I am new to python. I have a item list looks like this:
rank_item = [
(9, 0.99999999745996648),
(8, 0.99999996796861101),
(1, 0.99999996796861101),
(10, 0.0) ]
The question is how do i remove the item list with value 0.0
and return
[(9, 0.99999999745996648), (8, 0.99999996796861101), (1, 0.99999996796861101)]
Use list.remove
- which will modify your rank_item
list in-place :
rank_item.remove( (10, 0.0) )
rank_item
will contain:
[(9, 0.99999999745996648), (8, 0.99999996796861101), (1, 0.99999996796861101)]
Or, if you want to remove all tuples from your list, which have a the value 0.0
at position 1, you can try a list comprehension , eg:
[ x for x in rank_item if x[1] != 0.0 ]
[x for x in rank_item if x[1] != 0.0]
>>> [i for i in rank_item if i[1]]
[(9, 0.9999999974599665), (8, 0.999999967968611), (1, 0.999999967968611)]
For a more general case, you can use a list comprehension:
new_list = [(a, b) for a, b in old_list if b != 0]
This would only return the tuples where the second element is different from zero; you can adjust this behavior to fit your needs.
This approach favors a clear labeling of each element, as opposed to using an index.
Speed test:
>>> import timeit
>>> it = lambda : list(filter(lambda x: x[1]!=0, rank_item))
>>> timeit.timeit(it)
3.7716277663630535
>>> it2 = lambda: [x for x in rank_item if x[1] != 0.0]
>>> timeit.timeit(it2)
1.2550897693390652
>>> it3 = lambda: [i for i in rank_item if i[1]]
>>> timeit.timeit(it3)
1.147179730129892
>>> it4 = lambda: list(itertools.takewhile(lambda x: x[1] != 0, rank_item))
>>> timeit.timeit(it4)
3.8272935335999136
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