IndexError：列表索引超出范围-在Python中抛出错误？

Question

a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
l= []
for i in a:
    if a[i] % 2 == 0 :
        l.append(a[i])

The code above keeps throwing the error - "IndexError: list index out of range"

I cannot understand what to do ?

Answer 1

When you perform for i in a you are iterating over the elements of a , not the indices!

You are trying to access: a[1] and then a[4] and then a[9] and then a[16] -> Which is the one that is causing the IndexError .

If you want to iterate only over the indices try:

>>> for i in range(len(a)):
        if a[i] % 2 == 0 :
            l.append(a[i])


>>> print (l)
[4, 16, 36, 64, 100]

If you need both the value and index use for index, value in enumerate(a): .

Answer 2

When you are iterating over a , you are looking at the value rather than the position of that value in the list.

You could use just the value like so:

a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
l= []
for val in a:
    if val % 2 == 0 :
        l.append(val)

Or alternately, if you need both the position and the value then you can use the enumerate function, like so:

a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
l= []
for pos, val in enumerate(a):
    if a[pos] % 2 == 0 :
        l.append(a[pos])

Answer 3

There are some ways:

for e in a:
  if e % 2 == 0 :
    l.append(e)

or

for i in range(len(a)):
  if a[i] % 2 == 0 :
    l.append(a[i])

or

for i, e in enumerate(a):
  if e % 2 == 0 :
    l.append(e)

or

for i, e in enumerate(a):
  if a[i] % 2 == 0 :
    l.append(a[i])

Answer 4

You could use list comprehension for that:

a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
l = [i for i in a if i%2 == 0]

print(l)
[4, 16, 36, 64, 100]

Answer 5

列表组合可用于使其单线l=[x for x in a if x%2==0]并且如ather答案所述，您的问题是您使用列表项作为其索引。

Answer 6

You can use filter for your task:

a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
print filter(lambda x: x % 2 == 0, a)

Answer 7

as in other answers, i takes the values from a , which are [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
in the loop when i takes the value " 16 ", a[i] will be out of range!! ( 16 > len(a) )

for debugging i always suggest printing...
in this case if you print the value of i in the loop, you will find the problem yourself

Answer 8

a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
l= []
for i in a:
    if a[i] % 2 == 0 :
        l.append(a[i])

...
if a[1] % 2 == 0:
...
if a[4] % 2 == 0:
...
if a[9] % 2 == 0:
...
if a[16] % 2 == 0:

index error, because the biggest index is 9 in array a... so u have to use this:

a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
l= []
for i in a:
    if i % 2 == 0 :
        l.append(i)

or this example:

a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
l= []
for i in a:
    if i % 2 == 0 :
        l.append(i)

or the 1 line solution:

a = [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]
l= [x for x in a if x%2 == 0]