Python 从列表中删除每三个元素

Question

x = [1,2,3,4,5,6,7,8,9,10]
for i in range(len(x)):
    if i%3==0:
        x.pop(i)
print(x)

so i want to get this

[2,3,5,6,8,9]

instead i get error x.pop(i) out of range I know there is simpler way to do this, but i have to do this with for loop

Answer 1

Short and simple:

>>> x
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> del x[::3]
>>> x
[2, 3, 5, 6, 8, 9]

The problem with the loop is that after you call pop , the i th element isn't the same as before the call.

>>> x =  [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> x[3]
4
>>> x.pop(0)
>>> x[3]
5

You can mitigate this by working from the other end of the list.

for i in range(len(x)-1, -1, -1):
    if i % 3 == 0:
        x.pop(i)  # or del x[i]

Answer 2

You are getting that because you are resizing the x list while iterating over it, not advisable. You should create another list with the values filtered, like this:

x = [1,2,3,4,5,6,7,8,9,10]
output = []
for i, v in enumerate(x):
    if i % 3:
        output.append(v)

print(output)
>>> [2, 3, 5, 6, 8, 9]

Answer 3

As others have pointed out you're removing elements from the list whilst you are iterating over it.

Since you said you have to use a for loop I'd suggest a list comprehension:

l = [v for i, v in enumerate(x) if i % 3 != 0]
print(l)

Answer 4

This error is because of using len(x) instead of len(x)-1. But the code you wrote down doesn't output the desired result you look for.

Try this:

x = [1,2,3,4,5,6,7,8,9,10]
y = []
for i in range(len(x)-1):
    if i%3==0:
        continue;
    else:
        y.append(x[i])
print(y)

Answer 5

You are getting an error because your len(x) is decrementing each time you remove the element, whereas the value of i is not changing wrt the new len(x) .

You can check this by adding print() statements.

Adding those statements, your code is now :

x = [1,2,3,4,5,6,7,8,9,10]

for i in range(len(x)):
    if i%3==0:
        print("Before \n i:",i, "len : ",len(x))
        x.pop(i)
        print("After \n i:",i, "len : ",len(x))
print(x)

This is your stack trace :

Before 
 i: 0 len :  10
After 
 i: 0 len :  9
Before 
 i: 3 len :  9
After 
 i: 3 len :  8
Before 
 i: 6 len :  8
After 
 i: 6 len :  7
Before 
 i: 9 len :  7
Traceback (most recent call last):
  File "q.py", line 6, in <module>
    x.pop(i)
IndexError: pop index out of range

You can make the following change to the code to get your output :

x = [i for i in range(1,11)]
del x[::3]

print(x)