比较容易记忆的是用内置的
set l1 = [
‘
b
‘,
‘
c
‘,
‘
d
‘,
‘
b
‘,
‘
c
‘,
‘
a
‘,
‘
a
‘]

l2 = list(set(l1))

print l2   还有一种据说速度更快的，没测试过两者的速度差别 l1 = [
‘
b
‘,
‘
c
‘,
‘
d
‘,
‘
b
‘,
‘
c
‘,
‘
a
‘,
‘
a
‘]

l2 = {}.fromkeys(l1).keys()

print l2   这两种都有个缺点，祛除重复元素后排序变了： [‘a’, ‘c’, ‘b’, ‘d’]  
 
 
如果想要保持他们原来的排序：   用list类的sort方法 l1 = [
‘
b
‘,
‘
c
‘,
‘
d
‘,
‘
b
‘,
‘
c
‘,
‘
a
‘,
‘
a
‘]

l2 = list(set(l1))

l2.sort(key=l1.index)

print l2 也可以这样写 l1 = [
‘
b
‘,
‘
c
‘,
‘
d
‘,
‘
b
‘,
‘
c
‘,
‘
a
‘,
‘
a
‘]

l2 = sorted(set(l1),key=l1.index)

print l2     也可以用遍历 l1 = [
‘
b
‘,
‘
c
‘,
‘
d
‘,
‘
b
‘,
‘
c
‘,
‘
a
‘,
‘
a
‘]

l2 = []

for i
in l1:

    
if
not i
in l2:

        l2.append(i)

print l2 上面的代码也可以这样写 l1 = [
‘
b
‘,
‘
c
‘,
‘
d
‘,
‘
b
‘,
‘
c
‘,
‘
a
‘,
‘
a
‘]

l2 = []

[l2.append(i)
for i
in l1
if
not i
in l2]

print l2     这样就可以保证排序不变了： [‘b’, ‘c’, ‘d’, ‘a’]