python去除重复元素
Posted 落月_YU
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目的:让我们两个列表压缩后,去除重复的点;
1.set()方法:
蓦然回首,那人却在灯火阑珊处;
这个方法实现了我想要的压缩后的列表的重复删除;
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)]
到今天我才发现这种方法有个缺点是顺序是乱的 (时间:2020-04-1420:24:02),想要不乱的顺序:
2.np.unique()方法:
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
3.pd.Series()方法:
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","
4.遍历去除重复元素:
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","
5.列表推导式:
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。
6.再试试可恶的list7(list(zip)之后的):
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
#但是谨记缺点:乱序
参考地址:
https://www.cnblogs.com/zhangshuyang/p/8087895.html
https://jingyan.baidu.com/article/f3e34a12ebd701f5eb6535c6.html
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