# -*- coding: utf-8 -*-

__author__ = 'k22li'

from sets import Set

"""

题目: 如何去除list中的dup 元素

方案1: 列表解析方案

"""

# order preserving

aList = ['a', 'b', 'c', 'd', 'a', 'c']

[aList.remove(item) for item in aList if aList.count(item)>1] #返回值为None

print aList

"""

方案2： 利用字典的keys（）method可以自动去重

"""

#字典的keys（）method会自动去重

#NOT order preserving

def viaDict(seq):

temDic = {}

for value in seq:

temDic[value] = 1

return temDic.keys()

"""

方案3： 对原列表进行遍历

"""

#order preserving

def walkThrough(seq):

newList = []

for item in seq:

if not item in newList:

newList.append(item)

return newList

"""

python的set和其他语言类似, 是一个 基本功能包括关系测试和消除重复元素.

集合对象还支持union(联合), intersection(交), difference(差)和sysmmetric difference(对称差集)等数学运算.

from sets import Set

"""

#order perservings

def setList(seq):

set = Set(seq)

return list(set)

"""

"""

def listAndDict(seq, idFun = None):

if not len(seq):

return []

else:

if idFun == None:

def idFun(x): return x

tempDic = {}

refreshResult = []

for item in seq:

marker = idFun(item)

if not tempDic.has_key(marker):

tempDic[item] = 1

refreshResult.append(item)

else:

continue

return refreshResult

if __name__ == '__main__':

newList1 = viaDict(aList)

print newList1

# try with 'walkThrough'

newList2 = walkThrough(aList)

print newList2

# try with 'setList'

newList3 = setList(aList)

print newList3

# try with 'listAndDict

newList4 = listAndDict(aList,)

print newList4

# what's more supported by the 'listAndDic' methods,

"""

Clearly "listAndDict" is the "best" solution.

Not only is it really really fast;

it's also order preserving and supports an optional transform function which makes it possible to do this:

"""

bList = list('ABCdeDEFG')

newList5 = listAndDict(bList, lambda x : x.lower())

print newList5