list4 = [0,0,1,2,4,5,6,7,5,0,2]
list5 = [2,5,4,8,5,6,6,6,6,2,8]

list6 = zip(list4,list5)
list7 = list(list6)
print(list7)
a = list(set(list7))
print(a)

#结果：
[(0, 2), (0, 5), (1, 4), (2, 8), (4, 5), (5, 6), (6, 6), (7, 6), (5, 6), (0, 2), (2, 8)]
[(6, 6), (4, 5), (2, 8), (5, 6), (7, 6), (1, 4), (0, 5), (0, 2)]

import numpy as np

list1 = [0,0,1,2,4,5,6,7,5]
list2 = np.unique(list1)
print(list2)

list4 = [0,0,1,2,4,5,6,7,5,0,2]
list5 = [2,5,4,8,5,6,6,6,6,2,8]

list6 = zip(list4,list5)
list7 = list(list6)

print(np.unique(list7))

#结果：
[0 1 2 4 5 6 7]#这个结果是我们想要的
[0 1 2 4 5 6 7 8] #这个不是我们要的结果，相当于没把list(zip())中的小括号当整体对待

import pandas as pd

list1 = [0,0,1,2,4,5,6,7,5]

list3 = pd.Series(list1)
print(list3.unique())

list4 = [0,0,1,2,4,5,6,7,5,0,2]
list5 = [2,5,4,8,5,6,6,6,6,2,8]

list6 = zip(list4,list5)
list7 = list(list6)

list8 = pd.Series(list7)
print(list8.unique())

#结果：
[0 1 2 4 5 6 7] #这个是我们想要的
[(0, 2) (0, 5) (1, 4) (2, 8) (4, 5) (5, 6) (6, 6) (7, 6)]#这个也算是我们要的，但是就是()之间没有","

list1 = [0,0,1,2,4,5,6,7,5]

list9 = []
for i in list1:
    if not i in list9:
        list9.append(i)
print(list9)

#结果：
[0, 1, 2, 4, 5, 6, 7] #是我们想要的，就是中间多了","

list1 = [0,0,1,2,4,5,6,7,5]

list10 = []
[list10.append(i) for i in list1 if not i in list10]
print(list10)

#结果：
[0, 1, 2, 4, 5, 6, 7] #比上面简单多了，哈哈。

list4 = [0,0,1,2,4,5,6,7,5,0,2]
list5 = [2,5,4,8,5,6,6,6,6,2,8]

list6 = zip(list4,list5)
list7 = list(list6)

list9 = []
for i in list7:
    if not i in list9:
        list9.append(i)
print(list9)

list10 = []
[list10.append(i) for i in list7 if not i in list10]
print(list10)

#结果
[(0, 2), (0, 5), (1, 4), (2, 8), (4, 5), (5, 6), (6, 6), (7, 6)]
[(0, 2), (0, 5), (1, 4), (2, 8), (4, 5), (5, 6), (6, 6), (7, 6)]
#这样也行啊，但是我选择set(),一句话的事