Python 列表去重去除空字符的例子

            
# x = ['c b a',"e d f"]
# y = []
# for i in x:
# for ii in i:
# # print(ii)
# if ii == ' ':
# pass
# else:
# y.append(ii)


# print(y)

          

            
# list1 = ['122','2333','3444',' ','422',' ',' ','54',' ']
# 第一种方法会导致最后一个' '没有被移除掉['122', '2333', '3444', '422', '54', ' ']
# for x in list1:
# if x == ' ':
# list1.remove(' ')
# print(list1)


# 第二种方法：['122', '2333', '3444', '422', '54']
# for x in list1:
# if ' ' in list1:
# list1.remove(' ')
# print(list1)


# 第三种方法：遍历空格的个数，然后逐个删除
# for x in range(list1.count(' ')):
# list1.remove(' ')
# print(list1)


# 第四种方法：用了while和for一样
# while ' ' in list1:
# list1.remove(' ')
# print(list1)

          

            
# 第一种方法：使用replace,但是这种方法很笨，如果字符串中间有一万个空格怎么办，也要全部打出来吗
# a = 'hello world'
# b = a.replace(' ','')
# print(b)


# 第二种方法：
# a = 'hello world'
# a = list(a)
# for x in a:
# if ' ' in a:
# a.remove(' ')
# bb = ''.join(a)
# print(bb)




# a = 'hello wor ld'
# # aa = a.split()
# # print(aa)
# # print(''.join(aa))
# print(''.join(a.split()))




# list1 = ['122','2333','3444',' ','422',' ',' ','54',' ']
# for x in list1:
# for i,j in enumerate(list1):
# print(i,j)
# if x == ' ':
# list1.remove(' ')
# print(list1)
# print('***************************************')
# print(list1)

          

            
# 方法一： 对列表进行怕羞，从头到尾进行比较，遇到重复的元素就删除，否则指针向右移动一位
#方法1，逻辑复杂，临时变量保存值消耗内存，返回结果破坏了原列表顺序，效率最差
# def deleteDuplicatedElement(l):
# l.sort()
# length = len(l)
# firstItem = l[0]
# for x in range(1,length-1):
# # if x < length - 2:
# # if l[x] == l[x+1]:
# # l.remove(l[x])
# # return l
# currentItem = l[x]
# if firstItem == currentItem:
# l.remove(currentItem)
# else:
# firstItem = currentItem
# return l


# print(deleteDuplicatedElement(['d','d','1','2','1','4']))




# def deleteDuplicatedElement(l):
# l.sort()
# length = len(l)
# lastItem = l[length-1]
# for x in range(length-2,-1,-1):
# currentItem = l[x]
# if lastItem == currentItem:
# l.remove(currentItem)
# else:
# lastItem = currentItem
# return l


# print(deleteDuplicatedElement(['python','r','r','g','g','g','t','y','g','n']))




# 方法二：设一临时列表保存结果，从头遍历原列表，如临时列表中没有当前元素则追加：
#方法2，直接调用append方法原处修改列表，逻辑清晰，效率次之
# def deleteDuplicatedElement(l):
# ll = []
# for x in l:
# if x in ll:
# continue
# else:
# ll.append(x)
# return ll


# print(deleteDuplicatedElement(['python','r','r','g','g','g','t','y','g','n']))




# 方法三：利用Python中集合元素唯一性特点，将列表转换为集合，然后转换为列表输出即可
#方法3，极度简洁，使用python原生方法效率最高，但列表原有顺序被打乱
# def deleteDuplicatedElement(l):
# return sorted(list(set(l)),key=l.index)


# print(deleteDuplicatedElement(['python','r','r','g','g','g','t','y','g','n']))