问题
My code looks like this:
def nue(li):
for i in li:
li.remove(i)
print(li)
li = [1, 2, 3]
nue(li)
However, running this results in:
>>> [2]
More generally, how do I remove the i-th position on a list while iterating over the list (for reasons it failed some test in a nested loop or something like that)?
回答1:
you can do like this
def nue(li):
for i in li[:]:
li.remove(i)
print(li)
li = [1, 2, 3]
nue(li)
li[:] will clone original li
回答2:
Try list comprehension:
def remi(slist, i): # remi means REMove I
return [slist[j]
for j in range(len(slist))
if j != i]
print(remi ([1,2,3,4], 2))
Output:
[1, 2, 4]
回答3:
Why not just use a list comprehension instead of removing items from an existing list?
old_list = [1,2,3,4,5]
li = [1,2,3]
new_list = [i for i in old_list if i not in li]
If you don't want to use a list comprehension you can try the filter function
old_list = [1,2,3,4,5]
li = [1,2,3]
new_list = list(filter(lambda x: x not in li, old_list))
回答4:
You might want to consider going over the array in reverse. The following code demonstrates three functions that do the same thing. All are called prune and are interchangeable. By looking at an array in reverse order, you avoid changing the index of values you will examine in the future.
#! /usr/bin/env python3
def main():
array = list(range(20))
print(array)
prune(array, lambda value: value % 3)
print(array)
def prune(array, del_condition):
for index, value in reversed(tuple(enumerate(array))):
if del_condition(value):
del array[index]
def prune(array, del_condition):
for index in range(len(array) - 1, -1, -1):
value = array[index]
if del_condition(value):
del array[index]
def prune(array, del_condition):
for index in reversed(range(len(array))):
if del_condition(array[index]):
del array[index]
if __name__ == '__main__':
main()
来源:https://stackoverflow.com/questions/48327207/how-to-remove-an-item-while-iterating-over-a-list