TypeError: 'filter' object is not subscrip

I am receiving the error

TypeError: 'filter' object is not subscriptable

When trying to run the following block of code

bonds_unique = {}
for bond in bonds_new:
    if bond[0] < 0:
        ghost_atom = -(bond[0]) - 1
        bond_index = 0
    elif bond[1] < 0:
        ghost_atom = -(bond[1]) - 1
        bond_index = 1
    else: 
        bonds_unique[repr(bond)] = bond
        continue
    if sheet[ghost_atom][1] > r_length or sheet[ghost_atom][1] < 0:
        ghost_x = sheet[ghost_atom][0]
        ghost_y = sheet[ghost_atom][1] % r_length
        image = filter(lambda i: abs(i[0] - ghost_x) < 1e-2 and
                       abs(i[1] - ghost_y) < 1e-2, sheet)
        bond[bond_index] = old_to_new[sheet.index(image[0]) + 1 ]
        bond.sort()
        #print >> stderr, ghost_atom +1, bond[bond_index], image
    bonds_unique[repr(bond)] = bond

# Removing duplicate bonds
bonds_unique = sorted(bonds_unique.values())

And

sheet_new = [] 
bonds_new = []
old_to_new = {}
sheet=[]
bonds=[]

The error occurs at the line

bond[bond_index] = old_to_new[sheet.index(image[0]) + 1 ]

I apologise that this type of question has been posted on SO many times, but I am fairly new to Python and do not fully understand dictionaries. Am I trying to use a dictionary in a way in which it should not be used, or should I be using a dictionary where I am not using it? I know that the fix is probably very simple (albeit not to me), and I will be very grateful if someone could point me in the right direction.

Once again, I apologise if this question has been answered already

Thanks,

Chris.

I am using Python IDLE 3.3.1 on Windows 7 64-bit.

回答1:

filter() in python 3 does not return a list, but a iterable filter object. Call next() on it to get the first filtered item:

bond[bond_index] = old_to_new[sheet.index(next(image)) + 1 ]

There is no need to convert it to a list, as you only use the first value.

回答2:

Use list before filter condtion then it works fine. For me it resolved the issue.

For example

list(filter(lambda x: x%2!=0, mylist))

instead of

filter(lambda x: x%2!=0, mylist)

回答3:

image = list(filter(lambda i: abs(i[0] - ghost_x) < 1e-2 and abs(i[1] - ghost_y) < 1e-2, sheet))