查找列表中未包含的元素

我的原则是:

item = [0,1,2,3,4,5,6,7,8,9]
z = []  # list of integers


for item in z:
if item not in z:
print item

z包含一个整数列表。我想比较 item和 z，然后打印出与 item比较时不在 z中的数字。

我可以打印的元素是在 z的时候比较不是 item，但是当我尝试做相反的使用代码上面没有打印。

有人帮忙吗？

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小开 
list1 = [1,2,3,4]; list2 = [0,3,3,6]


print set(list2) - set(list1)
小开 
>> items = [1,2,3,4]
>> Z = [3,4,5,6]


>> print list(set(items)-set(Z))
[1, 2]
小开 
>>> item = set([0,1,2,3,4,5,6,7,8,9])
>>> z = set([2,3,4])
>>> print item - z
set([0, 1, 5, 6, 7, 8, 9])
小开

If you run a loop taking items from z, how do you expect them not to be in z? IMHO it would make more sense comparing items from a different list to z.

小开
最佳答案

Your code is not doing what I think you think it is doing. The line for item in z: will iterate through z, each time making item equal to one single element of z. The original item list is therefore overwritten before you've done anything with it.

I think you want something like this:

item = [0,1,2,3,4,5,6,7,8,9]


for element in item:
if element not in z:
print(element)

But you could easily do this like:

[x for x in item if x not in z]

or (if you don't mind losing duplicates of non-unique elements):

set(item) - set(z)
小开

No, z is undefined. item contains a list of integers.

I think what you're trying to do is this:

#z defined elsewhere
item = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]


for i in item:
if i not in z: print i

As has been stated in other answers, you may want to try using sets.

小开

Your code is a no-op. By the definition of the loop, "item" has to be in Z. A "For ... in" loop in Python means "Loop though the list called 'z', each time you loop, give me the next item in the list, and call it 'item'"

http://docs.python.org/tutorial/controlflow.html#for-statements

I think your confusion arises from the fact that you're using the variable name "item" twice, to mean two different things.

小开

You are reassigning item to the values in z as you iterate through z. So the first time in your for loop, item = 0, next item = 1, etc... You are never checking one list against the other.

To do it very explicitly:

>>> item = [0,1,2,3,4,5,6,7,8,9]
>>> z = [0,1,2,3,4,5,6,7]
>>>
>>> for elem in item:
...   if elem not in z:
...     print elem
...
8
9
小开

Using list comprehension:

print [x for x in item if x not in Z]

or using filter function :

filter(lambda x: x not in Z, item)

Using set in any form may create a bug if the list being checked contains non-unique elements, e.g.:

print item


Out[39]: [0, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9]


print Z


Out[40]: [3, 4, 5, 6]


set(item) - set(Z)


Out[41]: {0, 1, 2, 7, 8, 9}

vs list comprehension as above

print [x for x in item if x not in Z]


Out[38]: [0, 1, 1, 2, 7, 8, 9]

or filter function:

filter(lambda x: x not in Z, item)


Out[38]: [0, 1, 1, 2, 7, 8, 9]
小开

In the case where item and z are sorted iterators, we can reduce the complexity from O(n^2) to O(n+m) by doing this

def iexclude(sorted_iterator, exclude_sorted_iterator):
next_val = next(exclude_sorted_iterator)
for item in sorted_iterator:
try:
while next_val < item:
next_val = next(exclude_sorted_iterator)
continue
if item == next_val:
continue
except StopIteration:
pass
yield item

If the two are iterators, we also have the opportunity to reduce the memory footprint not storing z (exclude_sorted_iterator) as a list.

小开

If the lists are sorted and you know the elements of the checking list are in the base list - you can do a more optimal O(n) solution by using two pointers (where n will be the length of the base_list:

base_list = [0, 1, 2, 3, 4, 5, 6, 7, 8]
checking_list = [1, 3, 5]


expected_return = [0, 2, 4, 6, 7, 8]


j = 0
i = 0
elements_not_in_checking_list = []
while i < len(base_list):
if j < len(checking_list) and base_list[i] == checking_list[j]:
i += 1
j += 1
else:
elements_not_in_checking_list.append(base_list[i])
i += 1
小开

Many of the solutions already posted here will not preserve the original ordering of the elements (because sets are unordered) or are inefficient (because linear search in a list is slower than a lookup in a set).

You can make a set of elements to remove upfront, and then use a list comprehension to retain only the elements which aren't in the set:

items = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
z = [3, 4, 5, 6]


set_z = set(z)
result = [e for e in items if e not in set_z]

Then, result contains:

[0, 1, 2, 7, 8, 9]

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