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It instead makes two copies of lists (one from the start until the index but without it (a[:index]) and one after the index till the last element (a[index+1:])) and creates a new list object by adding both.
remove the item at index and return nothing. it can remove slices from a list or can clear the whole list. Benchmark: Worst case : deleting from the end of the list. yopy:-> python -m timeit "x=range(1000)" "x.pop(999)" 100000 loops, best of 3: 10 usec per loop. yopy:-> python -m timeit "x=range(1000)" "x.remove(999)"
First, peruse the elements of the list like so: print '%s: %s' % (x, yourlist[x]) Then, call this function with a list of the indexes of elements you want to pop. It's robust enough that the order of the list doesn't matter. result = [] itemstopop.sort() itemstopop = itemstopop[::-1] for x in itemstopop:
pop takes the index as a parameter and removes the element at that index. Unlike del, pop when called on list object returns the value at that index >>> a = [1, 5, 3, 4, 7, 8] >>> a.pop(3) # Will return the value at index 3 4 >>> a [1, 5, 3, 7, 8] For remove. remove takes the parameter value and remove that value from the list.
This implies that if you iterate the list back to front, if you remove an item at the current index, everything to it's right shifts left - but that doesn't matter, since you've already dealt with all the elements to the right of the current position, and you're moving left - the next element to the left is unaffected by the change, and so the ...
I stand by my question: this seems weird and contrived, and has nothing to recommend it. In fact, the first sentence is misleading, because you cannot remove the i'th element with list.remove(a[i]). With duplicate values, it may find an earlier element with the same value, and remove that one instead of the i'th. –
When current==4 is the 4 removed from the list because it has the value 4 or because it is in the fourth position (when counting from one)? Similarly, is the 3 removed because it is the smallest number less than 4, or because it's on the index preceding the removed item? –
python -m timeit -s "import numpy as np" -s "import itertools" -s "a = np.array(list(range(10000)))" -s "index=[i for i in range(10000) if i % 2 == 0]" "a = np.array(list(itertools.compress(a, [i not in index for i in range(len(a))])))" 10 loops, best of 3: 200 msec per loop python -m timeit -s "import numpy as np" -s "a = np.array(list(range ...
You lose the contiguity (single block) aspect of the "Python list", which gives you random access, i.e., the ability to access any element in constant time. In linked lists, the elements are scattered in memory and you store addresses to point to the next element.
I'm trying to figure out how to remove specific index from list, but I don't know how to fix my problem. Can someone help me to solve this? First, I have to get some numbers from users using input.