Python 类型错误：“过滤器”对象不可下标
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TypeError: 'filter' object is not subscriptable
提问by Christopher John Scott
I am receiving the error
我收到错误
TypeError: 'filter' object is not subscriptable
When trying to run the following block of code
尝试运行以下代码块时
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
和
sheet_new = []
bonds_new = []
old_to_new = {}
sheet=[]
bonds=[]
The error occurs at the line
错误发生在该行
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.
我很抱歉这种类型的问题已经发布了很多次，但我对 Python 相当陌生并且不完全理解字典。我是在尝试以不应该使用的方式使用字典，还是应该在不使用它的地方使用字典？我知道修复可能非常简单（尽管对我来说不是），如果有人能指出我正确的方向，我将不胜感激。
Once again, I apologise if this question has been answered already
如果这个问题已经得到回答，我再次道歉
Thanks,
谢谢，
Chris.
克里斯。
I am using Python IDLE 3.3.1 on Windows 7 64-bit.
我在 Windows 7 64 位上使用 Python IDLE 3.3.1。
回答by Martijn Pieters
filter()in python 3 does notreturn a list, but a iterable filterobject. Call next()on it to get the firstfiltered item:
filter()在python 3中不返回一个列表，而是一个可迭代的filter对象。调用next()它以获取第一个过滤的项目：
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.
无需将其转换为列表，因为您只使用第一个值。
回答by DKZ
image = list(filter(lambda i: abs(i[0] - ghost_x) < 1e-2 and abs(i[1] - ghost_y) < 1e-2, sheet))
回答by K Kotagaram
Use listbefore filtercondtion then it works fine. For me it resolved the issue.
list在filter条件之前使用然后它工作正常。对我来说，它解决了这个问题。
For example
例如
list(filter(lambda x: x%2!=0, mylist))
instead of
代替
filter(lambda x: x%2!=0, mylist)
