Python的魔力内置类型

在python中，仅通过python的内置类型就可以很容易地处理这样复杂的情况。

让我告诉你怎么做!

方法一:一般情况

方法(1行代码)删除重复的元素在列表中仍然保持排序顺序

line = [1, 2, 3, 1, 2, 5, 6, 7, 8]
new_line = sorted(set(line), key=line.index) # remove duplicated element
print(new_line)

你会得到结果的

[1, 2, 3, 5, 6, 7, 8]

方法二:特殊情况

TypeError: unhashable type: 'list'

处理不可哈希的特殊情况(3行代码)

line=[['16.4966155686595', '-27.59776154691', '52.3786295521147']
,['16.4966155686595', '-27.59776154691', '52.3786295521147']
,['17.6508629295574', '-27.143305738671', '47.534955022564']
,['17.6508629295574', '-27.143305738671', '47.534955022564']
,['18.8051102904552', '-26.688849930432', '42.6912804930134']
,['18.8051102904552', '-26.688849930432', '42.6912804930134']
,['19.5504702331098', '-26.205884452727', '37.7709192714727']
,['19.5504702331098', '-26.205884452727', '37.7709192714727']
,['20.2929416861422', '-25.722717575124', '32.8500163147157']
,['20.2929416861422', '-25.722717575124', '32.8500163147157']]

tuple_line = [tuple(pt) for pt in line] # convert list of list into list of tuple
tuple_new_line = sorted(set(tuple_line),key=tuple_line.index) # remove duplicated element
new_line = [list(t) for t in tuple_new_line] # convert list of tuple into list of list

print (new_line)

你会得到这样的结果:

[
  ['16.4966155686595', '-27.59776154691', '52.3786295521147'], 
  ['17.6508629295574', '-27.143305738671', '47.534955022564'], 
  ['18.8051102904552', '-26.688849930432', '42.6912804930134'], 
  ['19.5504702331098', '-26.205884452727', '37.7709192714727'], 
  ['20.2929416861422', '-25.722717575124', '32.8500163147157']
]

因为元组是可哈希的，你可以很容易地在列表和元组之间转换数据

不幸的是。这里的大多数答案要么不保持顺序，要么太长。这里有一个简单的、有序的答案。

s = [1,2,3,4,5,2,5,6,7,1,3,9,3,5]
x=[]

[x.append(i) for i in s if i not in x]
print(x)

这将得到x，删除重复项，但保留顺序。

Write a Python program to create a list of numbers by taking input from the user and then remove  the duplicates from the list. You can take input of non-zero numbers, with an appropriate  prompt, from the user until the user enters a zero to create the list assuming that the numbers  are non-zero.  
Sample Input: [10, 34, 18, 10, 12, 34, 18, 20, 25, 20]  
Output: [10, 34, 18, 12, 20, 25] 

 lst = []
print("ENTER ZERO NUMBER FOR EXIT !!!!!!!!!!!!")
print("ENTER LIST ELEMENTS  :: ")
while True:
    n = int(input())
    if n == 0 :
       print("!!!!!!!!!!! EXIT !!!!!!!!!!!!")
       break
    else :
        lst.append(n)
print("LIST ELEMENR ARE :: ",lst)
#dup = set()
uniq = []
for x in lst:
    if x not in uniq:
        uniq.append(x)
       # dup.add(x)
print("UNIQUE ELEMENTS IN LIST ARE :: ",uniq)

我认为转换为set是删除重复的最简单的方法:

list1 = [1,2,1]
list1 = list(set(list1))
print list1

今天，一位同事将接受的答案作为他的代码的一部分发给了我进行代码审阅。 虽然我很欣赏这个问题答案的优雅，但我对他的表现并不满意。 我尝试过这个解决方案(我使用set来减少查找时间)

def ordered_set(in_list):
    out_list = []
    added = set()
    for val in in_list:
        if not val in added:
            out_list.append(val)
            added.add(val)
    return out_list

为了比较效率，我使用了100个整数的随机样本，其中62个是唯一的

from random import randint
x = [randint(0,100) for _ in xrange(100)]

In [131]: len(set(x))
Out[131]: 62

这是测量结果

In [129]: %timeit list(OrderedDict.fromkeys(x))
10000 loops, best of 3: 86.4 us per loop

In [130]: %timeit ordered_set(x)
100000 loops, best of 3: 15.1 us per loop

如果把set从解中移除会发生什么?

def ordered_set(inlist):
    out_list = []
    for val in inlist:
        if not val in out_list:
            out_list.append(val)
    return out_list

结果并不像OrderedDict那样糟糕，但仍然是原始解决方案的3倍多

In [136]: %timeit ordered_set(x)
10000 loops, best of 3: 52.6 us per loop

可以使用Python set或dict.fromkeys()方法删除重复项。 dict.fromkeys()方法将一个列表转换为一个字典。字典不能包含重复的值，因此dict.fromkeys()将返回只有唯一值的字典。 集，像字典一样，不能包含重复的值。如果将列表转换为集合，则删除所有重复项。

方法一:幼稚法

mylist = [5, 10, 15, 20, 3, 15, 25, 20, 30, 10, 100]

uniques = []

for i in mylist:

    if i not in uniques:

       uniques.append(i)

print(uniques)

方法二:使用set()

mylist = [5, 10, 15, 20, 3, 15, 25, 20, 30, 10, 100]

myset = set(mylist)

print(list(myset))