon </code></pre> <p><strong>EDIT</strong></p> <p>I actually want to be able to do some comparisons in the loop. So basically I want to be able to take each element, one-by-one, remove that element for the list and compare it against all the other elements in the list. The comparison is a little complex so I don't want to use list comprehension. And I want to be reducing the list by one each time until the list is empty and all elements have been compared with each other.</p> <p>What is the best way to get each element, work with it and remove it without skipping elements in the list?</p> <p>Any help, much appreciated.</p> <p><strong>REDIT</strong></p> <p>Just to make clear - the real point of this is to go through each element, which is a string fragment and match it with other fragments which have overlapping sequences on either end, thereby building up a complete sequence. The element being processed should be removed from the list prior to looping so that it isn't compared with itself, and the list should shrink by 1 element each processing loop.</p> <p>In the case a better list example would be:</p> <pre><code>mylist = ['apples and or', 'oranges have', 'in common', 'e nothing in c'] </code></pre> <p>to give:</p> <pre><code>'apples and oranges have nothing in common' </code></pre> <p>Apologies for not being clear from the outset, but it was a specific part of this larger problem that I was stuck on.</p>
 

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